Does the British media have significant power Essay

Does the British media have significant power - Essay Example From this point of view, their power in promoting specific political, social and economic trends cannot be doubted; however, it should be examined whether this power is positive or negative. Most of the theorists who studied the specific issue – as the relevant literature is presented below – agree that the British media can have a key role in the improvement of the local political, social and economic rules and initiatives; from this point of view, the specific industry should be characterized not just as a tool for the promotion of the views of the leading political and economic forces but also as a means for securing social equality and fairness; it should be also regarded as a tool for criticizing and evaluating the plans of the local government; at this point, the value of British media could be significant ensuring the alignment of the governmental decisions with the local laws and ethics. 2. British media – role and power 2.1 British media – descrip tion and role In order to understand the power of media industry in Britain it would be necessary to refer primarily to the characteristics and the role of this industry; then, its power could be identified and evaluated – taking into consideration the local ethics and culture. In accordance with Stokes et al. (1999) the media industry in Britain is not similar to the media industries of other countries; reference is made though not to the content of the term but rather on its potential forms. More specifically, it is explained that in Britain, the term ‘media’ is used to describe the ‘books, newspapers, television programmes, films and music’ (Stokes et al. 1999, p.1); however, the content/ texts of British media is unique – being related to the country’s ethics (Stokes et al. 1999, p.1). This fact, leads to the following assumption: in Britain, the role of the media industry in the development of political and social decisions can be significant – reaching higher levels compared to other countries where there is no such close relationship and interaction between the media and the local culture – an issue also analyzed in section 2.2 below. A more descriptive definition of media is provided in the study of Oakland (2011); in accordance with the above researcher ‘the term ‘media’ may include any communication system by which people are informed, educated or entertained’ (Oakland 2011, p. 258). The above definition reveals the potentials of media in terms of supporting specific social or political trends; moreover, it proves that people of all ages are expected to be influenced by the media – being exposed to the media in different ways – either in the context of education, of entertainment or just of information/ professional support, as described above. In accordance with Garnett et al. (2007) in order to understand whether the British media interact or not wi th politics, it would be necessary to explain primarily whether such interaction would be positive or negative – referring to its effects on the country’s political framework. In any case, it is noted that the freedom of the British media should be secured – no matter whether the information provided to the public is well – justified (Garnett et al. 2007, p.62). At the next level, it is explained

Role of Political Parties in India Essay Example for Free

Role of Political Parties in India Essay As with any other democracy, political parties represent different sections among the Indian society and regions, and their core values play a major role in the politics of India. Both the executive branch and the legislative branch of the government are run by the representatives of the political parties who have been elected through the elections. Through the electoral process, the people of India choose which representative and which political party should run the government. Through the elections any party may gain simple majority in the lower house. Coalitions are formed by the political parties, in case no single party gains a simple majority in the lower house. Unless a party or a coalition have a majority in the lower house, a government cannot be formed by that party or the coalition. Indian state governments led by various political parties as of March 2009. India has a multi-party system, where there are a number of national as well as regional parties. A regional party may gain a majority and rule a particular state. If a party is represented in more than 4 states, it would be labelled a national party. Out of the 64 years of Indias independence, India has been ruled by the Indian National Congress (INC) for 51 of those years. The party enjoyed a parliamentary majority save for two brief periods during the 1970s and late 1980s. This rule was interrupted between 1977 to 1980, when the Janata Party coalition won the election owing to public discontent with the controversial state of emergencydeclared by the then Prime Minister Indira Gandhi. The Janata Dal won elections in 1989, but its government managed to hold on to power for only two years. Between 1996 and 1998, there was a period of political flux with the government being formed first by the nationalist Bharatiya Janata Party (BJP) followed by a left-leaning United Front coalition. In 1998, the BJP formed the National Democratic Alliance with smaller regional parties, and became the first non-INC and coalition government to complete a full five-year term. The 2004 Indian elections saw the INC winning the largest number of seats to form a government leading the United Progressive Alliance, and supported by left-parties and those opposed to the BJP. On 22 May 2004, Manmohan Singh was appointed the Prime Minister of India following the victory of the INC the left front in the 2004 Lok Sabha election. The UPA now rules India without the support of the left front. Previously, Atal Bihari Vajpayee had taken office in October 1999 after a general election in which a BJP-led coalition of 13 parties called the National Democratic Alliance emerged with a majority. Formation of coalition governments reflects the transition in Indian politics away from the national parties toward smaller, more narrowly based regional parties. Some regional parties, especially in South India, are deeply aligned to the ideologies of the region unlike the national parties and thus the relationship between the central government and the state government in various states has not always been free of rancor. Disparity between the ideologies of the political parties ruling the centre and the state leads to severely skewed allocation of resources between the states. Function of Opposition Party: The Opposition’s main role is to question the government of the day and hold them accountable to the public. The Opposition is equally responsible in upholding the best interests of the people of the country. They have to ensure that the Government does not take any steps , which might have negative implications on the people of the country. the role of the opposition in parliament is basically to check the excesses of the ruling or dominant party, and not to be totally antagonistic. There are actions of the ruling party which may be beneficial to the masses and oppositions are expected to support such things. In Parliament, Opposition Party should act firmly on behalf of common mass fighting for their common interest and grievances. They should raise immediate protest before a Bill passed, which is against the interest of common-men. Opposition legislators should always bear in mind that they are the representatives from each and every countryman fighting for justified demands and defending all unlawful and unfair practice.

The Silence Of The Lambs Essay -- Silence Lambs Cannibalism Essays

The Silence of The Lambs   Ã‚  Ã‚  Ã‚  Ã‚  Clarice Starling, a student preparing for a life in the FBI, hunts a serial killer by use of vague information given to her by an incarcerated psychologist. Hannibal â€Å"The Cannibal† Lecter relays information to Clarice in exchange for information about herself. The killer, known only as "Buffalo Bill", kidnaps large women, keeps them alive for a few days, and finally skins them. Clarice works against time as Buffalo Bill takes his newest victim, a U.S. Senator's daughter, and the countdown to death begins. "The Silence of the Lambs" was chosen for the title because it is Clarice Starling's ultimate goal for the bloodcurdling screams of the lambs in her nightmares to cease. When she was younger, she witnessed the slaughtering of a herd of lambs and to this day she awakens horrified some nights to escape the nightmares that she so longs to end. She believes deep down that if she catches Buffalo Bill herself she will sleep soundly in the silence of the lambs.   Ã‚  Ã‚  Ã‚  Ã‚  My first interpretation of Clarice was that she was very bright and observant. She reads people very well and is quick to make an accurate judgement, as in with Frederick Chilton, the prison warden. I believed that she was a very strong woman and was very careful to appear that way to others.   Ã‚  Ã‚  Ã‚  Ã‚  Clarice Starling did have a large change in herself. She began the story with a careful mentality; a risk would have been unheard of. She was always making sure she w...

The Philosophy of Gilles Deleuze and Félix Guattari :: Philosophy Philosophical Papers

The Philosophy of Gilles Deleuze and FÃ ©lix Guattari ABSTRACT: In academic philosophy the writings of Gilles Deleuze and FÃ ©lix Guattari are still treated as curiosities and their importance for philosophical discussions is not recognized. In order to remedy this, I demonstrate how the very concept of philosophy expounded by the two contributes to philosophical thinking at the end of the twentieth century while also providing a possible line of thought for the next millenium. To do this, I first emphasize the influence of Deleuze's thinking, while also indicating the impact Guattari had on him. This account will therefore show Deleuze's attempts before Guattari to concieve of a non-dialectic philosophy of becoming. I will turn to rethink this approach given the influence of Guattari and his anti-psychoanalytic analysis of territorial processes. The result is a conception of philosophical activity as an act of 'becoming minor'.(1) 1. Introduction In the following I would like to talk about a topic that has been treated very little in academic philosophy. The works of GILLES DELEUZE - and not to forget his co-author, FÉLIX GUATTARI - are still treated as 'curiosities' and their importance for philosophical discussions is not recognized. (2) In opposition to this, I will show what the very concept of philosophy means to these two thinkers. In doing this I will start with the more theoretical backround. As many others have already I will stress the decisive influence of DELEUZE'S thinking, but I will also try to indicate the impact GUATTARI had on him. This account will therefore show DELEUZE'S attempts - before GUATTARI - to concieve of a non-dialectic philosophy of becoming. After that I will turn to the rethinking of such an approach given the influence of GUATTARI and his anti-psychoanalytic analysis of territorial processes. The outcome will be the resulting conception of the philosophical activity as an act of 'becoming-minor'. 2. GILLES DELEUZE Philosophy of Difference - Against Dialectics GILLES DELEUZE'S early philosophy is dominated by the project of attaining a kind of philosophy that can be characterized best by naming its very enemy: dialectics. Whether as a 'school' of philosophy (including the leading figures in France, KOJÈVE and SARTRE) or as an ontological approach to the world itself, which implies - no matter if in the Hegelian or Platonic version - a fundamental dualism. (In PLATO the difference between the sensual and intellectual world, in HEGEL'S dialectics the 'sublation' [Aufhebung] of real differences in the world through the synthesizing faculty of the mind qua negation).

Dai Park Textbook

Stochastic Manufacturing & Service Systems Jim Dai and Hyunwoo Park School of Industrial and Systems Engineering Georgia Institute of Technology October 19, 2011 2 Contents 1 Newsvendor Problem 1. 1 Pro? t Maximization 1. 2 Cost Minimization . 1. 3 Initial Inventory . . 1. 4 Simulation . . . . . . 1. 5 Exercise . . . . . . . 5 5 12 15 17 19 25 25 27 29 29 31 32 33 34 39 39 40 40 42 44 46 47 48 49 51 51 51 52 54 55 57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Queueing Theory 2. 1 Introduction . . . . . . . 2. 2 Lindley Equation . . . . 2. 3 Tra? c Intensity . . . . . 2. 4 Kingman Approximation 2. 5 Little’s Law . . . . . . . 2. 6 Throughput . . . . . . . 2. 7 Simulation . . . . . . . . 2. 8 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Discrete Time Markov Chain 3. 1 Introduction . . . . . . . . . . . . . . . . . . . . 3. 1. 1 State Space . . . . . . . . . . . . . . . . 3. 1. 2 Transition Probability Matrix . . . . . . 3. 1. 3 Initial Distribution . . . . . . . . . . . . 3. 1. 4 Markov Property . . . . . . . . . . . . . 3. 1. 5 DTMC Models . . . . . . . . . . . . . . 3. 2 Stationary Distribution . . . . . . . . . . . . . 3. 2. 1 Interpretation of Stationary Distribution 3. 2. 2 Function of Stationary Distribution . . 3. 3 Irreducibility . . . . . . . . . . . . . . . . . . . 3. 3. 1 Transition Diagram . . . . . . . . . . 3. 3. 2 Accessibility of States . . . . . . . . . . 3. 4 Periodicity . . . . . . . . . . . . . . . . . . . . . 3. 5 Recurrence and Transience . . . . . . . . . . . 3. 5. 1 Geometric Random Variable . . . . . . 3. 6 Absorption Probability . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. 7 3. 8 3. 9 3. 0 Computing Stationary Distribution Using Cut Method Introduction to Binomial Stock Price Model . . . . . . Simulation . . . . . . . . . . . . . . . . . . . . . . . . . Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . 59 61 62 63 71 71 72 73 75 78 80 80 80 82 84 91 91 96 97 100 101 103 103 104 106 107 107 108 109 111 111 117 117 130 135 148 159 4 Poisson Process 4. 1 Exponential Distribution . . . . . . . 4. 1. 1 Memoryless Property . . . . 4. 1. 2 Comparing Two Exponentials 4. 2 Homogeneous Poisson Process . . . . 4. 3 Non-homogeneous Poisson Process . 4. Thinning and Merging . . . . . . . . 4. 4. 1 Merging Poisson Process . . . 4. 4. 2 Thinning Poisson Process . . 4. 5 Simulation . . . . . . . . . . . . . . . 4. 6 Exercise . . . . . . . . . . . . . . . . 5 Continuous Time Markov Chain 5. 1 Introduction . . . . . . . . . . . 5. 1. 1 Holding Times . . . . . 5. 1. 2 Generator Matrix . . . . 5. 2 Stationary Distribution . . . . 5. 3 M/M/1 Queue . . . . . . . . . 5. 4 Variations of M/M/1 Queue . . 5. 4. 1 M/M/1/b Queue . . . . 5. 4. 2 M/M/? Queue . . . . . 5. 4. 3 M/M/k Queue . . . . . 5. 5 Open Jackson Network . . . . . 5. 5. 1 M/M/1 Queue Review . 5. 5. 2 Tandem Queue . . . . . 5. 5. Failure Inspection . . . 5. 6 Simulation . . . . . . . . . . . . 5. 7 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Exercise Answers 6. 1 Newsvendor Problem . . . . . . . 6. 2 Queueing Theory . . . . . . . . . 6. 3 Discrete Time Markov Chain . . 6. 4 Poisson Process . . . . . . . . . . 6. 5 Continuous Time Markov Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Newsvendor Problem In this course, we will learn how to design, analyze, and manage a manufacturing or service system with uncertainty. Our ? rst step is to understand how to solve a single period decision problem containing uncertainty or randomness. 1. 1 Pro? t Maximization We will start with the simplest case: selling perishable items. Suppose we are running a business retailing newspaper to Georgia Tech campus. We have to order a speci? c number of copies from the publisher every evening and sell those copies the next day.One day, if there is a big news, the number of GT people who want to buy and read a paper from you may be very high. Another day, people may just not be interested in reading a paper at all. Hence, you as a retailer, will encounter the demand variability and it is the primary un certainty you need to handle to keep your business sustainable. To do that, you want to know what is the optimal number of copies you need to order every day. By intuition, you know that there will be a few other factors than demand you need to consider. †¢ Selling price (p): How much will you charge per paper? Buying price (cv ): How much will the publisher charge per paper? This is a variable cost, meaning that this cost is proportional to how many you order. That is why it is denoted by cv . †¢ Fixed ordering price (cf ): How much should you pay just to place an order? Ordering cost is ? xed regardless of how many you order. †¢ Salvage value (s) or holding cost (h): There are two cases about the leftover items. They could carry some monetary value even if expired. Otherwise, you have to pay to get rid of them or to storing them. If they have some value, it is called salvage value. If you have to pay, it is called 5 6 CHAPTER 1.NEWSVENDOR PROBLEM holding cost. Hence , the following relationship holds: s = ? h. This is per-item value. †¢ Backorder cost (b): Whenever the actual demand is higher than how many you prepared, you lose sales. Loss-of-sales could cost you something. You may be bookkeeping those as backorders or your brand may be damaged. These costs will be represented by backorder cost. This is per-item cost. †¢ Your order quantity (y): You will decide how many papers to be ordered before you start a day. That quantity is represented by y. This is your decision variable. As a business, you are assumed to want to maximize your pro? t. Expressing our pro? t as a function of these variables is the ? rst step to obtain the optimal ordering policy. Pro? t can be interpreted in two ways: (1) revenue minus cost, or (2) money you earn minus money you lose. Let us adopt the ? rst interpretation ? rst. Revenue is represented by selling price (p) multiplied by how many you actually sell. The actual sales is bounded by the realized dema nd and how many you prepared for the period. When you order too many, you can sell at most as many as the number of people who want to buy. When you order too few, you can only sell what you prepared. Hence, your revenue is minimum of D and y, i. . min(D, y) or D ? y. Thinking about the cost, ? rst of all, you have to pay something to the publisher when buying papers, i. e. cf +ycv . Two types of additional cost will be incurred to you depending on whether your order is above or below the actual demand. When it turns out you prepared less than the demand for the period, the backorder cost b per every missed sale will occur. The amount of missed sales cannot be negative, so it can be represented by max(D ? y, 0) or (D ? y)+ . When it turns out you prepared more, the quantity of left-over items also cannot go negative, so it can be expressed as max(y ? D, 0) or (y ? D)+ .In this way of thinking, we have the following formula. Pro? t =Revenue ? Cost =Revenue ? Ordering cost ? Holding c ost ? Backorder cost =p(D ? y) ? (cf + ycv ) ? h(y ? D)+ ? b(D ? y)+ (1. 1) How about the second interpretation of pro? t? You earn p ? cv dollars every time you sell a paper. For left-over items, you lose the price you bought in addition to the holding cost per paper, i. e. cv + h. When the demand is higher than what you prepared, you lose b backorder cost. Of course, you also have to pay the ? xed ordering cost cf as well when you place an order. With this logic, we have the following pro? t function. Pro? t =Earning ?Loss =(p ? cv )(D ? y) ? (cv + h)(y ? D)+ ? b(D ? y)+ ? cf (1. 2) 1. 1. PROFIT MAXIMIZATION 7 Since we used two di? erent approaches to model the same pro? t function, (1. 1) and (1. 2) should be equivalent. Comparing the two equations, you will also notice that (D ? y) + (y ? D)+ = y. Now our quest boils down to maximizing the pro? t function. However, (1. 1) and (1. 2) contain a random element, the demand D. We cannot maximize a function of random element if we all ow the randomness to remain in our objective function. One day demand can be very high. Another day it is also possible nobody wants to buy a single paper. We have to ? ure out how to get rid of this randomness from our objective function. Let us denote pro? t for the nth period by gn for further discussion. Theorem 1. 1 (Strong Law of Large Numbers). Pr g1 + g2 + g3 +  ·  ·  · + gn = E[g1 ] n>? n lim =1 The long-run average pro? t converges to the expected pro? t for a single period with probability 1. Based on Theorem 1. 1, we can change our objective function from just pro? t to expected pro? t. In other words, by maximizing the expected pro? t, it is guaranteed that the long-run average pro? t is maximized because of Theorem 1. 1. Theorem 1. 1 is the foundational assumption for the entire course.When we will talk about the long-run average something, it involves Theorem 1. 1 in most cases. Taking expectations, we obtain the following equations corresponding to (1. 1) and ( 1. 2). E[g(D, y)] =pE[D ? y] ? (cf + ycv ) ? hE[(y ? D)+ ] ? bE[(D ? y)+ ] =(p ? cv )E[D ? y] ? (cv + h)E[(y ? D)+ ] ? bE[(D ? y)+ ] ? cf (1. 4) (1. 3) Since (1. 3) and (1. 4) are equivalent, we can choose either one of them for further discussion and (1. 4) will be used. Before moving on, it is important for you to understand what E[D? y], E[(y? D)+ ], E[(D ? y)+ ] are and how to compute them. Example 1. 1. Compute E[D ? 18], E[(18 ? D)+ ], E[(D ? 8)+ ] for the demand having the following distributions. 1. D is a discrete random variable. Probability mass function (pmf) is as follows. d Pr{D = d} 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 Answer: For a discrete random variable, you ? rst compute D ? 18, (18 ? D)+ , (D ? 18)+ for each of possible D values. 8 d CHAPTER 1. NEWSVENDOR PROBLEM 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 Pr{D = d} D ? 18 (18 ? D)+ (D ? 18)+ 10 8 0 15 3 0 18 0 2 18 0 7 18 0 12 Then, you take the weighted average using corresponding Pr{D = d} for each possible D. 1 1 1 1 1 125 (10) + (15) + (18) + (18) + (18) = 4 8 8 4 4 8 1 1 1 1 1 19 + E[(18 ?D) ] = (8) + (3) + (0) + (0) + (0) = 4 8 8 4 4 8 1 1 1 1 1 + E[(D ? 18) ] = (0) + (0) + (2) + (7) + (12) = 5 4 8 8 4 4 E[D ? 18] = 2. D is a continuous random variable following uniform distribution between 10 and 30, i. e. D ? Uniform(10, 30). Answer: Computing expectation of continuous random variable involves integration. A continuous random variable has probability density function usually denoted by f . This will be also needed to compute the expectation. In this case, fD (x) = 1 20 , 0, if x ? [10, 30] otherwise Using this information, compute the expectations directly by integration. ? E[D ? 18] = ? 30 (x ? 18)fD (x)dx (x ? 18) 10 18 = = 10 18 1 dx 20 1 20 dx + 30 (x ? 18) x 10 dx + 18 30 (x ? 18) 1 20 dx 1 20 dx = = x2 40 1 20 + 18 x=18 x=10 18x 20 18 x=30 x=18 The key idea is to remove the ? operator that we cannot handle by separating the integration interval into two. The other two expectations can 1. 1. PROFIT MAXIMIZATION be computed in a similar way. 9 ? E[(18 ? D)+ ] = 30 (18 ? x)+ fD (x)dx (18 ? x)+ 10 18 = = 10 18 1 dx 20 1 20 1 20 +0 30 (18 ? x)+ (18 ? x) 10 x2 2 x=18 dx + 18 30 (18 ? x)+ 0 18 1 20 dx = dx + 1 20 dx 18x ? = 20 x=10 ? E[(D ? 18)+ ] = 30 (18 ? x)+ fD (x)dx (x ? 8)+ 10 18 = = 10 18 1 dx 20 1 20 30 (x ? 18)+ 0 10 x2 2 dx + 18 30 (x ? 18)+ 1 20 dx 1 20 dx = =0 + 1 20 dx + 18 x=30 (x ? 18) ? 18x 20 x=18 Now that we have learned how to compute E[D? y], E[(y? D)+ ], E[(D? y)+ ], we have acquired the basic toolkit to obtain the order quantity that maximizes the expected pro? t. First of all, we need to turn these expectations of the pro? t function formula (1. 4) into integration forms. For now, assume that the demand is a nonnegative continuous random variable. 10 CHAPTER 1. NEWSVENDOR PROBLEM E[g(D, y)] =(p ? cv )E[D ? y] ? (cv + h)E[(y ? D)+ ] ? bE[(D ? y)+ ] ? f ? =(p ? cv ) 0 (x ? y)fD (x)dx ? ? (cv + h) 0 ? (y ? x)+ fD (x)dx ?b 0 (x ? y)+ fD (x)dx ? cf y ? =(p ? cv ) 0 xfD (x)dx + y y yfD (x)dx ? (cv + h) 0 ? (y ? x)fD (x)dx ?b y (x ? y)fD (x)dx ? cf y y =(p ? cv ) 0 xfD (x)dx + y 1 ? 0 y y fD (x)dx xfD (x)dx ? (cv + h) y 0 y fD (x)dx ? 0 y ? b E[D] ? 0 xfD (x)dx ? y 1 ? 0 fD (x)dx ? cf (1. 5) There can be many ways to obtain the maximum point of a function. Here we will take the derivative of (1. 5) and set it to zero. y that makes the derivative equal to zero will make E[g(D, y)] either maximized or minimized depending on the second derivative.For now, assume that such y will maximize E[g(D, y)]. We will check this later. Taking the derivative of (1. 5) will involve di? erentiating an integral. Let us review an important result from Calculus. Theorem 1. 2 (Fundamental Theorem of Calculus). For a function y H(y) = c h(x)dx, we have H (y) = h(y), where c is a constant. Theorem 1. 2 can be translated as follows for our case. y d xfD (x)dx =yfD (y) dy 0 y d fD (x)dx =fD (y) dy 0 (1. 6) (1. 7) Also remember the relationship between cd f and pdf of a continuous random variable. y FD (y) = fD (x)dx (1. 8) 1. 1. PROFIT MAXIMIZATION Use (1. 6), (1. 7), (1. ) to take the derivative of (1. 5). d E[g(D, y)] =(p ? cv ) (yfD (y) + 1 ? FD (y) ? yfD (y)) dy ? (cv + h) (FD (y) + yfD (y) ? yfD (y)) ? b (? yfD (y) ? 1 + FD (y) + yfD (y)) =(p + b ? cv )(1 ? FD (y)) ? (cv + h)FD (y) =(p + b ? cv ) ? (p + b + h)FD (y) = 0 If we di? erentiate (1. 9) one more time to obtain the second derivative, d2 E[g(D, y)] = ? (p + b + h)fD (y) dy 2 11 (1. 9) which is always nonpositive because p, b, h, fD (y) ? 0. Hence, taking the derivative and setting it to zero will give us the maximum point not the minimum point. Therefore, we obtain the following result. Theorem 1. 3 (Optimal Order Quantity).The optimal order quantity y ? is the smallest y such that FD (y) = p + b ? cv ? 1 or y = FD p+b+h p + b ? cv p+b+h . for continuous demand D. Looking at Theorem 1. 3, it provides the following intuitions. †¢ Fixed cost cf does not a? ect the o ptimal quantity you need to order. †¢ If you can procure items for free and there is no holding cost, you will prepare as many as you can. †¢ If b h, b cv , you will also prepare as many as you can. †¢ If the buying cost is almost as same as the selling price plus backorder cost, i. e. cv ? p + b, you will prepare nothing. You will prepare only upon you receive an order.Example 1. 2. Suppose p = 10, cf = 100, cv = 5, h = 2, b = 3, D ? Uniform(10, 30). How many should you order for every period to maximize your long-run average pro? t? Answer: First of all, we need to compute the criterion value. p + b ? cv 10 + 3 ? 5 8 = = p+b+h 10 + 3 + 2 15 Then, we will look up the smallest y value that makes FD (y) = 8/15. 12 1 CHAPTER 1. NEWSVENDOR PROBLEM CDF 0. 5 0 0 5 10 15 20 25 30 35 40 D Therefore, we can conclude that the optimal order quantity 8 62 = units. 15 3 Although we derived the optimal order quantity solution for the continuous demand case, Theorem 1. applies to t he discrete demand case as well. I will ? ll in the derivation for discrete case later. y ? = 10 + 20 Example 1. 3. Suppose p = 10, cf = 100, cv = 5, h = 2, b = 3. Now, D is a discrete random variable having the following pmf. d Pr{D = d} 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 What is the optimal order quantity for every period? Answer: We will use the same value 8/15 from the previous example and look up the smallest y that makes FD (y) = 8/15. We start with y = 10. 1 4 1 1 3 FD (15) = + = 4 8 8 1 1 1 1 FD (20) = + + = 4 8 8 2 1 1 1 1 3 FD (25) = + + + = 4 8 8 4 4 ? Hence, the optimal order quantity y = 25 units.FD (10) = 8 15 8 < 15 8 < 15 8 ? 15 < 1. 2 Cost Minimization Suppose you are a production manager of a large company in charge of operating manufacturing lines. You are expected to run the factory to minimize the cost. Revenue is another person’s responsibility, so all you care is the cost. To model the cost of factory operation, let us set up variables in a slightly di? erent way. 1. 2. COST MINIMIZATION 13 †¢ Understock cost (cu ): It occurs when your production is not su? cient to meet the market demand. †¢ Overstock cost (co ): It occurs when you produce more than the market demand.In this case, you may have to rent a space to store the excess items. †¢ Unit production cost (cv ): It is the cost you should pay whenever you manufacture one unit of products. Material cost is one of this category. †¢ Fixed operating cost (cf ): It is the cost you should pay whenever you decide to start running the factory. As in the pro? t maximization case, the formula for cost expressed in terms of cu , co , cv , cf should be developed. Given random demand D, we have the following equation. Cost =Manufacturing Cost + Cost associated with Understock Risk + Cost associated with Overstock Risk =(cf + ycv ) + cu (D ? )+ + co (y ? D)+ (1. 10) (1. 10) obviously also contains randomness from D. We cannot minimize a random objective itself. Instead, based on Theorem 1. 1, we will minimize expected cost then the long-run average cost will be also guaranteed to be minimized. Hence, (1. 10) will be transformed into the following. E[Cost] =(cf + ycv ) + cu E[(D ? y)+ ] + co E[(y ? D)+ ] ? ? =(cf + ycv ) + cu 0 ? (x ? y)+ fD (x)dx + co 0 y (y ? x)+ fD (x)dx (y ? x)fD (x)dx (1. 11) 0 =(cf + ycv ) + cu y (x ? y)fD (x)dx + co Again, we will take the derivative of (1. 11) and set it to zero to obtain y that makes E[Cost] minimized.We will verify the second derivative is positive in this case. Let g here denote the cost function and use Theorem 1. 2 to take the derivative of integrals. d E[g(D, y)] =cv + cu (? yfD (y) ? 1 + FD (y) + yfD (y)) dy + co (FD (y) + yfD (y) ? yfD (y)) =cv + cu (FD (y) ? 1) + co FD (y) ? (1. 12) The optimal production quantity y is obtained by setting (1. 12) to be zero. Theorem 1. 4 (Optimal Production Quantity). The optimal production quantity that minimizes the long-run average cost is the smallest y such tha t FD (y) = cu ? cv or y = F ? 1 cu + co cu ? cv cu + co . 14 CHAPTER 1. NEWSVENDOR PROBLEM Theorem 1. can be also applied to discrete demand. Several intuitions can be obtained from Theorem 1. 4. †¢ Fixed cost (cf ) again does not a? ect the optimal production quantity. †¢ If understock cost (cu ) is equal to unit production cost (cv ), which makes cu ? cv = 0, then you will not produce anything. †¢ If unit production cost and overstock cost are negligible compared to understock cost, meaning cu cv , co , you will prepare as much as you can. To verify the second derivative of (1. 11) is indeed positive, take the derivative of (1. 12). d2 E[g(D, y)] = (cu + co )fD (y) dy 2 (1. 13) (1. 13) is always nonnegative because cu , co ? . Hence, y ? obtained from Theorem 1. 4 minimizes the cost instead of maximizing it. Before moving on, let us compare criteria from Theorem 1. 3 and Theorem 1. 4. p + b ? cv p+b+h and cu ? cv cu + co Since the pro? t maximization problem solved previously and the cost minimization problem solved now share the same logic, these two criteria should be somewhat equivalent. We can see the connection by matching cu = p + b, co = h. In the pro? t maximization problem, whenever you lose a sale due to underpreparation, it costs you the opportunity cost which is the selling price of an item and the backorder cost.Hence, cu = p + b makes sense. When you overprepare, you should pay the holding cost for each left-over item, so co = h also makes sense. In sum, Theorem 1. 3 and Theorem 1. 4 are indeed the same result in di? erent forms. Example 1. 4. Suppose demand follows Poisson distribution with parameter 3. The cost parameters are cu = 10, cv = 5, co = 15. Note that e? 3 ? 0. 0498. Answer: The criterion value is cu ? cv 10 ? 5 = = 0. 2, cu + co 10 + 15 so we need to ? nd the smallest y such that makes FD (y) ? 0. 2. Compute the probability of possible demands. 30 ? 3 e = 0. 0498 0! 31 Pr{D = 1} = e? 3 = 0. 1494 1! 32 ? Pr{D = 2} = e = 0. 2241 2! Pr{D = 0} = 1. 3. INITIAL INVENTORY Interpret these values into FD (y). FD (0) =Pr{D = 0} = 0. 0498 < 0. 2 FD (1) =Pr{D = 0} + Pr{D = 1} = 0. 1992 < 0. 2 FD (2) =Pr{D = 0} + Pr{D = 1} + Pr{D = 2} = 0. 4233 ? 0. 2 Hence, the optimal production quantity here is 2. 15 1. 3 Initial Inventory Now let us extend our model a bit further. As opposed to the assumption that we had no inventory at the beginning, suppose that we have m items when we decide how many we need to order. The solutions we have developed in previous sections assumed that we had no inventory when placing an order.If we had m items, we should order y ? ? m items instead of y ? items. In other words, the optimal order or production quantity is in fact the optimal order-up-to or production-up-to quantity. We had another implicit assumption that we should order, so the ? xed cost did not matter in the previous model. However, if cf is very large, meaning that starting o? a production line or placing an order i s very expensive, we may want to consider not to order. In such case, we have two scenarios: to order or not to order. We will compare the expected cost for the two scenarios and choose the option with lower expected cost.Example 1. 5. Suppose understock cost is $10, overstock cost is $2, unit purchasing cost is $4 and ? xed ordering cost is $30. In other words, cu = 10, co = 2, cv = 4, cf = 30. Assume that D ? Uniform(10, 20) and we already possess 10 items. Should we order or not? If we should, how many items should we order? Answer: First, we need to compute the optimal amount of items we need to prepare for each day. Since cu ? cv 1 10 ? 4 = , = cu + co 10 + 2 2 the optimal order-up-to quantity y ? = 15 units. Hence, if we need to order, we should order 5 = y ? ? m = 15 ? 10 items. Let us examine whether we should actually order or not. . Scenario 1: Not To Order If we decide not to order, we will not have to pay cf and cv since we order nothing actually. We just need to conside r understock and overstock risks. We will operate tomorrow with 10 items that we currently have if we decide not to order. E[Cost] =cu E[(D ? 10)+ ] + co E[(10 ? D)+ ] =10(E[D] ? 10) + 2(0) = $50 16 CHAPTER 1. NEWSVENDOR PROBLEM Note that in this case E[(10 ? D)+ ] = 0 because D is always greater than 10. 2. Scenario 2: To Order If we decide to order, we will order 5 items. We should pay cf and cv accordingly. Understock and overstock risks also exist in this case.Since we will order 5 items to lift up the inventory level to 15, we will run tomorrow with 15 items instead of 10 items if we decide to order. E[Cost] =cf + (15 ? 10)cv + cu E[(D ? 15)+ ] + co E[(15 ? D)+ ] =30 + 20 + 10(1. 25) + 2(1. 25) = $65 Since the expected cost of not ordering is lower than that of ordering, we should not order if we already have 10 items. It is obvious that if we have y ? items at hands right now, we should order nothing since we already possess the optimal amount of items for tomorrow’s op eration. It is also obvious that if we have nothing currently, we should order y ? items to prepare y ? tems for tomorrow. There should be a point between 0 and y ? where you are indi? erent between order and not ordering. Suppose you as a manager should give instruction to your assistant on when he/she should place an order and when should not. Instead of providing instructions for every possible current inventory level, it is easier to give your assistant just one number that separates the decision. Let us call that number the critical level of current inventory m? . If we have more than m? items at hands, the expected cost of not ordering will be lower than the expected cost of ordering, so we should not order.Conversely, if we have less than m? items currently, we should order. Therefore, when we have exactly m? items at hands right now, the expected cost of ordering should be equal to that of not ordering. We will use this intuition to obtain m? value. The decision process is s ummarized in the following ? gure. m* Critical level for placing an order y* Optimal order-up-to quantity Inventory If your current inventory lies here, you should order. Order up to y*. If your current inventory lies here, you should NOT order because your inventory is over m*. 1. 4. SIMULATION 17 Example 1. 6.Given the same settings with the previous example (cu = 10, cv = 4, co = 2, cf = 30), what is the critical level of current inventory m? that determines whether you should order or not? Answer: From the answer of the previous example, we can infer that the critical value should be less than 10, i. e. 0 < m? < 10. Suppose we currently own m? items. Now, evaluate the expected costs of the two scenarios: ordering and not ordering. 1. Scenario 1: Not Ordering E[Cost] =cu E[(D ? m? )+ ] + co E[(m? ? D)+ ] =10(E[D] ? m? ) + 2(0) = 150 ? 10m? 2. Scenario 2: Ordering In this case, we will order.Given that we will order, we will order y ? ?m? = 15 ? m? items. Therefore, we will start tomorrow with 15 items. E[Cost] =cf + (15 ? 10)cv + cu E[(D ? 15)+ ] + co E[(15 ? D)+ ] =30 + 4(15 ? m? ) + 10(1. 25) + 2(1. 25) = 105 ? 4m? At m? , (1. 14) and (1. 15) should be equal. 150 ? 10m? = 105 ? 4m? ? m? = 7. 5 units (1. 15) (1. 14) The critical value is 7. 5 units. If your current inventory is below 7. 5, you should order for tomorrow. If the current inventory is above 7. 5, you should not order. 1. 4 Simulation Generate 100 random demands from Uniform(10, 30). p = 10, cf = 30, cv = 4, h = 5, b = 3 1 p + b ? v 10 + 3 ? 4 = = p + b + h 10 + 3 + 5 2 The optimal order-up-to quantity from Theorem 1. 3 is 20. We will compare the performance between the policies of y = 15, 20, 25. Listing 1. 1: Continuous Uniform Demand Simulation # Set up parameters p=10;cf=30;cv=4;h=5;b=3 # How many random demands will be generated? n=100 # Generate n random demands from the uniform distribution 18 Dmd=runif(n,min=10,max=30) CHAPTER 1. NEWSVENDOR PROBLEM # Test the policy where we order 15 it ems for every period y=15 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 33. 4218 # Test the policy where we order 20 items for every period y=20 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 44. 37095 # Test the policy where we order 25 items for every period y=25 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 32. 62382 You can see the policy with y = 20 maximizes the 100-period average pro? t as promised by the theory. In fact, if n is relatively small, it is not guaranteed that we have maximized pro? t even if we run based on the optimal policy obtained from this section.The underlying assumption is that we should operate with this policy for a long time. Then, Theorem 1. 1 guarantees that the average pro? t will be maximized when we use the optimal ordering policy. Discrete demand case can also be simulated. Suppose the demand has the following distribution. All other parameters remain same. d Pr{D = d} 10 1 4 15 1 8 20 1 4 25 1 8 30 1 4 The theoretic optimal order-up-to quantity in this case is also 20. Let us test three policies: y = 15, 20, 25. Listing 1. 2: Discrete Demand Simulation # Set up parameters p=10;cf=30;cv=4;h=5;b=3 # How many random demands will be generated? =100 # Generate n random demands from the discrete demand distribution Dmd=sample(c(10,15,20,25,30),n,replace=TRUE,c(1/4,1/8,1/4,1/8,1/4)) # Test the policy where we order 15 items for every period y=15 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 19. 35 # Test the policy where we order 20 items for every period y=20 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 31. 05 # Test the policy where we order 25 items for every period 1. 5. EXERCISE y=25 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 26. 55 19There are other distributions such as triangular, normal, Poisson or binomial distributions available in R. When you do your senior project, for example, you will observe the demand for a departm ent or a factory. You ? rst approximate the demand using these theoretically established distributions. Then, you can simulate the performance of possible operation policies. 1. 5 Exercise 1. Show that (D ? y) + (y ? D)+ = y. 2. Let D be a discrete random variable with the following pmf. d Pr{D = d} Find (a) E[min(D, 7)] (b) E[(7 ? D)+ ] where x+ = max(x, 0). 3. Let D be a Poisson random variable with parameter 3.Find (a) E[min(D, 2)] (b) E[(3 ? D)+ ]. Note that pmf of a Poisson random variable with parameter ? is Pr{D = k} = ? k e . k! 5 1 10 6 3 10 7 4 10 8 1 10 9 1 10 4. Let D be a continuous random variable and uniformly distributed between 5 and 10. Find (a) E[max(D, 8)] (b) E[(D ? 8)? ] where x? = min(x, 0). 5. Let D be an exponential random variable with parameter 7. Find (a) E[max(D, 3)] 20 (b) E[(D ? 4)? ]. CHAPTER 1. NEWSVENDOR PROBLEM Note that pdf of an exponential random variable with parameter ? is fD (x) = ? e x for x ? 0. 6. David buys fruits and vegetables wholesal e and retails them at Davids Produce on La Vista Road.One of the more di? cult decisions is the amount of bananas to buy. Let us make some simplifying assumptions, and assume that David purchases bananas once a week at 10 cents per pound and retails them at 30 cents per pound during the week. Bananas that are more than a week old are too ripe and are sold for 5 cents per pound. (a) Suppose the demand for the good bananas follows the same distribution as D given in Problem 2. What is the expected pro? t of David in a week if he buys 7 pounds of banana? (b) Now assume that the demand for the good bananas is uniformly distributed between 5 and 10 like in Problem 4.What is the expected pro? t of David in a week if he buys 7 pounds of banana? (c) Find the expected pro? t if David’s demand for the good bananas follows an exponential distribution with mean 7 and if he buys 7 pounds of banana. 7. Suppose we are selling lemonade during a football game. The lemonade sells for $18 per g allon but only costs $3 per gallon to make. If we run out of lemonade during the game, it will be impossible to get more. On the other hand, leftover lemonade has a value of $1. Assume that we believe the fans would buy 10 gallons with probability 0. 1, 11 gallons with probability 0. , 12 gallons with probability 0. 4, 13 gallons with probability 0. 2, and 14 gallons with probability 0. 1. (a) What is the mean demand? (b) If 11 gallons are prepared, what is the expected pro? t? (c) What is the best amount of lemonade to order before the game? (d) Instead, suppose that the demand was normally distributed with mean 1000 gallons and variance 200 gallons2 . How much lemonade should be ordered? 8. Suppose that a bakery specializes in chocolate cakes. Assume the cakes retail at $20 per cake, but it takes $10 to prepare each cake. Cakes cannot be sold after one week, and they have a negligible salvage value.It is estimated that the weekly demand for cakes is: 15 cakes in 5% of the weeks, 1 6 cakes in 20% of the weeks, 17 cakes in 30% of the weeks, 18 cakes in 25% of the weeks, 19 cakes in 10% of the weeks, and 20 cakes in 10% of the weeks. How many cakes should the bakery prepare each week? What is the bakery’s expected optimal weekly pro? t? 1. 5. EXERCISE 21 9. A camera store specializes in a particular popular and fancy camera. Assume that these cameras become obsolete at the end of the month. They guarantee that if they are out of stock, they will special-order the camera and promise delivery the next day.In fact, what the store does is to purchase the camera from an out of state retailer and have it delivered through an express service. Thus, when the store is out of stock, they actually lose the sales price of the camera and the shipping charge, but they maintain their good reputation. The retail price of the camera is $600, and the special delivery charge adds another $50 to the cost. At the end of each month, there is an inventory holding cost of $25 fo r each camera in stock (for doing inventory etc). Wholesale cost for the store to purchase the cameras is $480 each. (Assume that the order can only be made at the beginning of the month. (a) Assume that the demand has a discrete uniform distribution from 10 to 15 cameras a month (inclusive). If 12 cameras are ordered at the beginning of a month, what are the expected overstock cost and the expected understock or shortage cost? What is the expected total cost? (b) What is optimal number of cameras to order to minimize the expected total cost? (c) Assume that the demand can be approximated by a normal distribution with mean 1000 and standard deviation 100 cameras a month. What is the optimal number of cameras to order to minimize the expected total cost? 10.Next month’s production at a manufacturing company will use a certain solvent for part of its production process. Assume that there is an ordering cost of $1,000 incurred whenever an order for the solvent is placed and the solvent costs $40 per liter. Due to short product life cycle, unused solvent cannot be used in following months. There will be a $10 disposal charge for each liter of solvent left over at the end of the month. If there is a shortage of solvent, the production process is seriously disrupted at a cost of $100 per liter short. Assume that the initial inventory level is m, where m = 0, 100, 300, 500 and 700 liters. a) What is the optimal ordering quantity for each case when the demand is discrete with Pr{D = 500} = Pr{D = 800} = 1/8, Pr{D = 600} = 1/2 and Pr{D = 700} = 1/4? (b) What is the optimal ordering policy for arbitrary initial inventory level m? (You need to specify the critical value m? in addition to the optimal order-up-to quantity y ? . When m ? m? , you make an order. Otherwise, do not order. ) (c) Assume optimal quantity will be ordered. What is the total expected cost when the initial inventory m = 0? What is the total expected cost when the initial inventory m = 700? 22 CHAPTER 1. NEWSVENDOR PROBLEM 11.Redo Problem 10 for the case where the demand is governed by the continuous uniform distribution varying between 400 and 800 liters. 12. An automotive company will make one last production run of parts for Part 947A and 947B, which are not interchangeable. These parts are no longer used in new cars, but will be needed as replacements for warranty work in existing cars. The demand during the warranty period for 947A is approximately normally distributed with mean 1,500,000 parts and standard deviation 500,000 parts, while the mean and standard deviation for 947B is 500,000 parts and 100,000 parts. (Assume that two demands are independent. Ignoring the cost of setting up for producing the part, each part costs only 10 cents to produce. However, if additional parts are needed beyond what has been produced, they will be purchased at 90 cents per part (the same price for which the automotive company sells its parts). Parts remaining at the end of the warr anty period have a salvage value of 8 cents per part. There has been a proposal to produce Part 947C, which can be used to replace either of the other two parts. The unit cost of 947C jumps from 10 to 14 cents, but all other costs remain the same. (a) Assuming 947C is not produced, how many 947A should be produced? b) Assuming 947C is not produced, how many 947B should be produced? (c) How many 947C should be produced in order to satisfy the same fraction of demand from parts produced in-house as in the ? rst two parts of this problem. (d) How much money would be saved or lost by producing 947C, but meeting the same fraction of demand in-house? (e) Is your answer to question (c), the optimal number of 947C to produce? If not, what would be the optimal number of 947C to produce? (f) Should the more expensive part 947C be produced instead of the two existing parts 947A and 947B. Why? Hint: compare the expected total costs.Also, suppose that D ? Normal( µ, ? 2 ). q xv 0 (x?  µ)2 1 e? 2? 2 dx = 2 q (x ?  µ) v 0 q (x?  µ)2 1 e? 2? 2 dx 2 + µ =  µ2 v 0 (q?  µ)2 (x?  µ)2 1 e? 2? 2 dx 2 t 1 v e? 2? 2 dt +  µPr{0 ? D ? q} 2 2 where, in the 2nd step, we changed variable by letting t = (x ?  µ)2 . 1. 5. EXERCISE 23 13. A warranty department manages the after-sale service for a critical part of a product. The department has an obligation to replace any damaged parts in the next 6 months. The number of damaged parts X in the next 6 months is assumed to be a random variable that follows the following distribution: x Pr{X = x} 100 . 1 200 . 2 300 . 5 400 . 2The department currently has 200 parts in stock. The department needs to decide if it should make one last production run for the part to be used for the next 6 months. To start the production run, the ? xed cost is $2000. The unit cost to produce a part is $50. During the warranty period of next 6 months, if a replacement request comes and the department does not have a part available in house, it has to buy a part from the spot-market at the cost of $100 per part. Any part left at the end of 6 month sells at $10. (There is no holding cost. ) Should the department make the production run? If so, how many items should it produce? 4. A store sells a particular brand of fresh juice. By the end of the day, any unsold juice is sold at a discounted price of $2 per gallon. The store gets the juice daily from a local producer at the cost of $5 per gallon, and it sells the juice at $10 per gallon. Assume that the daily demand for the juice is uniformly distributed between 50 gallons to 150 gallons. (a) What is the optimal number of gallons that the store should order from the distribution each day in order to maximize the expected pro? t each day? (b) If 100 gallons are ordered, what is the expected pro? t per day? 15. An auto company is to make one ? al purchase of a rare engine oil to ful? ll its warranty services for certain car models. The current price for the engine oil is $1 per g allon. If the company runs out the oil during the warranty period, it will purchase the oil from a supply at the market price of $4 per gallon. Any leftover engine oil after the warranty period is useless, and costs $1 per gallon to get rid of. Assume the engine oil demand during the warranty is uniformly distributed (continuous distribution) between 1 million gallons to 2 million gallons, and that the company currently has half million gallons of engine oil in stock (free of charge). a) What is the optimal amount of engine oil the company should purchase now in order to minimize the total expected cost? (b) If 1 million gallons are purchased now, what is the total expected cost? 24 CHAPTER 1. NEWSVENDOR PROBLEM 16. A company is obligated to provide warranty service for Product A to its customers next year. The warranty demand for the product follows the following distribution. d Pr{D = d} 100 . 2 200 . 4 300 . 3 400 . 1 The company needs to make one production run to satisfy the wa rranty demand for entire next year. Each unit costs $100 to produce; the penalty cost of a unit is $500.By the end of the year, the savage value of each unit is $50. (a) Suppose that the company has currently 0 units. What is the optimal quantity to produce in order to minimize the expected total cost? Find the optimal expected total cost. (b) Suppose that the company has currently 100 units at no cost and there is $20000 ? xed cost to start the production run. What is the optimal quantity to produce in order to minimize the expected total cost? Find the optimal expected total cost. 17. Suppose you are running a restaurant having only one menu, lettuce salad, in the Tech Square.You should order lettuce every day 10pm after closing. Then, your supplier delivers the ordered amount of lettuce 5am next morning. Store hours is from 11am to 9pm every day. The demand for the lettuce salad for a day (11am-9pm) has the following distribution. d Pr{D = d} 20 1/6 25 1/3 30 1/3 35 1/6 One lettu ce salad requires two units of lettuce. The selling price of lettuce salad is $6, the buying price of one unit of lettuce is $1. Of course, leftover lettuce of a day cannot be used for future salad and you have to pay 50 cents per unit of lettuce for disposal. (a) What is the optimal order-up-to quantity of lettuce for a day? b) If you ordered 50 units of lettuce today, what is the expected pro? t of tomorrow? Include the purchasing cost of 50 units of lettuce in your calculation. Chapter 2 Queueing Theory Before getting into Discrete-time Markov Chains, we will learn about general issues in the queueing theory. Queueing theory deals with a set of systems having waiting space. It is a very powerful tool that can model a broad range of issues. Starting from analyzing a simple queue, a set of queues connected with each other will be covered as well in the end. This chapter will give you the background knowledge when you read the required book, The Goal.We will revisit the queueing the ory once we have more advanced modeling techniques and knowledge. 2. 1 Introduction Think about a service system. All of you must have experienced waiting in a service system. One example would be the Student Center or some restaurants. This is a human system. A bit more automated service system that has a queue would be a call center and automated answering machines. We can imagine a manufacturing system instead of a service system. These waiting systems can be generalized as a set of bu? ers and servers. There are key factors when you try to model such a system.What would you need to analyze your system? †¢ How frequently customers come to your system? > Inter-arrival Times †¢ How fast your servers can serve the customers? > Service Times †¢ How many servers do you have? > Number of Servers †¢ How large is your waiting space? > Queue Size If you can collect data about these metrics, you can characterize your queueing system. In general, a queueing system can be denoted as follows. G/G/s/k 25 26 CHAPTER 2. QUEUEING THEORY The ? rst letter characterizes the distribution of inter-arrival times. The second letter characterizes the distribution of service times.The third number denotes the number of servers of your queueing system. The fourth number denotes the total capacity of your system. The fourth number can be omitted and in such case it means that your capacity is in? nite, i. e. your system can contain any number of people in it up to in? nity. The letter â€Å"G† represents a general distribution. Other candidate characters for this position is â€Å"M† and â€Å"D† and the meanings are as follows. †¢ G: General Distribution †¢ M: Exponential Distribution †¢ D: Deterministic Distribution (or constant) The number of servers can vary from one to many to in? nity.The size of bu? er can also be either ? nite or in? nite. To simplify the model, assume that there is only a single server and we have in? ni te bu? er. By in? nite bu? er, it means that space is so spacious that it is as if the limit does not exist. Now we set up the model for our queueing system. In terms of analysis, what are we interested in? What would be the performance measures of such systems that you as a manager should know? †¢ How long should your customer wait in line on average? †¢ How long is the waiting line on average? There are two concepts of average. One is average over time.This applies to the average number of customers in the system or in the queue. The other is average over people. This applies to the average waiting time per customer. You should be able to distinguish these two. Example 2. 1. Assume that the system is empty at t = 0. Assume that u1 = 1, u2 = 3, u3 = 2, u4 = 3, v1 = 4, v2 = 2, v3 = 1, v4 = 2. (ui is ith customer’s inter-arrival time and vi is ith customer’s service time. ) 1. What is the average number of customers in the system during the ? rst 10 minutes? 2 . What is the average queue size during the ? rst 10 minutes? 3.What is the average waiting time per customer for the ? rst 4 customers? Answer: 1. If we draw the number of people in the system at time t with respect to t, it will be as follows. 2. 2. LINDLEY EQUATION 3 2 1 0 27 Z(t) 0 1 2 3 4 5 6 7 8 9 10 t E[Z(t)]t? [0,10] = 1 10 10 Z(t)dt = 0 1 (10) = 1 10 2. If we draw the number of people in the queue at time t with respect to t, it will be as follows. 3 2 1 0 Q(t) 0 1 2 3 4 5 6 7 8 9 10 t E[Q(t)]t? [0,10] = 1 10 10 Q(t)dt = 0 1 (2) = 0. 2 10 3. We ? rst need to compute waiting times for each of 4 customers. Since the ? rst customer does not wait, w1 = 0.Since the second customer arrives at time 4, while the ? rst customer’s service ends at time 5. So, the second customer has to wait 1 minute, w2 = 1. Using the similar logic, w3 = 1, w4 = 0. E[W ] = 0+1+1+0 = 0. 5 min 4 2. 2 Lindley Equation From the previous example, we now should be able to compute each customerâ€℠¢s waiting time given ui , vi . It requires too much e? ort if we have to draw graphs every time we need to compute wi . Let us generalize the logic behind calculating waiting times for each customer. Let us determine (i + 1)th customer’s waiting 28 CHAPTER 2. QUEUEING THEORY time.If (i + 1)th customer arrives after all the time ith customer waited and got served, (i + 1)th customer does not have to wait. Its waiting time is 0. Otherwise, it has to wait wi + vi ? ui+1 . Figure 2. 1, and Figure 2. 2 explain the two cases. ui+1 wi vi wi+1 Time i th arrival i th service start (i+1)th arrival i th service end Figure 2. 1: (i + 1)th arrival before ith service completion. (i + 1)th waiting time is wi + vi ? ui+1 . ui+1 wi vi Time i th arrival i th service start i th service end (i+1)th arrival Figure 2. 2: (i + 1)th arrival after ith service completion. (i + 1)th waiting time is 0.Simply put, wi+1 = (wi + vi ? ui+1 )+ . This is called the Lindley Equation. Example 2. 2. Given the f ollowing inter-arrival times and service times of ? rst 10 customers, compute waiting times and system times (time spent in the system including waiting time and service time) for each customer. ui = 3, 2, 5, 1, 2, 4, 1, 5, 3, 2 vi = 4, 3, 2, 5, 2, 2, 1, 4, 2, 3 Answer: Note that system time can be obtained by adding waiting time and service time. Denote the system time of ith customer by zi . ui vi wi zi 3 4 0 4 2 3 2 5 5 2 0 2 1 5 1 6 2 2 4 6 4 2 2 4 1 1 3 4 5 4 0 4 3 2 1 3 2 3 1 4 2. 3. TRAFFIC INTENSITY 9 2. 3 Suppose Tra? c Intensity E[ui ] =mean inter-arrival time = 2 min E[vi ] =mean service time = 4 min. Is this queueing system stable? By stable, it means that the queue size should not go to the in? nity. Intuitively, this queueing system will not last because average service time is greater than average inter-arrival time so your system will soon explode. What was the logic behind this judgement? It was basically comparing the average inter-arrival time and the average serv ice time. To simplify the judgement, we come up with a new quantity called the tra? c intensity. De? nition 2. 1 (Tra? Intensity). Tra? c intensity ? is de? ned to be ? = 1/E[ui ] ? =  µ 1/E[vi ] where ? is the arrival rate and  µ is the service rate. Given a tra? c intensity, it will fall into one of the following three categories. †¢ If ? < 1, the system is stable. †¢ If ? = 1, the system is unstable unless both inter-arrival times and service times are deterministic (constant). †¢ If ? > 1, the system is unstable. Then, why don’t we call ? utilization instead of tra? c intensity? Utilization seems to be more intuitive and user-friendly name. In fact, utilization just happens to be same as ? if ? < 1.However, the problem arises if ? > 1 because utilization cannot go over 100%. Utilization is bounded above by 1 and that is why tra? c intensity is regarded more general notation to compare arrival and service rates. De? nition 2. 2 (Utilization). Utilization is de? ned as follows. Utilization = ? , 1, if ? < 1 if ? ? 1 Utilization can also be interpreted as the long-run fraction of time the server is utilized. 2. 4 Kingman Approximation Formula Theorem 2. 1 (Kingman’s High-tra? c Approximation Formula). Assume the tra? c intensity ? < 1 and ? is close to 1. The long-run average waiting time in 0 a queue E[W ] ? E[vi ] CHAPTER 2. QUEUEING THEORY ? 1 c2 + c2 a s 2 where c2 , c2 are squared coe? cient of variation of inter-arrival times and service a s times de? ned as follows. c2 = a Var[u1 ] (E[u1 ]) 2, c2 = s Var[v1 ] (E[v1 ]) 2 Example 2. 3. 1. Suppose inter-arrival time follows an exponential distribution with mean time 3 minutes and service time follows an exponential distribution with mean time 2 minutes. What is the expected waiting time per customer? 2. Suppose inter-arrival time is constant 3 minutes and service time is also constant 2 minutes. What is the expected waiting time per customer?Answer: 1. Tra? c intensity is ? = 1/E[ui ] 1/3 2 ? = = = .  µ 1/E[vi ] 1/2 3 Since both inter-arrival times and service times are exponentially distributed, E[ui ] = 3, Var[ui ] = 32 = 9, E[vi ] = 2, Var[vi ] = 22 = 4. Therefore, c2 = Var[ui ]/(E[ui ])2 = 1, c2 = 1. Hence, s a E[W ] =E[vi ] =2 ? c2 + c2 s a 1 2 2/3 1+1 = 4 minutes. 1/3 2 2. Tra? c intensity remains same, 2/3. However, since both inter-arrival times and service times are constant, their variances are 0. Thus, c2 = a c2 = 0. s E[W ] = 2 2/3 1/3 0+0 2 = 0 minutes It means that none of the customers will wait upon their arrival.As shown in the previous example, when the distributions for both interarrival times and service times are exponential, the squared coe? cient of variation term becomes 1 from the Kingman’s approximation formula and the formula 2. 5. LITTLE’S LAW 31 becomes exact to compute the average waiting time per customer for M/M/1 queue. E[W ] =E[vi ] ? 1 Also note that if inter-arrival time or service time distribution is deterministic, c2 or c2 becomes 0. a s Example 2. 4. You are running a highway collecting money at the entering toll gate. You reduced the utilization level of the highway from 90% to 80% by adopting car pool lane.How much does the average waiting time in front of the toll gate decrease? Answer: 0. 8 0. 9 = 9, =4 1 ? 0. 9 1 ? 0. 8 The average waiting time in in front of the toll gate is reduced by more than a half. The Goal is about identifying bottlenecks in a plant. When you become a manager of a company and are running a expensive machine, you usually want to run it all the time with full utilization. However, the implication of Kingman formula tells you that as your utilization approaches to 100%, the waiting time will be skyrocketing. It means that if there is any uncertainty or random ? ctuation input to your system, your system will greatly su? er. In lower ? region, increasing ? is not that bad. If ? near 1, increasing utilization a little bit can lead to a disaster. Atl anta, 10 years ago, did not su? er that much of tra? c problem. As its tra? c infrastructure capacity is getting closer to the demand, it is getting more and more fragile to uncertainty. A lot of strategies presented in The Goal is in fact to decrease ?. You can do various things to reduce ? of your system by outsourcing some process, etc. You can also strategically manage or balance the load on di? erent parts of your system.You may want to utilize customer service organization 95% of time, while utilization of sales people is 10%. 2. 5 Little’s Law L = ? W The Little’s Law is much more general than G/G/1 queue. It can be applied to any black box with de? nite boundary. The Georgia Tech campus can be one black box. ISyE building itself can be another. In G/G/1 queue, we can easily get average size of queue or service time or time in system as we di? erently draw box onto the queueing system. The following example shows that Little’s law can be applied in broade r context than the queueing theory. 32 CHAPTER 2. QUEUEING THEORY Example 2. 5 (Merge of I-75 and I-85).Atlanta is the place where two interstate highways, I-75 and I-85, merge and cross each other. As a tra? c manager of Atlanta, you would like to estimate the average time it takes to drive from the north con? uence point to the south con? uence point. On average, 100 cars per minute enter the merged area from I-75 and 200 cars per minute enter the same area from I-85. You also dispatched a chopper to take a aerial snapshot of the merged area and counted how many cars are in the area. It turned out that on average 3000 cars are within the merged area. What is the average time between entering and exiting the area per vehicle?Answer: L =3000 cars ? =100 + 200 = 300 cars/min 3000 L = 10 minutes ? W = = ? 300 2. 6 Throughput Another focus of The Goal is set on the throughput of a system. Throughput is de? ned as follows. De? nition 2. 3 (Throughput). Throughput is the rate of output ? ow from a system. If ? ? 1, throughput= ?. If ? > 1, throughput=  µ. The bounding constraint of throughput is either arrival rate or service rate depending on the tra? c intensity. Example 2. 6 (Tandem queue with two stations). Suppose your factory production line has two stations linked in series. Every raw material coming into your line should be processed by Station A ? rst.Once it is processed by Station A, it goes to Station B for ? nishing. Suppose raw material is coming into your line at 15 units per minute. Station A can process 20 units per minute and Station B can process 25 units per minute. 1. What is the throughput of the entire system? 2. If we double the arrival rate of raw material from 15 to 30 units per minute, what is the throughput of the whole system? Answer: 1. First, obtain the tra? c intensity for Station A. ?A = ? 15 = = 0. 75  µA 20 Since ? A < 1, the throughput of Station A is ? = 15 units per minute. Since Station A and Station B is linked in series, the throughput of Station . 7. SIMULATION A becomes the arrival rate for Station B. ?B = ? 15 = = 0. 6  µB 25 33 Also, ? B < 1, the throughput of Station B is ? = 15 units per minute. Since Station B is the ? nal stage of the entire system, the throughput of the entire system is also ? = 15 units per minute. 2. Repeat the same steps. ?A = 30 ? = = 1. 5  µA 20 Since ? A > 1, the throughput of Station A is  µA = 20 units per minute, which in turn becomes the arrival rate for Station B. ?B =  µA 20 = 0. 8 =  µB 25 ?B < 1, so the throughput of Station B is  µA = 20 units per minute, which in turn is the throughput of the whole system. 2. 7 SimulationListing 2. 1: Simulation of a Simple Queue and Lindley Equation N = 100 # Function for Lindley Equation lindley = function(u,v){ for (i in 1:length(u)) { if(i==1) w = 0 else { w = append(w, max(w[i-1]+v[i-1]-u[i], 0)) } } return(w) } # # u v CASE 1: Discrete Distribution Generate N inter-arrival times and service times = sample( c(2,3,4),N,replace=TRUE,c(1/3,1/3,1/3)) = sample(c(1,2,3),N,replace=TRUE,c(1/3,1/3,1/3)) # Compute waiting time for each customer w = lindley(u,v) w # CASE 2: Deterministic Distribution # All inter-arrival times are 3 minutes and all service times are 2 minutes # Observe that nobody waits in this case. 4 u = rep(3, 100) v = rep(2, 100) w = lindley(u,v) w CHAPTER 2. QUEUEING THEORY The Kingman’s approximation formula is exact when inter-arrival times and service times follow iid exponential distribution. E[W ] = 1  µ ? 1 We can con? rm this equation by simulating an M/M/1 queue. Listing 2. 2: Kingman Approximation # lambda = arrival rate, mu = service rate N = 10000; lambda = 1/10; mu = 1/7 # Generate N inter-arrival times and service times from exponential distribution u = rexp(N,rate=lambda) v = rexp(N,rate=mu) # Compute the average waiting time of each customer w = lindley(u,v) mean(w) > 16. 0720 # Compare with Kingman approximation rho = lambda/mu (1/mu)*(rho/(1-rho)) > 16. 33333 The Kingman’s approximation formula becomes more and more accurate as N grows. 2. 8 Exercise 1. Let Y be a random variable with p. d. f. ce? 3s for s ? 0, where c is a constant. (a) Determine c. (b) What is the mean, variance, and squared coe? cient of variation of Y where the squared coe? cient of variation of Y is de? ned to be Var[Y ]/(E[Y ]2 )? 2. Consider a single server queue. Initially, there is no customer in the system.Suppose that the inter-arrival times of the ? rst 15 customers are: 2, 5, 7, 3, 1, 4, 9, 3, 10, 8, 3, 2, 16, 1, 8 2. 8. EXERCISE 35 In other words, the ? rst customer will arrive at t = 2 minutes, and the second will arrive at t = 2 + 5 minutes, and so on. Also, suppose that the service time of the ? rst 15 customers are 1, 4, 2, 8, 3, 7, 5, 2, 6, 11, 9, 2, 1, 7, 6 (a) Compute the average waiting time (the time customer spend in bu? er) of the ? rst 10 departed customers. (b) Compute the average system time (waiting time plus service time) of the ? st 10 departed customers. (c) Compute the average queue size during the ? rst 20 minutes. (d) Compute the average server utilization during the ? rst 20 minutes. (e) Does the Little’s law of hold for the average queue size in the ? rst 20 minutes? 3. We want to decide whether to employ a human operator or buy a machine to paint steel beams with a rust inhibitor. Steel beams are produced at a constant rate of one every 14 minutes. A skilled human operator takes an average time of 700 seconds to paint a steel beam, with a standard deviation of 300 seconds.An automatic painter takes on average 40 seconds more than the human painter to paint a beam, but with a standard deviation of only 150 seconds. Estimate the expected waiting time in queue of a steel beam for each of the operators, as well as the expected number of steel beams waiting in queue in each of the two cases. Comment on the e? ect of variability in service time. 4. The arrival rate of customers to an ATM machi ne is 30 per hour with exponentially distirbuted in- terarrival times. The transaction times of two customers are independent and identically distributed.Each transaction time (in minutes) is distributed according to the following pdf: f (s) = where ? = 2/3. (a) What is the average waiting for each customer? (b) What is the average number of customers waiting in line? (c) What is the average number of customers at the site? 5. A production line has two machines, Machine A and Machine B, that are arranged in series. Each job needs to processed by Machine A ? rst. Once it ? nishes the processing by Machine A, it moves to the next station, to be processed by Machine B. Once it ? nishes the processing by Machine B, it leaves the production line.Each machine can process one job at a time. An arriving job that ? nds the machine busy waits in a bu? er. 4? 2 se? 2? s , 0, if s ? 0 otherwise 36 CHAPTER 2. QUEUEING THEORY (The bu? er sizes are assumed to be in? nite. ) The processing times fo r Machine A are iid having exponential distribution with mean 4 minutes. The processing times for Machine B are iid with mean 2 minutes. Assume that the inter-arrival times of jobs arriving at the production line are iid, having exponential distribution with mean of 5 minutes. (a) What is the utilization of Machine A?What is the utilization of Machine B? (b) What is the throughput of the production system? (Throughput is de? ned to be the rate of ? nal output ? ow, i. e. how many items will exit the system in a unit time. ) (c) What is the average waiting time at Machine A, excluding the service time? (d) It is known the average time in the entire production line is 30 minutes per job. What is the long-run average number of jobs in the entire production line? (e) Suppose that the mean inter-arrival time is changed to 1 minute. What are the utilizations for Machine A and Machine B, respectively?What is the throughput of the production system? 6. An auto collision shop has roughly 10 cars arriving per week for repairs. A car waits outside until it is brought inside for bumping. After bumping, the car is painted. On the average, there are 15 cars waiting outside in the yard to be repaired, 10 cars inside in the bump area, and 5 cars inside in the painting area. What is the average length of time a car is in the yard, in the bump area, and in the painting area? What is the average length of time from when a car arrives until it leaves? 7. A small bank is sta? d by a single server. It has been observed that, during a normal business day, the inter-arrival times of customers to the bank are iid having exponential distribution with mean 3 minutes. Also, the the processing times of customers are iid having the following distribution (in minutes): x Pr{X = x} 1 1/4 2 1/2 3 1/4 An arrival ? nding the server busy joins the queue. The waiting space is in? nite. (a) What is the long-run fraction of time that the server is busy? (b) What the the long-run average waiting tim e of each customer in the queue, excluding the processing time? c) What is average number of customers in the bank, those in queue plus those in service? 2. 8. EXERCISE (d) What is the throughput of the bank? 37 (e) If the inter-arrival times have mean 1 minute. What is the throughput of the bank? 8. You are the manager at the Student Center in charge of running the food court. The food court is composed of two parts: cooking station and cashier’s desk. Every person should go to the cooking station, place an order, wait there and pick up ? rst. Then, the person goes to the cashier’s desk to check out. After checking out, the person leaves the food court.The coo

What impressions of Silas Marner do you form early in the novel Essay Example

What impressions of Silas Marner do you form early in the novel Essay Example What impressions of Silas Marner do you form early in the novel Paper What impressions of Silas Marner do you form early in the novel Paper Essay Topic: Literature This essay plans to tell the reader about the early chapters in the book Silas Marner. Part I begins with the description and comparisons of the two settings that are introduced to us in this book, it then goes on to tell the reader about Silas himself, Silass physical drawbacks and his membership of the sect in the Lantern Yard. Part II then goes on to tell the reader about the more social aspects of Silass life and then about his settlement in Raveloe. In the opening chapters of this book we are introduced to two types of setting. Firstly we are introduced to Raveloe (pages 10 through to 12), this is quite a verdant village, a woody village tucked away in the back woods of the fresh English Midlands. The way Raveloe is placed geographically and its seclusion from the rest of the Midlands, the intellectual and spiritual confinement of its quite simple and its very misunderstood tastes to the countryside are premeditated. In this setting the writer George Eliot captures all the views of the bygone era, prompted by her indelible Warwickshire memories. The second setting that we are given is one that is not written in the text but one, which we have to look for. Just before the industrial revolution was due to begin is the books second setting. This is between the decades 1810 and 1840, when the first signs of industrialization were being seen in the urban areas, the first roads into the English countryside had begun to be put in place and land enclosures had fallen off the agenda and the peasants were being forced to move into the towns and cities to earn a decant living. Whilst this was proceeding, Raveloe village itself was remaining unaffected by this looming industrialisation and it continued to proceed with its rural backdrop still fertile and exuberant. The spinning wheels still hummed busily in the farmhouses, the bigger families lived in wealth and luxuries swearing by the thriving land, inherited estates and still using their ancestral tankards. Whilst these events are continue in the Raveloe village, Silass life as a weaver is starting to become quite easy for him and he is becoming quite a prosperous member of the village, but his fortune is still very secret and nobody knows the amount of money he has stored away apart from him. Silas is a very simple and extremely trusting man who lives for his work and works with pride. Silass has a Thread-bore physique; he appears to be very shabby and untidy. His body is not one thriving with muscles but a rather slender build. His eyes are very protruding and myopic (short sighted) and are quite baggy due to his heavy workload. Silas has a condition known as catalepsy, which is a suspension of the senses, all bodily powers and movements, with muscular rigidity and in some parts of the world called a trance. We first find out about the catalepsy around page twelve when Jem Rodney tells us that on coming up to him, I saw that Marners eyes were set like a dead mans, and he spoke to him, shook him, his limbs were stiff, and his hands clutched the bag as theyd been made of iron. The children of the village are terrified of Silas and after a while they begin to associate him with the devil. Jem Rodney then goes on to describe Silas as a dead man, come to life. Mr Macey believes that Silass soul leaves his body and goes into other peoples body to see what is going on in their lives. This plays a vital part in the early plot of the novel, while Silas is still a member of the Lantern Yard sect. The leader of the sect Deacon is suffering from a very bad illness; almost knocking on heavens door and it was Silass turn to watch him. Whilst he watches him he falls into a fit and William Dane knows this and takes advantage of it, by going in to the room and stealing the churches money from the cabinet where it is kept. He uses Silass knife, which he borrowed a few days before to cut some cloth, to pick the lock so he cannot be framed for the theft. Silas is obviously found guilty and banished from the sect; he does not only lose his religion but his best friend and fiancà ¯Ã‚ ¿Ã‚ ½e too. In this sect at Lantern Yard Silas has a major to play in the running and organization inside of the sect. He does his fair share of the work for it and is very highly thought in the sect. He is a man of exemplary life and he has a great faith in god and its surrounding faith. He was very trusting and thought that his best friend and fiancà ¯Ã‚ ¿Ã‚ ½e would not abandon him in his moment of need and when they did he was ashamed to have even known them and upon losing them, he lost his life and everything that meant anything to him. He felt that he had been betrayed by everyone and especially God because in the end he begins to call God The God of Lies. William Dane was one of Silass best friends during his time at the sect, but when it comes to the end of Silass time at the sect he stabs him in the back by not supporting him and then going on to marry his fiancà ¯Ã‚ ¿Ã‚ ½e. Dane had very narrow slanting eyes but that was not his only unusual facial feature, his lips seemed to be always very compressed. This gives us an impression that he is quite a sadistic man and sometimes can be evil. Dane describes Silass fit as a visitation of Satan only after Silas has moved on out of the sect and they are no longer friends. For Silas to be found guilty of the theft the leader has to draw lots out of a black bag and if the number drawn is even he is innocent but if it is an odd number drawn he is guilty. This is done on the basis that the leader of the sect puts them in but God takes the power of his hand as he puts them in. After the sect find him guilty of theft and they turn their backs on him, he loses the framework of his life, his friends, his belief in god and then goes on to resent God in a way that can not be described in words but not only does he lose all of this he also loses the one person in the whole world which he loves dearly. When we first come into contact with Raveloe it has been fifteen years since Silas had first come to settle there. There are three reasons given in the book to why Silas chose to settle there. It begins by telling us that believes that he is safe and is disconnected from the rest of the world and from God. It is a low wooded area and is quite populated so there is enough custom to ply his trade and earn a good living, and also that this village is very different from the sect at Lantern Yard and this will help him to forget all the bad memories he posses from there. When Silas arrives in Raveloe for the first time the villages form impressions on Silas because not only are they both intellectually and physically slow they arse also very superstitious. This new person moving into the village also intrigues them. They find the way he works at his loom all day long with few breaks quite mysterious and not moving to scare away peeping scoundrels just to stare at them without a mumble of a word, they suspected that this stare gave the children or even adults a disease that would paralyse them or even kill them. They already knew that Silas had learnt about herbal remedies and this struck both fascination and fear into them. The ladies in Raveloe know that he is in not interested in them or he never lets anyone over his doorstep to talk or never goes for a quiet drink in the pub. Silas sits at his loom working all day because he has nothing left to do, while doing this he becomes quite prosperous. The gold that he earns replaces everything that he had from Lantern Yard. The rhythm and framework has gone from his life ever since he was banished from the sect. He lost all his belief in God, lost and any routine that he once might have had but now all he is left with is his loom and gold. The routine that he had is now replace by the counting of his gold, he enjoys doing this because it is warm and soft to touch, another reason we are given for his counting of the gold is he thinks that gold cannot betray him like his old friends did. George Eliot uses two references to natural imagery in the early chapters of the book. She explains about Silass weaving pattern by saying, He seemed to weave, like the spider, from pure impulse, without a minuets reflection. He used the weavers chair as a comfort zone where he relaxed and freed himself of bad feelings. She described his life after the sect to be like a rivulet. An old drying up, small stream that is leading to nowhere or not coming from anywhere.

Why Did British Men Enlist in the British Army in 1914 Essay Example

Why Did British Men Enlist in the British Army in 1914 Essay Example Why Did British Men Enlist in the British Army in 1914 Essay Why Did British Men Enlist in the British Army in 1914 Essay is an example of making soldiers feeling guilty, as they had feelings of fear and embarrassment holding them back from participating in the war. The poster shows two children asking their father, who is sitting on a chair, if he had taken part in the war. This implies that the children look up to the father. The father has a look of guilt in his eyes, and viewers, especially young men, would not want this to happen to them. As a result, men joined the army in large numbers. Also like the White Feather method people also sent abusive letters to people who had not joined the army.. A taunting letter forwarded to a railway porter who had not yet enlisted It reads: Dear Mr. E. A. Brookes seeing that you cannot be a man not to Join the army. We offer you an invitation to join our Girl Scouts as washer up, .. Scout mistress Bath Girl Scouts. The objective of all this pressure was to push the people receiving the intimidation to feel guilty and join up right away; these men were probably less patriotic if patriotic at all because they refused to help the country. It is clear there are many reasons why the British forces needed more men, but through unforeseen circumstances the British underestimated the German forces. The reasons why the British men joined the army included: patriotism, adventure, guilt, more money, peer pressure and many more.

Wedding Toast Essays

Wedding Toast Essays Wedding Toast Paper Wedding Toast Paper When Valerie first told me note that I was told, not asked! That I was making a speech, I thought it was her way of getting back at me for anything I had ever done to her while growing up. Lately I realized that it is actually an honor to be asked to address YOU. Look to your left. Look to your right. (Pause) YOU are the people Valerie and Christian love the most. They’ve invited you because this day would not be the same without you. And you have made them a priority in your lives by celebrating with them.You have honored them with your presence and made their day so special and unforgettable. On behalf of my family, I would like to thank you all for coming. As sister of the bride, I have a unique view of Valerie. She was the little sister I had always wanted. She was a happy, precocious child I remember her ability to get into all sorts of trouble and then proclaim quite indignantly NOT ME! One of my favorite memories of her childhood was how much w e loved rainy day we would get a tire and make it into a boat.Flash forward to a charming pre-teen in braces that got permission to come and stay with me in New York at my first apartment. She was willing to go along with my crazy schemes. Then as she got older, we shared a room for a few months while I stayed in New York. She had learned that little-sister trick of appropriating and wearing things that were not hers. And, every single night before we went to sleep I was subjected to the same boy-band song Valerie has grown into a beautiful woman.She is no longer my little sister, but a graceful, smart, generous, caring equal. It is an honor to know the person she has become. She is a great friend as you know loyal, patient and generous. And she loves a challenge and this is where Christian comes in. Christian, you are one lucky guy! Valerie you look stunning and Christian. Well you look stunned! So Christian, here’s some advice for the future: She is always right She always needs some new clothes She never looks fat in anythingIf she’s right say so, and if she isn’t -say nothing! So here we are their wedding. They have made it. a DOZEN years together and stronger every day. High school sweethearts who have defied the odds. Leave it to them to be defiant in the face of anything! Today is a celebration of not only where they are now, but the road that has brought them here. They have built a beautiful life together. I think you all will join me in this wish: Valerie. Christian. May you live forever in love, and may love live forever in you.

Nutrition for a childs health and welfare Essay Example | Topics and Well Written Essays - 750 words

Nutrition for a childs health and welfare - Essay Example Proper dieting and good nutrition for the age group is ensured by food safety and sanitation, food storage, food preparation, proper food handling and presentation. Safe food for consumption is well accounted for in terms of its stipulated expiry dates when purchasing. In case of purchasing meat ensure that the meat is free from bad or strange oduor/smell this possess a high risk of food poisoning to an individual .Proper food preservation is key: any well maintained and food fit for human consumption, one should ascertain that food especially highly perishable is well preserved in a cool dry place. In addition meat should be separated from fruits and vegetables as this would lead to contamination. Secondly food storage is the delicate point that sees long lasting foods and perishables well-kept before consumption. Basic rules do indicate that perishable foods should be stored in a cool place and even refrigerated to prevent them from going stale. Selecting food products whose labeling clearly indicates the mode of refrigeration for example: keep refrigerated, should be in a refrigerator and cold to touch. Again food storage is backed up by the clean environment at all times. Proper food handling and sanitation is key to healthy food practices. Hygiene should begin from an individual handling the food to the place where the food is being prepared. In case of eggs make sure they are cooked fully i.e. the yolk as well as the whites are firm.

Cosmopolitanism Essay Example | Topics and Well Written Essays - 2000 words - 1

Cosmopolitanism - Essay Example While in Europe he integrated with Romany people that his English contemporaries. His integration with the people enabled him to acquire Gypsy wandering culture. For instance while he was in Marseilles, France he learned Modern Greek, French, Italian, and Neapolitan dialect. Those languages came handy when the family traversed Europe especially France and Italy before settling in Britain. He learned an intermediate language between French and Spanish called BÃ ©arnaise in addition to classical Greek and Latin. I think Burton was an exemplary linguistic and a great European swordsperson that saw him expelled at Britain’s Oxford college since he had contravened one of its by-laws against mixing with other races. He challenged a fellow Oxford college student that had heard of his prowess with the Saber that ridiculed his walrus moustache. Burton later traveled to Asia to work for the East India Company in Sindth, India. He learned Gujarati, Marathi, Persian, and Afghan. He deepened his Arabic and Persian languages that he had started learning while in England. He traveled to the Arabian Peninsula in 1853 and admitted to Mecca and Medina as a pilgrim while he posed as a Pathan from India’s Northwest Frontier Province. Later on in 1858, he traveled to Africa together with his colleague John Hanning Speke. Historically, they were the first Europeans to see Lake Tanganyika. . Other places he visited in Africa included Somalia, Sierra Leone, Ghana, Lagos, and Cape Coast.

Appreciative Inquiry of David Cooperrider Essay - 2

Appreciative Inquiry of David Cooperrider - Essay Example The questions asked are affirmative in nature and it focuses on the topic valuable to the people who are involved and directed at topics and issues for the success of the organization. This inquiry identifies â€Å"the area† where the organization needs to change. The change can be of different types which are as follows:   When an organization has to adapt to external factors, it may go through a major strategic change. Strategic organizational changes are usually quite transformative in nature as they include major adjustments and complete upheavals of the present way the organization operates.   People changes can be of large-scale or incremental. Large-scale people changes include replacement of the top executives with new employees in order to change the entire organization’s culture. Smaller-scale or incremental people changes include sending of management workers to team-building workshops and classes. These changes can be planned or unplanned, which may impact the employee’s attitudes towards work, behaviors of the individual and their performances.   Process changes define to be an attempt to improve the overall workflow efficiency and productivity of the organization. Organizations implementing these types of changes are highly successful when the new innovative process is applied to employee groups and the outcome of the new process work (Sharma, 2008).   In Silkeborg Council (Denmark) the problem in every department was absenteeism. The average number of working days that are lost over the first six months of 2001 was 9.25. One of the departments in which absenteeism was very high was the elderly care department. At this stage, the employees and the employers are inquired about the reason for absenteeism and then they are informed about the type of change which should be undertaken by them.   

How did Bill Gates became successful Essay Example | Topics and Well Written Essays - 1000 words

How did Bill Gates became successful - Essay Example â€Å"Bill Gates: Profile of a Digital Entrepreneur†). Bill Gates was born in an upper-middle class family of Seattle on 28th October, 1955. His father was a reputed lawyer in the town and his mother being one of the Board of Directors in the Interstate Bank of Seattle, played a vital role in many other significant and dominant institutions of the then society. Being monetarily and psychologically quite secured, Bill Gates developed his interest towards technology and especially computers. Gradually he became passionate about developing software which later emerged as a significant reason for his friendship with Paul Allen who afterward was recognized as his companion in personal as well as professional life. At an early age of only 13 years, Bill Gates and his companion Paul Allen were encouraged by the Computer Centre Corporation for the development of advanced software languages (Lockwood, B. â€Å"Bill Gates: Profile of a Digital Entrepreneur†). In his later years, Bill Gates improved his competitiveness. It was during this course of time that Bill Gates depicted his enthusiasm and early entrepreneurial skills supported with his desire for complete power to control. It is worth mentioning that his competitiveness and strong desire to innovate as well as to control the world of technology at a very early age, followed by a few denials along with supports and several arguments led to the recognition of the co-founder of Microsoft Corporation (Guth, â€Å"Raising Bill Gates†). Bill Gates is currently known as one of the wealthiest peoples in the world, an excellent software titan, the co-founder and present CEO of the multinational information technology giant company, Microsoft Incorporation. He is also an investor, author and philanthropist in his personal life. Many people may consider this as the success of Bill Gates. However, from an in-depth perspective, it is unequivocally observable that the main achievement of Bill

Why the Diagnosogenic theory of stuttering onset has been dispelled by Essay

Why the Diagnosogenic theory of stuttering onset has been dispelled by recent literature - Essay Example Neurophysiology- Recent research has shown that people who stutter process speech and language in different areas of the brain than those who do not stutter. Family dynamics-High expectations and fast-paced lifestyles can contribute to stuttering. Technically known as dysphemia, it has sometimes been attributed to an underlying personality disorder. Brain scans of stutterers have found higher than normal activity in brain areas that coordinate conscious movement, suggesting that in people who stutter speech occurs less automatically than it does in most people. In 1939, a controversial study, on the possibility of "creating a stutterer", was conducted by University of Iowa speech pathologist, Wendell Johnson and his graduate student Mary Tudor. The study tried to create stutterers over the course of 4 months, using 22 unwitting orphans from the Soldiers and Sailors Orphan's Home in Davenport, Iowa. Ethically acceptable at the time, it was designed to induce stuttering in normally fluent children and to test out Johnson's "Diagnosogenic theory" a theory suggesting that negative reactions to normal speech disfluencies cause stuttering in children. The study divided the orphans into 3 groups. 6 normally fluent orphans would be given negative evaluations and criticisms regarding their speech, another group of 5 orphans who allegedly already stuttered would also receive that treatment, and the remaining 11 would be treated neutrally. The study concluded that the children given negative evaluative labeling went on to develop persistent, permanen t stutters. The study was influential at the time, with many speech pathologists and child-health and educational professionals accepting Johnson's theory. In 1988, Silverman first reported the results of this study in the Journal of Fluency Disorders and labeled it "The Monster Study". In June 2001, the San Jose Mercury News revealed this study to the public for the first time, leading to widespread controversy and debate about scientific ethics. Soon after, University of Illinois professors Nicoline Ambrose and Ehud Yairi wrote a paper discrediting the 1939 study, revealing flaws in data collection and method, as well as pointing out that none of the orphans actually did develop a permanent stutter. The relevance of the Ambrose-Yairi study 63 years later is that the authors conclude, in effect, that the 1939 thesis did not prove the theory with which it is credited. In other words, the researcher did not, and could not have, "caused stuttering" in the subjects. For this and many other reasons the authors also conclude that most all of the ethical criticisms of the study are misplaced and unjustified. While criticism of a developing child's speech can certainly make a present stutter worse, it does not create a stutter. The major findings, as have been reported over the last several years in the Journal of Speech, Language, and Hearing Research (JSLHR), question longstanding concepts about the onset and developmental trends of early childhood stuttering. Like most other speech disorders, stuttering onset was gradual and occurred under uneventful circumstances, that early symptoms included only easy repetition of syllables and words, and that parents helped create the problem by reacting negatively to normal disfluencies. Stuttering onset was sudden in at least one-third of the children, was severe in nature,

How did Bill Gates became successful Essay Example | Topics and Well Written Essays - 1000 words

How did Bill Gates became successful - Essay Example â€Å"Bill Gates: Profile of a Digital Entrepreneur†). Bill Gates was born in an upper-middle class family of Seattle on 28th October, 1955. His father was a reputed lawyer in the town and his mother being one of the Board of Directors in the Interstate Bank of Seattle, played a vital role in many other significant and dominant institutions of the then society. Being monetarily and psychologically quite secured, Bill Gates developed his interest towards technology and especially computers. Gradually he became passionate about developing software which later emerged as a significant reason for his friendship with Paul Allen who afterward was recognized as his companion in personal as well as professional life. At an early age of only 13 years, Bill Gates and his companion Paul Allen were encouraged by the Computer Centre Corporation for the development of advanced software languages (Lockwood, B. â€Å"Bill Gates: Profile of a Digital Entrepreneur†). In his later years, Bill Gates improved his competitiveness. It was during this course of time that Bill Gates depicted his enthusiasm and early entrepreneurial skills supported with his desire for complete power to control. It is worth mentioning that his competitiveness and strong desire to innovate as well as to control the world of technology at a very early age, followed by a few denials along with supports and several arguments led to the recognition of the co-founder of Microsoft Corporation (Guth, â€Å"Raising Bill Gates†). Bill Gates is currently known as one of the wealthiest peoples in the world, an excellent software titan, the co-founder and present CEO of the multinational information technology giant company, Microsoft Incorporation. He is also an investor, author and philanthropist in his personal life. Many people may consider this as the success of Bill Gates. However, from an in-depth perspective, it is unequivocally observable that the main achievement of Bill