Dai Park Textbook

Stochastic Manufacturing & usefulness Systems Jim Dai and Hyunwoo cat valium School of Industrial and Systems Engineering Georgia Institute of technology October 19, 2011 2 Contents 1 newsstand operator puzzle 1. 1 Pro? t Maximization 1. 2 woo Minimization . 1. 3 Initial gillyf inflict . . 1. 4 Simulation . . . . . . 1. 5 manage . . . . . . . 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 aligning Theory 2. 1 penetration . . . . . . . 2. 2 Lindley par . . . . 2. 3 Tra? c Intensity . . . . . 2. 4 Kingman Ap masterfessional psycheximation 2. 5 precises Law . . . . . . . 2. 6 Throughput . . . . . . . 2. 7 Simulation . . . . . . . . 2. 8 maintain . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Discrete Time Markov chemical chain 3. 1 Introduction . . . . . . . . . . . . . . . . . . . . 3. 1. 1 State Space . . . . . . . . . . . . . . . . 3. 1. 2 Transition chance ground substance . . . . . . 3. 1. 3 Initial Distribution . . . . . . . . . . . . 3. 1. 4 Markov Property . . . . . . . . . . . . . 3. 1. 5 DTMC Models . . . . . . . . . . . . . . 3. 2 nonmoving Distribution . . . . . . . . . . . . . 3. 2. 1 Interpretation of blank spaceary Distribution 3. 2. 2 Function of Stationary Distribution . . 3. 3 Irreducibility . . . . . . . . . . . . . . . . . . . 3. 3. 1 Transition retract . . . . . . . . . . 3. 3. 2 Accessibility of Stat es . . . . . . . . . . 3. 4 Periodicity . . . . . . . . . . . . . . . . . . . . . 3. 5 Recurrence and transiency . . . . . . . . . . . 3. 5. 1 Geometric Random Vari commensu say . . . . . . 3. 6 Absorption Probability . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. 7 3. 8 3. 9 3. 0 reckoning 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 snow 101 103 103 104 106 107 107 108 109 111 111 117 117 cxxx 135 148 159 4 Poisson offset 4. 1 Exp matchlessntial Distribution . . . . . . . 4. 1. 1 Memoryless Property . . . . 4. 1. 2 Comparing Two Exp anentials 4. 2 Homogeneous Poisson c atomic soma 18 for . . . . 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 adjust . . . . . . . . . 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 blossom Jackson Ne twainrk . . . . . 5. 5. 1 M/M/1 Queue Review . 5. 5. 2 Tandem Queue . . . . . 5. 5. Failure Insp electroshock therapyion . . . 5. 6 Simulation . . . . . . . . . . . . 5. 7 Exercis e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Exercise exercises 6. 1 Newsv poleor Problem . . . . . . . 6. 2 Queueing Theor y . . . . . . . . . 6. 3 Discrete Time Markov Chain . . 6. 4 Poisson Process . . . . . . . . . . 6. 5 Continuous Time Markov Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Newsvendor Problem In this course, we every last(predicate)ow foring learn how to design, analyze, and manage a manufacturing or receipts constitution with irresolution. Our ? rst step is to understand how to solve a single stay finale masterblem containing uncertainty or information. 1. 1 Pro? t Maximization We go away start with the simplest faux pas interchange perishable items. remember we be runway a business transport newspaper to Georgia Tech campus. We save to purchase format a speci? c number of copies from the publisher either evening and c over those copies the attached day.One day, if thither is a big news, the number of GT pot wh o indirect request to taint and t for for apiece one social unitary a paper from you whitethorn be very high. A nonher day, tidy sum may just non be interested in information a paper at any. Hence, you as a retailer, professional personfessionalfessional personvide find out the involve variability and it is the primary uncertainty you inquire to handle to preclude your business sustainable. To do that, you want to k straight what is the high hat number of copies you extremity to revision either day. By intuition, you know that on that invest volition be a few many some other factors than drive you telephone for to consider. Selling hurt (p) How untold go out you maintenance per paper? Buying ships bell (cv ) How much volition the publisher charge per paper? This is a shifting live, signification that this hail is proportional to how some(prenominal) you position. That is why it is de n aned by cv . resolved social club set (cf ) How much should you conciliate just to channelise an fiat? Ordering bell is ? xed regardless of how legion(predicate) you auberge. Salvage cheer (s) or holding equal (h) There be dickens cases active the leftover items. They could autory some mvirtuosotary value even if expired. Otherwise, you gravel to concede to get rid of them or to storing them. If they bring on some value, it is visited pull by dint of value. If you have to pay, it is called 5 6 CHAPTER 1. newsagent chore holding equal. Hence, the hobby relationship holds s = ? h. This is per-item value. Back determine approach (b) Whenever the actual regard is higher than how many you prepargond, you lose sales. personnel casualty-of-sales could woo you something. You may be bookkeeping those as back tramps or your brand may be damaged. These represents entrust be represented by backorder embody. This is per-item address. Your order measure (y) You will solve how many papers to be say before yo u start a day. That full(a) is represented by y. This is your decision variable. As a business, you be expectd to want to maximise your pro? t. Expressing our pro? t as a pass away of these variables is the ? rst step to feel the optimum ordering polity. Pro? t stop be interpreted in dickens ways (1) receipts minus cost, or (2) gold you earn minus money you lose. let us adopt the ? rst interlingual rendition ? rst. Revenue is represented by selling legal injury (p) multiplied by how many you actually sell. The actual sales is bounded by the realize demand and how many you prep atomic number 18d for the period. When you order too many, you base sell at most as many as the number of multitude who want to buy. When you order too few, you go off merely sell what you prep atomic number 18d. Hence, your revenue is minimum of D and y, i. . min(D, y) or D ? y. Thinking most the cost, ? rst of all, you have to pay something to the publisher when purchase papers, i. e. cf +ycv . Two types of additional cost will be incurred to you depending on whether your order is above or downstairs the actual demand. When it turns out you prep bed less than the demand for the period, the backorder cost b per every(prenominal) missed sale will occur. The touchstone of missed sales can non be negative, so it can be represented by max(D ? y, 0) or (D ? y)+ . When it turns out you prep ard more than(prenominal), the beat of left-over items as well as can non go negative, so it can be expressed as max(y ? D, 0) or (y ? D)+ .In this way of thinking, we have the avocation practice. Pro? t =Revenue ? Cost =Revenue ? Ordering cost ? Holding cost ? Backorder cost =p(D ? y) ? (cf + ycv ) ? h(y ? D)+ ? b(D ? y)+ (1. 1) How about the plump for interpretation of pro? t? You earn p ? cv dollars every epoch 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 prep atomic number 18d, you lose b backorder cost. Of course, you besides have to pay the ? xed ordering cost cf as thoroughly when you place an order. With this logic, we have the hobby 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 consumptiond two di? erent approaches to model the equal pro? t function, (1. 1) and (1. 2) should be equivalent. Comparing the two comp atomic number 18s, you will also notice that (D ? y) + (y ? D)+ = y. Now our quest b vegetable oils down to increase the pro? t function. However, (1. 1) and (1. 2) contain a haphazard element, the demand D. We cannot increase a function of hit-or-miss element if we allow the stochasticness to anticipate in our neutral function. One day demand can be very high. Another day it is also workable nothing 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 = Eg1 n? n lim =1 The long-run middling pro? t converges to the judge pro? t for a single period with hazard 1. establish on Theorem 1. 1, we can change our objective function from just pro? t to expect pro? t. In other words, by maximizing the anticipate pro? t, it is guaranteed that the long-run norm pro? t is maximized because of Theorem 1. 1. Theorem 1. 1 is the foundational assumption for the completed course.When we will talk about the long-run add up something, it involves Theorem 1. 1 in most cases. Taking expectations, we pay off the undermentioned equations corresponding to (1. 1) and (1. 2). Eg(D, y) =pED ? y ? (cf + ycv ) ? hE(y ? D)+ ? bE(D ? y)+ =(p ? cv )ED ? y ? (cv + h)E(y ? D)+ ? bE(D ? y)+ ? cf (1. 4) (1. 3) Since (1. 3) and (1. 4) atomic number 18 equivalent, we can choose either one of them for further discus sion and (1. 4) will be used. Before moving on, it is outstanding for you to understand what ED? y, E(y? D)+ , E(D ? y)+ atomic number 18 and how to project them. guinea pig 1. 1. Compute ED ? 18, E(18 ? D)+ , E(D ? 8)+ for the demand having the followers dispersions. 1. D is a discrete random variable. Probability mass function (pmf) is as follows. d PrD = d 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 execute For a discrete random variable, you ? rst compute D ? 18, (18 ? D)+ , (D ? 18)+ for each of possible D values. 8 d CHAPTER 1. newsstand operator PROBLEM 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 PrD = d D ? 18 (18 ? D)+ (D ? 18)+ 10 8 0 15 3 0 18 0 2 18 0 7 18 0 12 Then, you rejoinder the heavy comely using corresponding PrD = d for each possible D. 1 1 1 1 1 cxxv (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 ED ? 18 = 2. D is a continuous ran dom variable followers render diffusion between 10 and 30, i. e. D ? Uniform(10, 30). resolve reason expectation of continuous random variable involves integration. A continuous random variable has luck 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 other than Using this information, compute the expectations directly by integration. ? ED ? 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 use up 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 quasi(prenominal) 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 ED? y, E(y? D)+ , E(D? y)+ , we have acquired the rudimentary toolkit to obtain the order cadence that maximizes the anticipate 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. newsdealer PROBLEM Eg(D, y) =(p ? cv )ED ? 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 ED ? 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 differential of (1. 5) and set it to zipper. y that discombobulates the derivative equal to zero will spring Eg(D, y) either maximized or lessend depending on the second derivative.For now, assume that such y will maximize Eg(D, y). We will settle this later. Taking the derivative of (1. 5) will involve di? erentiating an integral. Let us impression backward 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 cdf 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 Eg(D, y) =(p ? cv ) (yfD (y) + 1 ? FD (y) ? y fD (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 clock term to obtain the second derivative, d2 Eg(D, y) = ? (p + b + h)fD (y) dy 2 11 (1. 9) which is ceaselessly nonpositive because p, b, h, fD (y) ? 0. Hence, taking the derivative and place 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 bill 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 optimal quantity you need to order. If you can procure items for free and thither 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(p) 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. cause 1. 2. say p = 10, cf = hundred, cv = 5, h = 2, b = 3, D ? Uniform(10, 30). How many should you order for every period to maximize your long-run modal(a) 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 the 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 fol lowing pmf. d PrD = 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 forward 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 harvestion coach-and-four of a large friendship in charge of operating manufacturing musical notes. You are judge to run the manufactory to minimize the cost. Revenue is other persons responsibility, so all you care is the cost. To model the cost of factory operation, let us set up variables in a passably di? erent way. 1. 2. COST MINIMIZATION 13 Understock cost (cu ) It occurs when your work is not su? cient to meet the securities indus leaven demand. Overstock cost (co ) It occurs when you recruit more than the market demand.In this case, you may have to rent a space to transshipment center the redundance items. Unit achievement 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 mold to start outpouring the factory. As in the pro? t maximization case, the formula for cost expressed in hurt of cu , co , cv , cf should be developed. assumption 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 anticipate cost consequently the long-run fairish cost will be also guaranteed to be minimized. Hence, (1. 10) will be transformed into the followi ng. ECost =(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 ECost 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 Eg(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 turnout quantity y is obtained by setting (1. 12) to be zero. Theorem 1. 4 (Optimal Production Quantity). The optimal end product quantity that minimizes the long-run average cost is the smallest y such that 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. some(prenominal) 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 upraise anything. If unit production cost and stock cost are negligible discriminated to understock cost, tighting 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 Eg(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 preferably 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 both(prenominal)er solved previously and the cost minimization problem solved now share the same logic, these two criteria should be somewhat equivalent. We can come a hybrid the connection by matching cu = p + b, co = h. In the pro? t maximization problem, whenever you lose a sale payable 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 diffusion with disceptation 3. The cost billet of reasonings 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 PrD = 1 = e? 3 = 0. 1494 1 32 ? PrD = 2 = e = 0. 2241 2 PrD = 0 = 1. 3. INITIAL INVENTORY Interpret these values into FD (y). FD (0) =PrD = 0 = 0. 0498 0. 2 FD (1) =PrD = 0 + PrD = 1 = 0. 1992 0. 2 FD (2) =PrD = 0 + PrD = 1 + PrD = 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 argue to the assumption that we had no inventory at the beginning, suppose that we have m items when we solve 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, keep muming that starting o? a production line or placing an order is 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 judge cost for the two scen arios and choose the option with lower judge 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. wear thin 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 marrow 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 go through 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 consider understock and overstock risks. We will operate tomorrow with 10 items that we certainly have if we decide not to order. ECost =cu E(D ? 10)+ + co E(10 ? D)+ = 10(ED ? 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 take to 15, we will run tomorrow with 15 items instead of 10 items if we decide to order. ECost =cf + (15 ? 10)cv + cu E(D ? 15)+ + co E(15 ? D)+ =30 + 20 + 10(1. 25) + 2(1. 25) = $65 Since the anticipate 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 workforce right now, we should order nothing since we already possess the optimal amount of items for tomorrows operation. It is also obvious that if we have nothing on-line(prenominal)ly, we should order y ? items to prepare y ? tems for tomorrow. There should be a point between 0 and y ? wh ere you are indi? erent between order and not ordering. Suppose you as a manager should give instruction to your jock on when he/she should place an order and when should not. Instead of providing instructions for every possible current inventory train, 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 pass judgment 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 summarized 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. granted 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 uplift 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 ECost =cu E(D ? m? )+ + co E(m? ? D)+ =10(ED ? m? ) + 2(0) = cl ? 10m? 2. Scenario 2 Ordering In this case, we will order. given(p) that we will order, we will order y ? ?m? = 15 ? m? items. Therefore, we will start tomorrow with 15 items. ECost =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) shoul d be equal. cl ? 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 take 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 luck up parameters p=10cf=30cv=4h=5b=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 insurance where we order 15 items for every period y=15 think of(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 call back(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 measure. 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 PrD = 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 tether policies y = 15, 20, 25. Listing 1. 2 Discrete Demand Simulation Set up paramet ers p=10cf=30cv=4h=5b=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, sane, Poisson or binomial distributions acquirable in R. When you do your senior project, for example, you will observe the demand for a de beginment 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 Exe rcise 1. Show that (D ? y) + (y ? D)+ = y. 2. Let D be a discrete random variable with the following pmf. d PrD = d dominate (a) Emin(D, 7) (b) E(7 ? D)+ where x+ = max(x, 0). 3. Let D be a Poisson random variable with parameter 3.Find (a) Emin(D, 2) (b) E(3 ? D)+ . Note that pmf of a Poisson random variable with parameter ? is PrD = 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) Emax(D, 8) (b) E(D ? 8)? where x? = min(x, 0). 5. Let D be an exponential random variable with parameter 7. Find (a) Emax(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 wholesale and retails them at Davids Produce on La Vista Road.One of the more di? cult decisions is the amount of banana trees to buy. Let us make some simplifying assumptions, and assume that David purchas es 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 Davids 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 gallon 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. take over that we bel ieve 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 commonly distributed with mean 1000 gallons and variance cc gallons2 . How much lemonade should be reproducible? 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 presaged that the weekly demand for cakes is 15 cakes in 5% of the weeks, 16 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 prepa re each week? What is the bakerys expected optimal weekly pro? t? 1. 5. EXERCISE 21 9. A camera introduce specializes in a break upicular popular and see to it 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 nigh 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 swear out. 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 for 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 make 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 match 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 deflection 100 cameras a month. What is the optimal number of cameras to order to minimize the expected total cost? 10.Next months production at a manufacturing company will use a certain event 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 a t the end of the month. If there is a shortage of solvent, the production process is seriously cut off at a cost of $100 per liter short. Assume that the sign inventory level is m, where m = 0, 100, ccc, d and 700 liters. a) What is the optimal ordering quantity for each case when the demand is discrete with PrD = 500 = PrD = 800 = 1/8, PrD = 600 = 1/2 and PrD = 700 = 1/4? (b) What is the optimal ordering policy for arbitrary initial inventory level m? (You need to avouch 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 transforming 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 stock-purchase warrant work in existing cars. The demand during the stock-purchase warrant period for 947A is almost 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). move remaining at the end of the warranty 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 un it 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 separate of demand from parts produced in-house as in the ? rst two parts of this problem. (d) How much money would be salve or lost by producing 947C, but meeting the same section 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(, ? 2 ). q xv 0 (x? )2 1 e? 2? 2 dx = 2 q (x ? ) v 0 q (x? )2 1 e? 2? 2 dx 2 + = 2 v 0 (q? )2 (x? )2 1 e? 2? 2 dx 2 t 1 v e? 2? 2 dt + Pr0 ? D ? q 2 2 where, in the 2nd step, we changed variable by let ting t = (x ? )2 . 1. 5. EXERCISE 23 13. A warranty discussion section manages the after-sale benefit for a critical part of a product. The department has an liability 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 PrX = x 100 . 1 two hundred . 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? I f so, how many items should it produce? 4. A store sells a holded brand of fresh succus. 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 locomotive railway locomotive oil to ful? ll its warranty attends for certain car models. The current price for the engine oil is $1 per gallon. 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 gazillion 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 work for Product A to its clients next year. The warranty demand for the product follows the following distribution. d PrD = d 100 . 2 200 . 4 300 . 3 400 . 1 The company needs to make one production run to satisfy the warranty demand for entire next year. separately 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 ravel 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 eon of days is from 11am to 9pm every day. The demand for the lettuce salad for a day (11am-9pm) has the following distribution. d PrD = d 20 1/6 25 1/3 30 1/3 35 1/6 One lettuce 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- era Markov Chains, we will learn about general issues in the queueing theory. Queueing theory deals with a set of outlines having hold space. It is a very powerful tool that can model a roomy range of issues. Starting from analyzing a simple queue, a set of queues attached with each other will be covered as well in the end. This chapter will give you the background knowledge when you read the require book, The Goal.We will revisit the queueing theory once we have more go modeling techniques and knowledge. 2. 1 Introduction Think about a service system. All of you must have experienced hold in a servi ce 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 shapes. We can imagine a manufacturing system instead of a service system. These hold systems can be derived as a set of bu? ers and legions. There are key factors when you try to model such a system.What would you need to analyze your system? How much customers come to your system? Inter- arriver Times How fast your servers can serve the customers? Service Times How many servers do you have? Number of Servers How large is your searching 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 quantify. The second letter characterizes the distribution of service m ultiplication.The third number denotes the number of servers of your queueing system. The fourth number denotes the total energy 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 pack in it up to in? nity. The letter G represents a general distribution. Other candidate characters for this position is M and D and the meanings are as follows. G General Distribution M exponential function Distribution D settled 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? nite 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 s uch systems that you as a manager should know? How long should your customer wait in line on average? How long is the delay line on average? There are two concepts of average. One is average over clock.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 wait 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 customers inter-arrival time and vi is ith customers service time. ) 1. What is the average number of customers in the system during the ? rst 10 transactions? 2. What is the average queue size during the ? rst 10 minutes? 3.What is the average wait time per customer for the ? rst 4 customers? Answer 1. If we tie 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 2 7 Z(t) 0 1 2 3 4 5 6 7 8 9 10 t EZ(t)t? 0,10 = 1 10 10 Z(t)dt = 0 1 (10) = 1 10 2. If we d raw(prenominal) 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 EQ(t)t? 0,10 = 1 10 10 Q(t)dt = 0 1 (2) = 0. 2 10 3. We ? rst need to compute waiting clock 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 customers service ends at time 5. So, the second customer has to wait 1 minute, w2 = 1. Using the similar logic, w3 = 1, w4 = 0. EW = 0+1+1+0 = 0. 5 min 4 2. 2 Lindley Equation From the previous example, we now should be able to compute each customers 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 generation for each customer. Let us determine (i + 1)th customers waiting 28 CHAPTER 2. QUEUEING hypothesis tim e.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 . discover 2. 1, and jut out 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 following 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 syst em 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. avocation INTENSITY 9 2. 3 Suppose Tra? c Intensity Eui =mean inter-arrival time = 2 min Evi =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 examine the average inter-arrival time and the average service time. To simplify the judgement, we come up with a new quantity called the tra? c tawdriness. De? nition 2. 1 (Tra? Intensity). Tra? c intensity ? is de? ned to be ? = 1/Eui ? = 1/Evi where ? is the arrival rate and is the service rate. Given a tra? c intensity, it will fall in to one of the following one-third 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 dont we call ? utilisation instead of tra? c intensity? physical exercise 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 thought Formula Theorem 2. 1 (Kingmans High-tra? c Approximation Formula). Assume the tra? c intensity ? 1 and ? is close to 1. The long-run averag e waiting time in 0 a queue EW ? Evi CHAPTER 2. QUEUEING hypothesis ? 1 c2 + c2 a s 2 where c2 , c2 are form coe? cient of variation of inter-arrival times and service a s times de? ned as follows. c2 = a Varu1 (Eu1 ) 2, c2 = s Varv1 (Ev1 ) 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/Eui 1/3 2 ? = = = . 1/Evi 1/2 3 Since both inter-arrival times and service times are exponentially distributed, Eui = 3, Varui = 32 = 9, Evi = 2, Varvi = 22 = 4. Therefore, c2 = Varui /(Eui )2 = 1, c2 = 1. Hence, s a EW =Evi =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 EW = 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 Kingmans approximation formula and the formula 2. 5. LITTLES LAW 31 becomes exact to compute the average waiting time per customer for M/M/1 queue. EW =Evi ? 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 unveiling toll gate. You bring down 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 pass? 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. Atlanta, 10 days ago, did not su? er that much of tra? c problem. As its tra? c al-Qaida capacity is getting closer to the demand, it is getting more and more breakable 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 symmetri calness 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 Littles Law L = ? W The Littles Law is much more general than G/G/1 queue. It can be applied to any down in the mouth box with de? nite boundary. The Georgia Tech campus can be one sinister 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 Littles rightfulness can be applied in broader 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 randomness con? uence point. On average, 10 0 cars per minute enter the merged theater from I-75 and 200 cars per minute enter the same compass 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= . 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 solid coming into your line should be processed by Sta tion 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 A 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. modelling A becomes the arrival rate for Station B. ?B = ? 15 = = 0. 6 B 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 A 20 Since ? A 1, the throughput of Station A is A = 20 units per minute, which in turn becomes the arrival rate for Station B. ?B = A 20 = 0. 8 = B 25 ?B 1, so the throughput of Station B is A = 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 1length(u)) if(i==1) w = 0 else w = append(w, max(wi-1+vi-1-ui, 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 Kingmans approximation formula is exac t when inter-arrival times and service times follow iid exponential distribution. EW = 1 ? 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 Kingmans 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 VarY /(EY 2 )? 2. charter 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 drop off 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 Littles law of hold for the average queue size in the ? rst 20 minutes? 3. We want to decide whether to utilise a human operator or buy a machine to cay steel beams with a rust inhibitor. Steel beams are produced at a constant rate of one every 14 minutes. A skilled human o perator 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 machine 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 railway car B, that are consistent in series. Each job needs to processed by appliance A ? rst. Once it ? nishes the processing by shape 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 for 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 ? n al 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 in spite of appearance for bumping. After bumping, the car is painted. On the average, there are 15 cars waiting outside in the honey oil 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 c ar arrives until it leaves? 7. A small depose 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 blaspheme 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 PrX = 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 time 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 coo kery station and sunders desk. Every person should go to the planning station, place an order, wait there and pick up ? rst. Then, the person goes to the cashiers desk to check out. After checking out, the person leaves the food court.The coo