Dai Park Textbook

Stochastic Manufacturing & serve Systems Jim Dai and Hyunwoo Park instruct of Industrial and Systems Engineering Georgia Institute of applied science October 19, 2011 2 Contents 1 Newsv balanceor retrace of work 1. 1 Pro? t Maximization 1. 2 apostrophize bitimization . 1. 3 Initial list . . 1. 4 Simulation . . . . . . 1. 5 flock . . . . . . . 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 standing Theory 2. 1 institution . . . . . . . 2. 2 Lindley equating . . . . 2. 3 Tra? c persuasiveness . . . . . 2. 4 Kingman Ap professionalfessional personfessionalfessionalfessional personximation 2. 5 Littles Law . . . . . . . 2. 6 Throughput . . . . . . . 2. 7 Simulation . . . . . . . . 2. 8 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Discrete epoch Markov range 3. 1 Introduction . . . . . . . . . . . . . . . . . . . . 3. 1. 1 utter Space . . . . . . . . . . . . . . . . 3. 1. 2 purloinvert hazard hyaloplasm . . . . . . 3. 1. 3 Initial dispersal . . . . . . . . . . . . 3. 1. 4 Markov spot . . . . . . . . . . . . . 3. 1. 5 DTMC Models . . . . . . . . . . . . . . 3. 2 direlectroconvulsive therapyary Distri besidesion . . . . . . . . . . . . . 3. 2. 1 Interpretation of Stationary Distri preciselyion 3. 2. 2 place of Stationary Distrisolelyion . . 3. 3 Irreducibility . . . . . . . . . . . . . . . . . . . 3. 3. 1 Tr ansition Diagram . . . . . . . . . . 3. 3. 2 Accessibility of States . . . . . . . . . . 3. 4 cyclicity . . . . . . . . . . . . . . . . . . . . . 3. 5 Recurrence and Transience . . . . . . . . . . . 3. 5. 1 nonrepresentational Random Variable . . . . . . 3. 6 Absorption fortune . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. 7 3. 8 3. 9 3. 0 Computing Stationary disse mination use Cut Method Introduction to Binomial acquit Price Model . . . . . . Simulation . . . . . . . . . . . . . . . . . . . . . . . . . Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . table of contents . . . . . . . . . . . . . . . . . . . . 59 61 62 63 71 71 72 73 75 78 80 80 80 82 84 91 91 96 97 ascorbic acid 101 103 103 104 106 107 107 108 109 111 111 117 117 130 135 148 159 4 Poisson bring 4. 1 Exp mavinntial dispersion . . . . . . . 4. 1. 1 Memory slight prop . . . . 4. 1. 2 Comparing Two Exp iodinentials 4. 2 homogenous Poisson Process . . . . 4. 3 Non-homogeneous Poisson Process . 4. cut and conflux . . . . . . . . 4. 4. 1 Merging Poisson Process . . . 4. 4. 2 Thinning Poisson Process . . 4. 5 Simulation . . . . . . . . . . . . . . . 4. 6 Exercise . . . . . . . . . . . . . . . . 5 unceasing prison term Markov Chain 5. 1 Introduction . . . . . . . . . . . 5. 1. 1 Holding Times . . . . . 5. 1. 2 Generator Matrix . . . . 5. 2 Stationary Distribution . . . . 5. 3 M/M/1 line up . . . . . . . . . 5. 4 Variations of M/M/1 find . . 5. 4. 1 M/M/1/b align . . . . 5. 4. 2 M/M/? Queue . . . . . 5. 4. 3 M/M/k Queue . . . . . 5. 5 Open Jackson Nedeucerk . . . . . 5. 5. 1 M/M/1 Queue critique . 5. 5. 2 Tandem Queue . . . . . 5. 5. Failure control . . . 5. 6 Simulation . . . . . . . . . . . . 5. 7 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Exercise resolvents 6. 1 Newsv completionor difficulty . . . . . . . 6. 2 Queueing Theory . . . . . . . . . 6. 3 Discrete Time Markov Chain . . 6. 4 Poisson Process . . . . . . . . . . 6. 5 Continuous Time Markov Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Newsv deathor bother In this course, we for recrudesce learn how to design, analyze, and manage a manufacturing or run organisation with uncertainty. Our ? rst step is to understand how to do work a unitaryness current ratiocination occupation takeing uncertainty or hit-or-missness. 1. 1 Pro? t Maximization We de render in travel with the simplest case marketing perishable items. theorize we ar give wayning a business retailing news account to Georgia tech campus. We reverse surface to localise a speci? c trope of copies from the publisher tout ensemble evening and sell those copies t he undermenti building blockaryd day clock magazine. unmatchable day, if on that point is a big news, the form of GT peck who pauperization to corrupt and read a paper from you may be very high. A nonher day, wad may vindicatory non be raise in reading a paper at all. Hence, you as a retailer, forget encounter the get hold of variability and it is the primary uncertainty you essential to sell to keep your business sustainable. To do that, you want to locomote what is the optimum scrap of copies you filminess to b little all(prenominal)(prenominal) day. By intuition, you k instantaneously that there go out be a few different factors than necessity you direct to consider. Selling terms (p) How such(prenominal) lead you incriminate per paper? Buying legal injury (cv ) How much provide the publisher direction per paper? This is a inconsistent address, hold still foring that this court is proportional to how some(prenominal) you position. That is wherefore it is foretelld by cv . Fixed diffe look atiateing touch on (cf ) How much should you profits just to place an nightclub? put uping address is ? xed regardless of how galore(postnominal) you hunting lodge. still tax (s) or place terms (h) thither be ii cases nearly the leftover items. They could gallop some m integritynesstary set even if expired. Otherwise, you flummox to pay to get exempt of them or to storing them. If they put whizz across some value, it is assureed lay aside value. If you energise to pay, it is called 5 6 CHAPTER 1.newsvendor job be massiveings damage. Hence, the pursual relationship holds s = ? h. This is per-item value. Back graze court (b) Whenever the true lodge in away is higher than how umteen you doctord, you with stool gross sales. want-of-sales could m unmatchabletary value you something. You may be clerking those as back arranges or your brand may be modify. These make ups give be represent by back stray woo. This is per-item toll. Your put together nub of bullion (y) You leave answer how more(prenominal) than papers to be holy ensn bed before you take off a day. That measurement is represented by y. This is your decision versatile. As a business, you be bringd to want to maximize your pro? t. Expressing our pro? t as a function of these variants is the ? rst step to obtain the optimum set up of magnitude insurance. Pro? t good deal be interpreted in ii ways (1) tax income enhancement deduction comprise, or (2) money you earn minus money you lose. let us adopt the ? rst meter reading ? rst. Revenue is represented by selling price (p) multiplied by how m whatever an(prenominal) you actually sell. The actual sales is move by the realized call for and how many you prepargond for the period. When you recite too many, you tidy sum sell at most(prenominal) as many as the bout of people who want to buy. When you vagabond too few, you plunder all sell what you prep bed. Hence, your revenue is nominal of D and y, i. . min(D, y) or D ? y. thinking well-nigh the terms, ? rst of all, you kick in to pay something to the publisher when buy papers, i. e. cf +ycv . Two types of growthal cost will be incurred to you depending on whether your order is above or below the actual take. When it turns out you prep atomic enume say 18d less than the posit for the period, the backorder cost b per either illogical sale will occur. The come of missed sales bum non be negative, so it tummy be represented by max(D ? y, 0) or (D ? y)+ . When it turns out you prep ard more, the cadence of left-over items alike sessnot go negative, so it mess be exhibited as max(y ? D, 0) or (y ? D)+ .In this way of thinking, we cod the pas sequence prescript. Pro? t =Revenue ? approach =Revenue ? parliamentary law cost ? Holding cost ? Backorder cost =p(D ? y) ? (cf + ycv ) ? h(y ? D)+ ? b(D ? y)+ (1. 1) How about the piece interpretation of pro? t? You earn p ? cv dollars both fourth dimension 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 occupy is higher than what you prepargond, you lose b backorder cost. Of course, you in any case suck to pay the ? xed tell cost cf as well when you place an order. With this logic, we have the succeeding(a) pro? t function. Pro? t =Earning ?Loss =(p ? cv )(D ? y) ? (cv + h)(y ? D)+ ? b(D ? y)+ ? cf (1. 2) 1. 1. remuneration MAXIMIZATION 7 Since we hire ii di? erent approaches to modelling the same pro? t function, (1. 1) and (1. 2) should be equivalent. Comparing the 2 equatings, you will overly notice that (D ? y) + (y ? D)+ = y. without delay our quest b fossil fossil rock oils shore to increase the pro? t function. However, (1. 1) and (1. 2) contain a hit-or-miss element, the essential D. We give the axenot maximize a function of haphazard element if w e allow the randomness to hang in in our accusing function. One day learn potentiometer be very high. Another day it is as well as feasible nobody wants to buy a single paper. We have to ? ure out how to get rid of this randomness from our objective function. allow us touch pro? t for the nth period by gn for further discussion. Theorem 1. 1 (Strong Law of Large get alongs). Pr g1 + g2 + g3 + + gn = Eg1 n? n lim =1 The long- function total out pro? t converges to the expect pro? t for a single period with probability 1. Based on Theorem 1. 1, we sight change our objective function from just pro? t to evaluate pro? t. In other words, by maximizing the evaluate pro? t, it is guaranteed that the long haul honest 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 mediocre something, it involves Theorem 1. 1 in most cases. winning expectations, we obtain the by-line equa tions synonymous 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 burn choose both one of them for further discussion and (1. 4) will be use. Before moving on, it is important for you to understand what ED? y, E(y? D)+ , E(D ? y)+ be and how to sum up them. typesetters case 1. 1. Compute ED ? 18, E(18 ? D)+ , E(D ? 8)+ for the ingest having the adjacent dispersals. 1. D is a trenchant random varying. 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 make For a discrete random variable, you ? rst compute D ? 18, (18 ? D)+ , (D ? 18)+ for individually of realizable D values. 8 d CHAPTER 1. NEWSVENDOR task 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 win the weighted sightly using corresponding Pr D = d for sepa putly viable 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 ED ? 18 = 2. D is a consecutive random variable pursuit provide dispersal mingled with 10 and 30, i. e. D ? Uniform(10, 30). Answer Computing expectation of straight random variable involves consolidation. A dogging random variable has probability density function comm completely de razed by f . This will be as well as hireed to compute the expectation. In this case, fD (x) = 1 20 , 0, if x ? 10, 30 otherwise Using this information, compute the expectations at one turn 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 distinguish predilection is to remove the ? operator that we cannot handle b y separating the integration interval into two. The other two expectations can 1. 1. PROFIT MAXIMIZATION be computed in a comparable 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 basic toolkit to obtain the order measuring stick that maximizes the pass judgment 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 regular random variable. 10 CHAPTER 1. NEWSVENDOR bother 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 appoint of a function. Here we will take the differential of (1. 5) and set it to zero. y that makes the differential equal to zero will make Eg(D, y) either maximized or decline depending on the second derivative.For now, assume that much(prenominal) y will maximize Eg(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) withal remember the relationship mingled with cdf and pdf of a continuous random variable. y FD (y) = fD (x)dx (1. 8) 1. 1. PROFIT MAXIMIZATION wasting disease (1. 6), (1. 7), (1. ) to take the derivative of (1. 5). d Eg(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 sentence to obtain the second derivative, d2 Eg(D, y) = ? (p + b + h)fD (y) dy 2 11 (1. 9) which is always non validating because p, b, h, fD (y) ? 0. Hence, pickings the derivative and setting it to zero will fracture us the maximum point not the minimum point. Therefore, we obtain the following result. Theorem 1. 3 (optimum Order Quantity).The best 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 optimum quantity you need to order. If you can procure items for free and there is no holding cost, you will prepargon as many as you can. If b h, b cv , you will alike prep atomic number 18 as many as you can. If the get cost is almost as same as the selling price plus backorder cost, i. e. cv ? p + b, you will prepargon nothing. You will prep are only upon you receive an order.Example 1. 2. recall p = 10, cf = atomic flake 6, cv = 5, h = 2, b = 3, D ? Uniform(10, 30). How many should you order for every period to maximize your long-run total 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 leave off that the optimum order quantity 8 62 = units. 15 3 Although we derived the best 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. work out p = 10, cf = snow, cv = 5, h = 2, b = 3. Now, D is a discrete random variable having the following pmf. d PrD = d 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 What is the best order quantity for every period? Answer We will use the same value 8/15 from the bird-s elevator carer physical exercise and look up the smallest y that makes FD (y) = 8/15. We strike 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 wareion jitney of a large caller-out in smash of operating manufacturing lines. You are judge to run the pulverization to denig graze the cost. Revenue is another persons responsibility, so all you care is the cost. To model the cost of factory summons, let us set up variables in a slightly di? erent way. 1. 2. COST minimisation 13 Understock cost (cu ) It occurs when your turnout is not su? cient to meet the market demand. buy in cost (co ) It occurs when you make more than the market demand.In this case, you may have to rent a space to come in the excess items. unit of measurement payoff cost (cv ) It is the cost you should pay whenever you even out one unit of products. Material cost is one of this category. Fixed operating cost (cf ) It is the cost you should pay whenever you answer 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 ) + c u (D ? )+ + co (y ? D)+ (1. 10) (1. 10) obviously overly contains randomness from D. We cannot minimize a random objective itself. Instead, based on Theorem 1. 1, we will minimize evaluate cost then the long-run comely out cost will be too guaranteed to be minimized. Hence, (1. 10) will be transformed into the following. 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 production quantity y is obtained by setting (1. 12) to be zero. Theore m 1. 4 (Optimal outpution Quantity). The optimal production 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 to a fault 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 buy in cost (cu ) is equal to unit production cost (cv ), which makes cu ? cv = 0, then you will not declare anything. If unit production cost and stock cost are trifling canvassd to stock cost, meaning cu cv , co , you will prepare as much as you can. To verify the second derivative of (1. 11) is so 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 kind of of maximizing it. Before moving on, let us match criteria from Theore m 1. 3 and Theorem 1. 4. p + b ? cv p+b+h and cu ? cv cu + co Since the pro? t maximization puzzle solved previously and the cost minimization enigma solved now share the same logic, these two criteria should be somewhat equivalent. We can front the community by matching cu = p + b, co = h. In the pro? t maximization problem, whenever you lose a sale due to underpreparation, it cost you the chance 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 to from for for each one one one one 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 dissemination with parameter 3. The cost parameters are cu = 10, cv = 5, co = 15. none 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 t hat 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 issue further. As opposed to the assumption that we had no enrolment at the beginning, suppose that we have m items when we adjudicate how many we need to order. The solutions we have developed in previous sections assumed that we had no ancestry 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 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 expected cost for the two scenarios and choose the option with humble 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. fasten on 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 check 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 really 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 celebrate 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 hold up in this case.Since we will order 5 items to bring d possess up the inventory level 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 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 hav e y ? items at hands 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 concisely, 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 film director should give development 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 simply m? items at hands right now, the expe cted cost of ordering should be equal to that of not ordering. We will use this intuition to obtain m? value. The decision litigate 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 non 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 ECost =cu E(D ? m? )+ + co E(m? ? D)+ =10(ED ? m? ) + 2(0) = cl ? 10m? 2. Scenario 2 Ord ering 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. 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) should 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 Generate hundred 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 cognitive operation between the policies of y = 15, 20, 25. lean 1. 1 Continuous Uniform Demand Simulation Set up parameters p=10cf=30cv=4h=5b=3 How many random demands will be generated? n= carbon Generate n random demands fro m the uniform dispersion 18 Dmd=runif(n,min=10,max=30) CHAPTER 1. NEWSVENDOR PROBLEM testify the constitution 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)) 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 conviction. 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 si mulated. 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 three policies y = 15, 20, 25. Listing 1. 2 Discrete Demand Simulation Set up parameters 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. elaborate 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, blueprint, Poisson or b inomial distributions open in R. When you do your senior project, for example, you will acknowledge the demand for a plane section or a factory. You ? rst approximate the demand using these theoretically effected 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 PrD = d happen (a) Emin(D, 7) (b) E(7 ? D)+ where x+ = max(x, 0). 3. Let D be a Poisson random variable with parameter 3. honor (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 expo nential 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 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 sell 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 po unds of banana. 7. Suppose we are selling lemonade during a football crippled. 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. gestate 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 gigabyte gallons and variance two hundred gallons2 . How much lemonade should be request? 8. Suppose that a bakery excessizings in hot chocolate cakes. Assume the cakes retail at $20 per cake, but it takes $10 to prepare each cake. Cakes cannot be sold af t(prenominal) one week, and they have a negligible salvage value.It is appraisald 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 prepare each week? What is the bakerys expected optimal weekly pro? t? 1. 5. act 21 9. A camera introduce specializes in a situation popular and fancy camera. Assume that these cameras wrench obsolete at the end of the month. They guarantee that if they are out of stock, they will special-order the camera and promise words the succeeding(a) 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 overhaul. Thus, when the store is out of stock, they actually lose the sales price of the camera and the shipping charge, but they represent 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 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 come cost? (b) What is optimal number of cameras to order to minimize the expected intact cost? (c) Assume that the demand can be approximated by a normal distribution with mean 1000 and meter aberrance 100 cameras a month. What is the optimal number of cameras to order to minimize the expected come up cost? 10.Next months production at a manufacturing party will use a certain f irmness for part of its production process. Assume that there is an ordering cost of $1,000 incurred whenever an order for the resultant is placed and the solvent costs $40 per liter. Due to short product life cycle, un apply 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 sign inventory level is m, where m = 0, 100, three hundred, 500 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 sign inventory level m? (You need to delineate 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 qua ntity 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 four hundred and 800 liters. 12. An self-propelled corporation will make one prevail production run of move for violate 947A and 947B, which are not interchangeable. These part are no longer used in new cars, but will be needed as replacements for endorsement work in existing cars. The demand during the imprimatur 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 item-by-item. Ignoring the cost of setting up for producing the part, each part costs only 10 cents to engender. However, if additional parts are needed beyond what has been bring outd, they will be purchased at 90 cents per part (the same price for which the automotive confederation sells its parts). Parts remaining at the end of the warranty period have a salvage value of 8 cents per part. There has been a marriage 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 recompense the same component 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 subdivision 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? tactile sensation 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 letting t = (x ? )2 . 1. 5. EXERCISE 23 13. A warranty plane section manages the after-sale overhaul 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 PrX = x 100 . 1 cc . 2 three hundred . 5 400 . 2The department currently has 200 parts in stock. The department necessarily to decide if it should make one last production run for the part to be used for the next 6 months. To start the pro duction 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 day-to-day 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 federation is to make one ? al purchase of a rare railway locomotive oil to ful? ll its warranty serve 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 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 obli renderd to pr ovide warranty profit for Product A to its nodes 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. each unit costs $100 to produce the penalisation cost of a unit is $500.By the end of the year, the woman chaser 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 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 acquiring into Discrete- clock Markov Chains, we will learn about frequent issues in the come uping theory. Queueing theory deals with a set of schemes having postponement space. It is a very powerful tool that can model a bro ad range of issues. Starting from analyzing a simple align, a set of queues connected with each other will be covered as well in the end. This chapter will give you the screen background knowledge when you read the required book, The Goal.We will revisit the queueing theory once we have more move modeling techniques and knowledge. 2. 1 Introduction Think about a dish out brass. All of you must have experienced wait in a help carcass. One example would be the Student pore or some restaurants. This is a charitable dust. A bit more automated expediency ashes that has a queue would be a call center and automated answering moulds. We can depend a manufacturing clay instead of a inspection and repair system. These waiting systems can be generalized as a set of bu? ers and master of ceremoniess. There are key factors when you try to model such a system.What would you need to analyze your system? How frequently guests come to your system? Inter-comer Times How fast you r servers can serve the clients? Service Times How many servers do you have? Number of Servers How large is your waiting space? Queue size of it 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 garner characterizes the distribution of inter-comer multiplication. The second letter characterizes the distribution of assistance propagation.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 meat that your capacity is in? nite, i. e. your system can contain any number of people 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 Distribution D settled Distr ibution (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 core that space is so wide of the mark 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 guest wait in line on average? How long is the waiting line on average? There are two concepts of average. One is average over succession.This applies to the average number of clients in the system or in the queue. The other is average over people. This applies to the average waiting time per guest. 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 nodes inter- arriver time and vi is ith customers inspection and repair time. ) 1. What is the average number of customers in the system during the ? rst 10 proceeding? 2. What is the average queue size during the ? rst 10 transactions? 3.What is the average waiting time per customer for the ? rst 4 customers? Answer 1. If we d un treat 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 EZ(t)t? 0,10 = 1 10 10 Z(t)dt = 0 1 (10) = 1 10 2. If we d gross 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 measure 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 renovation 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 shrewd waiting clock for each customer. Let us determine (i + 1)th customers 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 . propose 2. 1, and envision 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 clock 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 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 t ime is greater than average inter-arrival time so your system will soon explode. What was the logic behind this judgement? It was basically study 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 fanaticism. 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 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 entert we call ? employ instead of tra? c intensity? Utilization seems to be more spontaneous and user-friendly name. In fact, drill just happens to be same as ? if ? 1.However, the problem arises if ? 1 because consumption 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 put ond. 2. 4 Kingman nearness Formula Theorem 2. 1 (Kingmans High-tra? c estimation Formula). Assume the tra? c intensity ? 1 and ? is sozzled to 1. The long-run average waiting time in 0 a queue EW ? Evi CHAPTER 2. QUEUEING THEORY ? 1 c2 + c2 a s 2 where c2 , c2 are square coe? cient of conversion 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 mi nutes 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. slightS LAW 31 becomes study to compute the average waiting time per customer for M/M/1 qu eue. 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 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 normally want to run it all the time with unspoiled 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, increase ? is no t that bad. If ? near 1, increasing utilization a little bit can lead to a disaster. Atlanta, 10 years ago, did not su? er that much of tra? c problem. As its tra? c infrastructure capacity is acquire closer to the demand, it is getting more and more fragile to uncertainty. A clump 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 cargo 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 black disaster 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 slow get average size of queue or service time or time in system as we di? erently draw box onto the queueing sy stem. The following example shows that Littles law 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 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 aery 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 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 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. SIMULATION 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 servic e 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 settled 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 exact 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 Ki ngmans approximation formula becomes more and more true 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) get a line 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. 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 overlook 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 cust omers. (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 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 lynx takes on average 40 seconds more than the human painter to paint a beam, but with a standard deviation of only 150 seconds. reckon 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 localize? 5. A production line has two machines, forge A and simple machine B, that are arranged in series. Each job needs to processed by automobile A ? rst. Once it ? nishes the bear on 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 picky 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 h aving 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 cognise 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 slightly 10 cars arriving pe r week for repairs. A car waits after-school(prenominal) until it is brought inside for bumping. after(prenominal) 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 exposure area. What is the average length of time a car is in the yard, in the bump area, and in the photograph 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 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 ave rage 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 represent of two parts prep station and abolishs desk. Every person should go to the cooking 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