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

Communication and Its Importance in Creating Conflict Essay

Conflict is essential in all works of fiction, whether it might a simple disagreement in a marriage or internally trying to deal with guilt. The real life situations of Interpreter of Maladies connect the problems with communication with conflict that will arise. In Interpreter of Maladies, communication problems in the short stories â€Å"A Temporary Matter†, â€Å"This Blessed House†, and â€Å"Interpreter of Maladies† resulted in conflict. In â€Å"A Temporary Matter†, the conflict in was the game that Shoba and Shukamar played, referring to the revealing one truth about oneself when there were not any lights, and the game where they were hiding from each other, hoping to avoid awkward encounters. This was an internal conflict from Shukumar’s perspective because Shukumar was thinking of how to play his next move in order to effectively restore the relationship. †Now he had to struggle to say something that interested he, something that made her look up from her plate, or from proofreading files.† Shukumar thought the purpose of the game where the couple would share previously unknown facts was that it would restore their relationship. Shukumar thought the game was used in order to reconcile and reform the marriage. But he was gravely mistaken. Shoba used this game to tell him that she was moving out. â€Å"It sickened Shukumar, knowing that she had spent these past evening s preparing for a life without him.† (21). The internal conflict in Shukumar was a result of miscommunication on Shoba’s part. She rarely communicated her thoughts and feelings after the stillbirth, and Shukumar took this as a sign to not talk. This resulted in their relationship slowing fading, and finally to the point of avoidance. In â€Å"This Blessed House† Sanjeev and Twinkle had many fights throughout the story because of the Christian artifacts that were laid throughout the house. Sanjeev believed it was impractical, since both he and Twinkle were not Christian. When he kept on insisting on eradicating the religious objects, Twinkle insisted on keeping them. While Sanjeev had a practical and logical reason of why, he never effectively compromised with Twinkle, and kept on believing that he was right. And without the proper compromise and communication, it resulted in Sanjeev not understanding Twinkle’s emotions. He was quite surprised when he saw Twinkle cry in the bath. â€Å"Oh God, please, Twinkle, I didn’t mean it.† But Finally in the end, they reached a compromise and managed to continue on to the party. But the conflict arose because of Sanjeev’s inconsideration and not communicating his thoughts along with listening to Twinkle’s approach. â€Å"In the end they settled on a compromise: the statue would be placed in a recess at the side of the house, so that it wasn’t obvious to passerby, but was still clearly visible to all who came.† Even though they come to a compromise, it was after Sanjeev made Twinkle cry, which is a poor example of effective communication. In â€Å"Interpreter of Maladies†, Mr. Karpasi was under a delusion that Mrs. Das was interested in him. The conflict was Mr. Karpasi’s internal struggle interpreting Mrs. Das’s actions and determining whether she liked him or not. â€Å"As he stole glances in the rear view mirror, wrapping elastic bands around Tina’s hair, he wondered how he might make this tour a little longer.† He stated that he was an interpreter in the story. Mrs. Das inferred that he was some sort of psychologist or therapist that would cure her of her guilt that she had been holding in. However she was mistaken, Mr Karpasi essentially was a translator for a doctor. â€Å"†¦But many people do not speak Gujarati in this area, including the doctor. And so the doctor asked me to work in his office interpreting what the patients say.† He did not process any knowledge in the field of which Mrs. Das was interested in. When Mrs. Das was asked by Mr. Karpasi about why she told him about the illegitimate child, Mrs. Das said, â€Å"Well don’t you have anything to say? I thought that it as your job.† Mr. Karpasi responded with, â€Å"My job is to give tours, Mrs. Das.† Mrs. Das continued with, â€Å"Not that. Your other job. As an interpreter.† Mrs. Karpasi answered with, â€Å"But we do not face a language barrier. What need is there for an interpreter?† Mrs. Das completely misunderstood Mr. Karpasi in what he meant as an interpreter. She was only interested in Mr. Karpasi because she believed that Mr. Karpasi would cure of her guilt. On the contrary, Mr. Karpasi thought that Mrs. Das was interested in him in an attracting fashion. In Interpreter of Maladies, communication problems in the short stories â€Å"A Temporary Matter†, â€Å"This Blessed House†, and â€Å"Interpreter of Maladies† resulted in conflict. In â€Å"A Temporary Matter†, Shoba gave Shukumar the impression that she was trying to reform their marriage when her true purpose was to end it. She also is responsible for their avoidance with each other. In â€Å"This Blessed House†, Sanjeev was oblivious to Twinkle’s connections to the Mary statue, and kept on pressing his belief. This resulted in a fight in which Sanjeev finally realized his misdemeanor. In â€Å"Interpreter of Maladies†, Mr. Karpasi miscommunicated what his job was, which then resulted Mrs. Das giving him a problem that he could not solve. Mrs. Das regarded Mr. Karpasi with interest because she believed that Mr. Karpasi can help her. However this resulted in her humiliation when she told Mr. Karpasi her biggest secret that she believed he could resolve. Communication is essential to the everyday life of humans. A small mistake in communication can destroy a relationship or start an unwanted situation. It is essential that people would communicate in order to create a more stable relationship.

Lies and Deceit Paper

Lies and Deceit on Animal Farm/ the Real World: Speech There have been many examples of dictatorship across the world. Some examples of these are Hitler of Germany, Castro of Cuba, and Mussolini of Italy; they all have been corrupted by power. Joseph Stalin of communist Russia is the most accurate example because he is like Napoleon. People, especially leaders, use lies and deceit to maintain power; Napoleon is like Stalin because they are both corrupted by power.People often tell lies when it is convenient; the pigs told the other animals many lies some of which include, when the pigs took the milk and apples, when they lied about Boxer’s death, and when they said that the farm would be so much better off without Jones, but everything was just as bad if not worse. People tell lies as simple as trying to protect others’ feelings or something really complicated to try and get out of trouble either way the bottom line is people lie.The first lie of many that the Napoleon and the other pigs told is when Major is saying that Jones the farmer is treating the animals poorly and is trying to convince them to rebel against him. He tells them, â€Å"The life of an animal is misery and slavery; that is just the plain truth† The animals didn’t know that the pigs were just planning to take over and be just as bad as Jones was, or even worse. People tend to lie and deceive others when it is convenient for them, to make it easier on themselves.I’m pretty sure that all of us, at one point or another lied to our parents because it was what seemed easier at the time; For example, a child may lie because they broke something on accident, or a teenager stayed out past curfew and are lying to get out of it. After Old Major’s speech to the animals, Snowball tries to convince Mollie that Jones was bad; she was one of the rare few who liked him. Snowball was being corrupted by power, and he is a very convicing liar which is why Mollie believe s him. Then once the pigs got rid of Jones, they were taking all of the good apples and milk and saying it was for the health of their brains.Humans in the real world lie in their favor if we want something bad enough we will lie to get it. Boxer starts to work himself to death when they build the mill and he works day and night for weeks straight. Napoleon then names the mill after himself even though Boxer and the other animals did all of the work. People also lie or stretch the truth when we want to take credit for an achievement that we may not have even done ourselves. Then Napoleon sold wood to a human without consulting the group which made everyone a little angry.Humans do things without asking first. Squealer lied to the others after that by telling them Boxer was going to the vet instead of the knackers. â€Å"The van had been previously owned by the knackers and then bought by a vet and they have not yet re-painted the van† He said. When it was all said and done an d the transformation was complete one dictator was switched for another the animals were in the same place they had previously been. Napoleon would stand on his hind legs and even carry around a whip with him to keep the others in line.Then he tried to lie and say that the humans aren’t so bad after all. Nothing could have been further from the truth. In chapter three, the pigs’ deceit the other animals into thinking everything was okay when in fact it most definitely was not. The pigs were so clever that they could think of a way out of any and every difficulty. To recapitulate, people lie all the time and it is similar to the book because the pigs told the others the milk and apples were just a sacrifice, that Boxer’s death was an unfortunate accident, and that the pigs didn’t help anything.As a consequence the farm was worse then before, Boxer and many of the animals were killed, the mill was torn down and the pigs took over and made everything worse. Telling the truth may be worse at first, but in the long run it is the better decision. ————————————————- ————————————————- ————————————————- ————————————————- Top of Form Powered by Bottom of Form

Write an essay on theme Example | Topics and Well Written Essays - 750 words

Write an on theme - Essay Example ing repulsion for the flavorless, colorless mass of everyday existence† (Cather 453) and the boring classroom with "bare floors and naked walls," (Cather 458) and the teachers in "dull gowns" with "shrill voices"(Cather 458) who felt that he was a little odd. The novel opens with a situation where Paul is suspended from school for reasons which the teachers cannot pinpoint. At the same time he has a somewhat ostentatious appearance with "something of the dandy about him"(Cather 448). The inner feeling of Paul makes him appear to be on drugs with dilated, unusually brilliant and large eyes. However, Paul is not on drugs but he lives in an altered condition of consciousness. The theme of the novel is surreal because the boy is not attracted to his life at home and school, instead he finds happiness in the theatrical environment where he holds the job of an usher and spends time on Sundays. The author has depicted several occasions that reveal Paul’s emotional fulfillment when he is engaged in the theatre setting. Cather says "It was very much as though these were a great reception and Paul were the host" (Cather 451). The author points out the happiness of Paul while working at the theatre by saying that "as though it were his greatest pleasure in life" (Cather 451). Paul’s eagerness to live a flashy lifestyle is depicted in the line which says; this was Paul’s fairy tale, and it had for him all the allurement of a secret love. The moment he inhaled the gassy, painty, dusty odor behind the scenes, he breathed like a prisoner set free, and felt within him the possibility of doing or saying splendid, brilliant things. (Cather 457) Paul visualizes real life at the theatre which is furnished with garish satins, diamonds and rhinestones. Paul views the setting of the theatre and the setting of his home and school to be at different extremes of the pleasant and unpleasant. Paul is depicted as a sensitive artist who loves the stage and the attractive

Water and Its Importance in Broward Countys Environment Term Paper

Water and Its Importance in Broward Countys Environment - Term Paper Example Pollutants taken by this run-off can be released into pathways, ponds, and other systems water with negative ecological effects. In Broward Nation, a major resource of stormwater contamination is scenery run-off. The plant foods, weed killers, and bug sprays we use to keep our lawns looking their best can actually end up doing more damage than good. When it downpours, these pollutants along with other contaminants like fuel and oil can run off from our qualities and into our local storage areas, pathways, and larger systems water like the Everglades and Seaside Ocean. Flower foods, in particular, can cause some serious issues for our water resources. Flower foods promote the development of plankton and marine fresh mushrooms because they contain high levels of nitrogen and phosphorus, two nutritional values that encourage plant development. The out of control development of plankton and unwanted marine vegetation (like hydrilla and cattails) and Water hyacinth damages our ocean often . Once fertilizers get into our waterways, they can cause the rapid development of plankton. Algal biomass atmosphere the water and blocks sunshine, causing stress to marine vegetation. When that vegetation passes away, it sinks to the bottom and is decomposed by parasites. As the parasites feed off the vegetation, it reduces oxygen from the water that seafood and other creatures need to live, and can outcome in large seafood destroys. Plant biomass can also cram pathways, reducing discharge. This often results in the need for chemical treatment, which is not only costly but also plays a role in the contamination of our water.   Excess fertilizers in our water can also cause changes in established surroundings. For instance, cattails can grow so heavy that they press out all local vegetation, allowing exotic and obtrusive vegetation to take hold. This can limit looking by creatures and eventually cause local varieties to seek other places for the environment.

Form a short story base on this conversation Essay

Form a short story base on this conversation - Essay Example He said it was just to see how the fields were coming along, but they both knew it was because she enjoyed the tractor rides. But a few months ago Uncle Vern hadn’t woken her up when he left on foot to check the fields. When she asked him about it he replied, â€Å"Oh I thought you might be getting too big for tractor rides.† Not only that, they had stopped asking about her homework. Aunt Marie always knew when report cards came out too, but this time she hadn’t asked to see it. A week later when Annelise was starting to wonder if she had truly become invisible she asked her Aunt Marie if something was wrong, because she hadn’t asked to see her report card. All she said was, â€Å"Well honey, you’re almost eighteen, you are responsible for your own grades.† Now she had begun to worry, The Eighteenth Birthday, the birthday when everything changes was here. Would she be kicked out of her home and expected to fend for herself? Was she no longer wanted now that she was â€Å"older?† As she walked down the stairs she pushed those thoughts aside and tried her hardest to smell coffee cake. Nothing. She saw her Aunt and Uncle reading the paper and drinking coffee. They looked up and said â€Å"happy birthday,† but then went back to reading. She made a bowl of cereal and stared at the Cheerios floating around in the milk. She began to wonder if things would have been different if her parents had lived. If the car crash had never happened would she be happier or in worse circumstances. Aunt Marie looked up from the paper. â€Å"Is something wrong Annelise?† â€Å"Well,† Annelise began, she knew an explanation was necessary, â€Å"it has kind of seemed like since I’ve been getting older the little things that I used to expect no longer happen.† All of the little things that had changed poured out of her at once, and her fears about what would happen after today tumbled out as well. When everything was out

Peer response Article Example | Topics and Well Written Essays - 250 words - 6

Peer response - Article Example Second, the topic and subject of research that was shared was relevant and seen to be applicable in current settings: homeschooling or distance education. With the prolific use of the internet, one of the significant applications is the access to knowledge through the online medium. One of the initial comments was the dismal regret for the inability to locate more scholarly research on the subject of â€Å"how do homeschooled students do compared with traditional-schooled students†. A quick look at scholarly journals generated only 11 results; which validated the comment. As such, one strongly believes that the lack of studies on the subject matter should be a motivating factor to pursue the study. There are vast options to discern the most appropriate research methodology and could be applied using a nearby university that offers distance education courses in additional to the traditional classroom-based

The International Marketing Mix Case Study Example | Topics and Well Written Essays - 3000 words - 4

The International Marketing Mix - Case Study Example In terms of product, PepsiCo utilizes a wide variety of different positioning strategies for each product brand. However, as each food and beverage product is being distributed to different multinational consumers and a high volume of competition exists in this market, PepsiCo must use its marketing focus to create a connection with consumers through various branding activities. For example, if the typical Aquafina (bottled water) drinker is the athletic and motivated consumer fitting the psychographic profile of social and recreational enthusiast, PepsiCo must position Aquafina as a brand which befits lifestyle and utilizes various marketing communications (as an integrated bundle of marketing messages) to appeal to this demographic. In some developing countries, Aquafina might be more of a staple drink in an environment where clean, city water systems are not common in foreign neighborhoods. This would provide PepsiCo with the opportunity to position Aquafina as a reliant brand suc h as emphasizing its quality or mineral content. For instance, PepsiCo, in an attempt to appeal to the international consumer with a dedication for environmental advocacy, created a partnered marketing strategy with Sam’s Club (a Wal-Mart company) entitled Return the Warmth to promote recycling and other environmental initiatives (Annual Report, 2007). This indicates how the company utilizes various social trends and a desire to make brands appeal to different demographic segments as a means to give PepsiCo brands competitive advantage and boost international profitability. Â   Â   Â   Â  The company also utilizes creative marketing practices, such as new logo design and product packaging visuals, to appeal to the diverse multinational consumer. With the recent decline in global demand for PepsiCo products, the company is attempting to reinvigorate the Pepsi brand by changing cans. In an attempt to appeal to the foreign, contemporary U.S. consumer, the Pepsi brand designed the cans to reflect various emotions which are used in countries where text messaging is popular (Birchall, 2008).

Demonstrative Speech (3-4 mins) Essay Example | Topics and Well Written Essays - 250 words

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Management Concept Essay Example | Topics and Well Written Essays - 1000 words

Management Concept - Essay Example The fact is, with onset of new era of globalisation, the workforce is increasingly varied and faces problems due to geographical constraints, multilingual offices, changing organisational cultures and shifting domains as seen in case of IT organisations. The need of the hour is to develop a strategy to project a unified organisational culture which defines the whole character of the organisation and fosters an attitude of comradeship across employees, verticals, units and businesses thus creating a potential for organic and inorganic organisational growth. The new model on motivation that combines tenets of neuroscience, biology and evolutionary psychology points towards four basic needs (or drives) that guide motivation levels in people. (Nohria Groysberg & Lee 2008) These four drives, namely, drive to acquire, bond, comprehend and defend underlies every thought and action of employees and it is only by satiating these drives that an organisation can boost employee morale, enhance c reativity and extract higher productivity which leads to wholesome growth. The article emphasises on the need to focus on all four tangents for the desired results as there is tendency among Managers to overlook one or all aspects of these drives with disastrous consequences in terms of attrition percentages evoking high levels of loss n terms f experienced resources and business accordingly. The tunnel vision of handful of managers can have grievous consequences according by squelching creativity and life out of the organisation and promoting a stagnant organisational culture.(Amabile 1998) The root cause of loss of creative inputs are attributed to various factors that affect the three pillars of creativity in resources which include expertise, creative thinking skills and motivation. Thus from the ideas of these researches it is imperative to develop a theory that would promote the motivation levels and help in retaining the creativity of the employees. The organisational culture reflects the degree of interrelationship between the employees and the set of values and ideals that join them. Organizations with clearly codified and enforced cultures enjoy great employee and customer loyalty, in large part because they are effective in either altering ineffective behaviours or disengaging from values-challenged employees in a timely manner. ( Hessket Sasser and Wheeler 2008) Thus it is imperative that there is a positive culture that facilitates individual and organisational growth. The company can begin by promoting a clear, well organized mission statement which reflects the values and ideals stands up for and distinguishes it from its peers by reflecting for what it stands against. (Talbot 2003) The mission statement is important as it differentiates the company and lays down a clear line of ethics to be followed according to the organizations core values. The â€Å"IBMer† values that are instilled in every employee of IBM remind them of the core valu es of the organisation and its dedication to excellence in balancing personal and organisational goals. IBM markets itself as one of the best places to work due to the fact that it believes every employee is â€Å"special† and this is also propagated in its recruitment ad campaigns under the tag-â€Å"What makes you special?† Thus the company promotes a healthy image and strong organisation culture of mutual respect and camaraderie. The employees are given proper induction

INTELLECTUAL PROPERTY LAWS Essay Example | Topics and Well Written Essays - 500 words

INTELLECTUAL PROPERTY LAWS - Essay Example This is before utilizing a person’s image or voice for commercial gain, which in this case is Tiger Woods’ picture. Therefore, â€Å"Right of Publicity† ensures corporations, firms and businesspersons do not exploit people’s persona either famous or not to benefit themselves. Hence, protect them and compel those who wish to use these pictures to obtain a legal consent from the owners (Fishman, 2010). States usually recognize people’s rights by granting them free will to either allow or disallow usage of their images especially for commercial purposes (Fishman, 2010). Hence, in this case Tiger Woods has a sole responsibility of deciding if Wheaties cereal firm is liable of using his image or not by agreement. In this case, Tiger woods ought to exercise the â€Å"Right of Publicity† in controlling how Wheaties cereal’s firm utilizes his pictures and others that may intend to do so (Fishman, 2010). This is via granting them consent besides deciding the required compensation based on how long the firms will have to use his reputation in advertising their products. Tiger Woods owns the â€Å"Right of Publicity† similar to any other people in ensuring firms do not exploit his reputation and pictures. Hence, he has the authority of demanding compensation in return, which is in form of cash though this is as per the agreement with the corporate firms. Based on the â€Å"Right of Publicity†, owners have the mandate of prosecuting anyone who might have used their images without their consent (Fishman, 2010). This is especially through advertising or through other forms like reputation riding when the owners are not aware. However, the right also grants owners like Tiger Woods freedom to negotiate with the potential businesspersons, firms and corporations if they are ready to pay for his pictures in advertising their

Everyday Drugs Essay Example for Free

Everyday Drugs Essay The essay â€Å"Everyday Drugs† written by Adam Smith reminds me that government plays an important role in our lives. The government is an important element for the existence of a state. Without which, a particular society will be in chaos. Anarchy will govern. There will be no order because people are not guided by any rules. Due to this, people will always think that every action they make is right. But since there is government, a world full of chaos is prevented. Every person is guided by laws that govern our action. Each government prescribes what is right and what is wrong. In my opinion, government is very important because it produces harmony and tranquility in a particular society. Sometimes, however, the presence of government restricts our actions too much. It sometimes hinders our freedom to act. Also, government in different countries perceive certain action to be right while some government perceives the same act as illegal. For example, in Holland, people are allowed to smoke cannabis in public places without being apprehended by the authorities. This is because smoking of cannabis is allowed in their country. However, if a person is seen smoking cannabis the United States, he or she will be apprehended because that particular act is proscribed in that country. There is no substantial distinction between the people in Holland and the people in America which will give the people in the former be allowed to smoke cannabis. The only difference between them is that the government perceives such act to be permissible in Holland while the other perceive it otherwise. In the essay, Smith associates government as a drug which prescribes to its people what are the action that are legal and what are those illegal. The government tries to manipulate what our action should be. I believe that it is alright for us to be guided by rules for as long as our freedom to act is not infringed.

The Multi Store Model And The Reconstructive Model Psychology Essay

The Multi Store Model And The Reconstructive Model Psychology Essay In this essay there are two memory models what are compared. The memory models are the Multi Store Model and the Reconstructive Model. It shows how the models work, some studies which have been done experimenting the model, advantages and disadvantages to the models. The multi store model is a theory by Atkinson and Shiffrin (1968). They say that memory uses a number of different processes and that there is actually more than one kind of memory store. There is three parts to the multi store model: sensory memory, short term memory and long term memory. The process starts at sensory memory where you use your senses: Sight, smell, hearing, touch and taste to receive and store information. The information here doesnt last very long from milliseconds to two seconds, just long enough for it to transfer to short term memory. To get to short term memory you use the process, encoding. Encoding means taking the information and making a memory trace. There are three types of encoding: Acoustic, visual and semantic. Acoustic is repeating the topic; you would say or sing the words in your head. Visual is where you would try to see the topic in your head and try to picture it. Semantic encoding is what it means to you e.g. a favourite birthday present, someth ing that has a personal meaning. If the information is not attended to from sensory memory to short term memory this can be lost. The information that has been attended to then rehearsed will go to store in long term memory. (AS Level Psychology) Peterson and Peterson (1959) did a study that supported the multi store model for the duration of short term memory; they named the study Trigram Retention Experiment also known as TRE. The study was highly controlled and the people all started on the same base level. Their aim of the study was to see what the duration of short term memory would be if you did no rehearsal. When they tested this, the people who partook in the trigram, were read three letters. These three constanants were randomly picked and had no meaning e.g. CWT. immediately after been given this the people had to count backwards in threes starting with a very large three digit number. This was called the distractor task they had to do it for a specified time and it was designed so that you could not rehearse the TRE. The time period was called The retention interval and this varied from 3 seconds to 18 seconds. Then when told, the people had to recall the trigram. The letters had to be said exactly the same. They f ound that nearly all the people who recalled after a three second retention interval were successful. Then when they did it at 18 seconds, only very few people could recall the TRE. Doing this study found out that without rehearsing things the duration of short term memory is very short. So this shows that without rehearsal memory is very limited. There were criticisms to this theory though, it was said that the distractor task did prevent rehearsal, but it also meant that the people were doing an additional processing task so this may have affected their recall scores. (Essential AS Psychology for AQA, Richard Gross and Geoff Rolls) Shallice and Warrington (1970) also did a study which supported the multi store model they reported a case where someone had, had a motorbike accident. This resulted in him now having extremely poor short term memory. However his short term memory what he couldnt remember was only for verbal information. His visual and acoustic was normal. This suggests there is more than one type of short term memory, not incorporated in the multi store model. He could only remember one or two digits. Yet his long term memory after the accident was normal. This is in support with the multi store model, because it provides evidence to show that short term memory and long term memory stores are separate. The reconstructive model of memory was a theory by Bartlett (1932). He said that memory is unlike a camera, that its not accurate and the interpretations are based on each persons memories and previous experiences. (Essential Psychology). People tend to be selective about what bit of event they remember. This can be a problem though when needing an accurate recall. Schemas are a major part in reconstructive memory. Schemas are our own ready made expectation of a scenario. They fill the gaps in our own knowledge. Schemas can lead to memory distortions when information doesnt already fit in to the existing schemas we have. There is also a lot of influence to memory from stereo-typing and cultural expectations. (Essential AS Psychology). The eye witness testimony also known as EWT is for criminal incidents that people have seen. Eye witness testimony may not always be accurate; it can be influences by a lot of different things like anxiety, age and even leading questions. If you were to witness a crime it would make you anxious. There are advantages to knowing that the eye witness testimony is not always accurate, this warns us and helps us to know we do have to treat with care. Bartlett (1932) did a study on the reconstructive memory called War of the Ghosts. He did this study to investigate the effect of previous experiences and expectations have on memory recall. A group of people were shown a drawing or a short story, which was the war of the ghosts, to look and read through. The group of people then had to reproduce the story or drawing after different time periods. The time periods ranged from fifteen minutes to years later, this was called a repeated reproduction technique. Over time they found that the group of people produced shorter more clear and carefully thought out versions, more ordinary. They had made the story sound more normal e.g. instead of using the word canoes they used the word boats. They only recalled the basic storyline. In conclusion to this study they found that an explanation or opinion plays a major role in remembering. An active process of reconstruction is remembering, Bartlett called this the effort after meaning, that is mak ing the past more ordinary and normal so that it fits in to our existing knowledge or schemas. This study in psychology is regarded as a classic. Although there are criticisms to this study, It was said that it was a poorly controlled study that Bartlett may have been biased in the interpretations of the different stories. There was also another argument saying the story was very different and unusual. There have been other studies which have done real life everyday memories, which over time were successful. The study was by Wynn and Logie (1998). There is also the study of the unreliability of the eye witness testimony which was done by Loftus and Palmer (1974). They studied smash/ contacted car speed study. The aim of this study was to investigate an immediate recall from the effect of language the group of people use after being asked leading questions. The group of people were shown some slides of an accident which involved two cars. The group were split into two groups, by asking some about how fast were the cars going when they smashed in to each other? for the other set of people they were asked the same but instead of using the word smashed they said hit, bumped and contacted. From doing this study they found that from using different words peoples opinions of the speed were very different. The difference of using the word smashed was higher than the ones who were asked bumped, hit and contacted. So this proves there is an effect when using leading questions. The reconstructive model does provide us with a good explanation of everyday memory. But that we dont have perfect memories. There are disadvantages to the reconstructive memory model there is no information on how we can improve our memory. Also it is more opinion based, because this model of memory assumes that memories are so complicated. Some predictions cannot be made, so we cannot predict what information will be and wont be remembered, as we dont know what a persons schemas are. In conclusion to this, Atkinson and Schiffrins model of the multi store model is one of the best known models of memory, but this doesnt necessarily mean its the best theory. Its a very simple and straight forward theory, but in some ways its too simplistic. It shows that rehearsal doesnt always work, and that you cannot rehearse smells and sights. The reconstructive model is more reliable as its based more on everyday life. It explains that the eye witness testimony is not always accurate and not to fully trust it as much. It also states that we dont have perfect memories. The reconstructive model also helps us to understand how our previous knowledge effects our interpretations of memory.