SIMULATION 19.1 INTRODUCTION " edu) a i thods (or procedures) to provide r fers we have discussed a number of analytical me an optima Pl a aiven problem. However, there are certain real world problems which although orhematical in nature involve variables whose values are determined by chance. Thus solution to problems f expected (maximum or minimum) pay-off value. in such cases is obtained in terms of i 1 u " ‘Simulation is a numerical solution method that seeks optimal alternatives (strategies) through a trial and error process. The simulation approach can be used to study almost any problem that involves uncertainty, ie. problems where probability distribution of variables is known in advance or specified can be solved by this technique. However, simulation approach requires an analogous physical model to represent ‘mathematical and logical relationship among variables of the problem under study. After having constructed the desired model, the simulation approach evaluates each altemative (measure of performance) by generating a series of values of random variables on paper over a period of time within the given set of conditions or criteria. This process of generating series of values one after another to understand the behaviour of the system (operational informations) is called executing (running or experimenting) model on computers. It is the availability of computers which makes it possible to deal with an extraordinarily large quantity of details which can be incorporated into a model and the ability to manipulate the model over many experiments (‘e. replicating all the passihilities that may he imbedded in the external world and events wonld seem to recur). The use of the word simulation can be traced to the mathematicians Von Neumann and Ulam in the late 19403 whieu dey developed the te1m Monte Carlo analysis while trying first to break the casino at Monte Carlo and subsequently, applying it to solution of nuclear shiclding problems that were either to expensive for physical experimentation or too complicated for treatment by known mathematical An agreed definition for the world simulation has not been reached so far, however few definitions are stated as: 4 r voprenation of a system or an organism is the operation of a model or simulator which is @ ha a Of the system or organism. The model is amenuble to manipulation which would af the model nan ba eens OF Unpractcal to perform on the entity it portrays. The operation can be inferred, ted ane for tt, properties concerning the behaviour of the actual system This definition is ae fon is broad enough to i i els ete. In thie view rae eels equally to military war games, business games, economic sal computer using itera een es logical and mathematical canstrcts that ean be manipulated © Simuton Or successive trials, this model for - ees of designing eriterion oF soy ne PUPOSE Of under: t Of criteria) Sor the oy Scanned with CamScanner— Simulation ous simulation is a numericé ' cal technique for conduct involves certain types of mathemar ing. experiments ji ? leat f Is on a digital computer, which ir and and logical relat ‘pehaviour and structure of a complex real-world sy ye ombipe eestor’ fo describe the cover extended periods of time. yew other definitions of simulation are as under: — Naylor ef. al. “y simulated ¥" is true if and onty if () X and ¥ are formal systems, i) ¥ is taken {0 be the real system, Gi) X taken fo be an approximation to the ret system, and fin) The rules of validity in are non-errrree, eerwise X wil become the rel system «Simulation is the use of a system mode ;odel that has the desi istic sinvvoducethe csrence. of actoal operaiton asthe designed characteristics af realty in order For operations research, simulation is a problem solving technique which. uses a computer-aided experimental approach to study problems that cannot be analysed using direct and formal analytical, trehods. As a result simulation can be thought of as a lat resort technique. It is not a technique which Thould be applied in all cases. However, Table 19.1 highlights what simulation is and what it is not. Table 19.1 Simulation what it is/not Tis It is not + a technique which uses computers. © an analytical technique which provides exact solution. «an approach for reproducing the processes J* & progiamming language but it could be progra- by which events of chance dnd change are fammed into a set of commands which can form a created in a computer. Janguage to facilitate the programming of simulation. + a procedure for testing and experimenting on models to answer What if... then s0 and s0...Lypes of questions. Simmation tb a fast and eloively inexpensive method of pertoming “experiments! on the complies Forexample, 1. tn inventory control, the problem of detemining the optimal eplenislnent policy arises —— probabilistic atochastic) nature of demand and Teed Tot ns, nsted of manly ng ou the thee replensheren atennatves for each fevel of demand and cd ine fr a perio fae year and he ng the best oe, we process data called experimen) °° ea mputer and obtain the results im very sh time at a very small cost. of walling against the enst of idle time of 2. In queuing.theory, the proble im of balancing the cos gaint the 0 : sevice facilities in the system arises due to the probit mate of i feral is so seen ra, sce to the customer. THUS ata ey der aa we et cing system, We process the data on ae sod aba be expect d i rervers, average waiting time, Value of various sinc Sei ‘of the queuing system such as idle tim 7 . Meus length, etc. Unlike various analytic! metho HE Simulation models. Each application ‘of simulation is o written and fixed rules to guide the formulation of different from the other and ad hoc to a large extent. Scanned with CamScanner19.2 STEPS OF SIMULATION PROCESS ‘The process of simulating a system consists of following steps: Step 1 Identify the problem then the problem may concern the determination of the invent tem i mulated, If any inventory system is being simulate: tls up to the reorder level size of order (number of units to be ordered) when inventory level fal ). Step2 (a) Identify the decision variables (b) Decide the performance criterion (objective) and decision rules In the context of the above defined inventory problem, the demand (consumption ratc), lead time and safety stock are identified as decision variables. These variables shall be responsible to measure the performance of the system in terms of total inventory enst under the decision rule—when to order. Step 3 Construet a numerical model A numerical model is constructed to be analysed on the computer. Sometimes the model is written in a partigular simulation language which is suited for the problem under analysis. Step 4 Validate the model Validation of the model is necessary to ensure whether i and the results will be reliable. is truly representing the system being analysed Step 5 Design the experiments Conduct experiments with the simulation model by listing specific values of vari courses of activ fur testing) at cach trial (run). les to be tested (i. list Step 6 Run the simulation model Run the model on the computer to get the results in the form’ of operating characteristics, Step 7 Examine the results Examine the results of problem as well as their reliability and correctness. If the simulation process is complete, then select the best course of action (or alternative) otherwise make desired changes in model decision variables, parameters or design, and return to Step 3. The steps of simulation process are also.shown in Fig, 19. 19.3. ADVANTAGES AND DISADVANTAGES OF SIMULATION Advantages 4, a spoons is suitable to analyse lrg and complex rele problems which canna he solved by i ies varus on the tn perme fdr de ee eect of cane Sif ssn Ta fee es aa ea Ce aad oki be used to experiment on a mode! ofa real situation without incurng the costs of operating Scanned with CamScannerTeulfy the problem Tdeniy decision variables, performance criterion and decision rules ‘Construct simulation model ee | Vatdate the model eee | Design experiments (specify values cf decision variables to be tested) Moaify the model by changing the input data, ie. values of decision variables Run or conduct the simulation ¥ No Is simulation process completed > Examine the results and select the beat course of action Fig, 19.1 . Steps of Simulation Process 4. Simulation can be used as a pre-service test (o try out new policies and decision rules for operating a system before running the risk of experi entation in the real system. “when all other techniques become intractable or fail s a * The only ‘remaining tol” Disadvantages 1. Sometimes simulation models are expensive and take a long time 10 develop. For example, a corporate bienng model may take a tong time to develop and prove expensive 2e : ‘ Tis the trial and error approach that produce different solutions in repeated runs. This means it does "I generate optimal solutions to problems. Each application of simulation is ad hoc to a great exter fe ge Smlation model does not produce answers by itself © solutions which he wants to examine: ‘The user has to provide all the constraints Scanned with CamScanner19.4 STOCHASTIC SIMULATION AND RANDOM NUMBERS : : . ‘i in a sample space In simulation, probability distributions are used to define mam soa fl coin, the fame a assigning a probability to each of pussible outcomes. For examp de Some probability, reflecting the {H, T} is the set of possible outcomes. Each outcome can occur f rhonce, In satstin, these element of chance. A random variable assigns a number to this element of chancr, 1 Sus, the numbers are estimated to assess the uncertainty inherent in the model, Dur in simulation Tess varies are controlled numerically and used to mimic these elements of uncertainty whick an le cs el. ‘This is done by generating (using the computer) outcomes with same frequency as ose eet in the process being mimicked (simulated). In this manner many experiments (also called simulation runs) can be performed, and leading to a collection of outcomes that have a frequency (probability) distribution similar to that ot the model you wish to study. To use simulation, it is first necessary that you Icarn how to generate the sample random events that make up complex models. Once this is done, it is possible to use the computer to reproduce the process through which chance is generated in real life. In this manner a problem can be evaluated that involves ‘many interrelationships for their aggregate behaviour and assess this behaviour as a function of a set of given parameters. Thus process geneiation (simulating chance processes) and modelling are the two fundamental techniques that we need in simulation The most elementary and important type of process is the random process, which requires for its simulation the selection of samples (or events) drawn from a given distribution so that repetition of this selection process will yield a. frequency distribution of sample values that faithfully matches the original distribution. When these samples are generated through some mechanical or electronic means, they are pseudo random numbers (for they are not really random since they are generated by a machine). Alternately it is possible to use table of Random Numbers where the selection of number in any consistent manner will yield numbers that behave as if they were drawn trom a uniform distribution, _ There are several ways of generating random numbers such as; Random numbers generator (which are inbuilt feature of spread sheets and many computer languages) tables (cee appendin), a roulette wheel, ete. Random numbers between 00 and 99 are used to obtain values of random variables that have a known diserete probability distribution in which the random variable of interest can assume one of a finite number of different values. In some applications, however the random variables are continuoes, thar We ther eat assume any real value according to a continuous probability distribution. For example, a queine peo applications, the amount of time a server spends with a customer is such a random torble ee ae. follow an exponential distribution, in. variable whieh rola’ 19.4.1 Monte Carlo Simulation n technique is representative of the given system under Probability distribution and then drawing random samples in terms of standard probability distribution such ag isace, i$ not possible to describe a sy mn " = i iri Probability distribution can be constructed Tah Polson, exponent gama, etc, an empiri ‘The Monte Carlo simulation technique consists of followi ing, steps: lables to be analysed, n for each random variable, appropriate s¢ dom variable °F "840M numbers to represent value oF ( Setting up a probability distribution for var Gi) Building a cumulative probability distributio, Gi) Generate random numbers. Assign an ange (interval) of values for each ran Scanned with CamScannerro) Conduct the simulation experiment by (i) Repeat Step 4 until the required numb (i) Design and implement a course of a gaz Random Number Generation means of random sampling, er of simulation runs has been generated. in and maintain control. oxte Carlo simulation requires the generation of a sequence of random numbers. This sequence of random fombers help in choosing random observations (samples) from the probability distribution. «) Arihmetic computation The nth random number ry consi wiiplicative congruential method is given by 1g of kdigits generated by using Th = Paty (modulo m) where p and m are positive integers, pp D P = prob. of success; n = number of trials: Poisson Reka where ‘5 Hlogn oy of. log ily ms 0). Cominious Pax halen A= mean arrival rate per unit of time Uniform 2 4b =at(ba)r Exponential 2=CIA)logr. ; 4, usu Nonnal Hod,b = 9 ha - 8 6 2 3 es : : 8 3 1 1 8 = 5 ~ Boos sues Ss Scanned with CamScanner19.6 when the inventory of books Since 8 books have been ordered three times a5 showin Table 19,6 when {Moen Ose at the beginning of the day plus orders outstanding is less than 8. Rs 3 x 10) = Rs 30. : ; Closing stock of 10 days is of 45 (=5 +9 +6+3+5+3 +1 +8 + 5) books. Therefore holding cost, Re 0.5 per book per day is Rs (45 * 0.5) = Rs’22.50. Total cost for 10 days = Ordering cost + Holding cost = Rs, 52.50. Since option B has lower total cost than option A, therefore manager should choose, option B. Example 19.4 XYZ spare parts company wishes to determine the levels-of stock it should carry for - items in its range. Demand is not certain and there is a lead time for stock replenishment. For one item X, the following information is obtained: Demand (units/day) 2 3 4 5 6 7 Probability a 010 0.20 030 0.30 0.10 Carrying cost (per unit/day) Rs2 Ordering cost (per order) + Rs 80 Lead time for replenishment : 3 days Stock on hand at the beginning of the simulation exercise was 20 units. Carry out a simulation run over a period of 10 days with the objective of evaluating the inventory rule: Order 15 units when present inventory plus any outstandirig order falls below 15 units. The sequence of random numbers to be used is: 0, 9, 1, 1, 5, 1, 8 6 3, 5, 7, 1, 2, Onsing the first number for day one. Solution Let us begin simulation hy assuming that orders are placed at the end of the day and received after 3 days at the end of the day. (i) back orders are accumulated in case of short supply and are supplied when stock is available. The cumulative probab Table 19.7. distribution and the random number range for daily demand is shown in Table 19.7 Daily Demand Distribution Daily demand Probability Cumulative prahability Random number range ; 0.10 0.10 00 4 020 é 030 01-02 : es a0 03 -05 6 030 0.90 06 — 08 7 010 1.00 The results of the simulation experiment conducted are shown in Table 19.8 Scanned with CamScannerTable 19.8 Simulation Experiments pening Random . Resulting Closi one losing Order Order ‘Average stock stock number demand stock placed delivered _in the evening —— 0 fees act 3 tah wens = a 2 0 ° ye 10 15 - 135 3 10 1 4 6 6 z 3 4 6 1 4 2 : - : 5 2 5 3 oes)" 1s 1s 1 6 7 i 4 8 4 = 10 1 8 8 6 2 ve 6 i 2 eI 6 o(-4ye 5 15 1 9 " 3 5 6 a a 35 0 6 5 3 i a = a ‘ Negative figure indicates back wider ‘Average ending stock = 78/10 = 7.8 units/day (Cost of placing one order) * (Number of orders placed per day) = 50 *3 = Rs 150 ‘ost of carrying one unit for’ one day) * (Average ending stock) 2% 7.8 = Rs 15.60 Daily ordering cost + Daily camying cost = 150 + 15.60 = Rs 165.60. Total daily inventory cost — xample 19.5 The manager of a warehouse is interested in designing an inventory control system for one afte products in stock, The demand for the product comes from numerous retail outlets and orders arrive ona weekly basis, The warehouse receives its stock from the factory but the lead time is not constant. The manager wants to determine the best time to release orders to the factory so that stockouts are minimised yet inventory holding costs are at acceptable levels. Any order from retailers not supplicd ou | agiven day constitute lost demand. Based on @ sampling, study, the following data are available ‘Demand per week Probability Lead time « Probabiliiy (in thousand) 0 020 2 030 1 040 3 040 | 2 030 4 030 3 0.10 The manager of the warehouse has determined the following cost parameters: ordering cost (Ca) per equals Rs 50, carrying cost (Ci) equals Rs 2 per thousand units per week, and shortage cost (C.) “als Rs 10 per tho r yusand units. . . | The objective of inventory analysis is t determine the optimal size of an order and the best time to Place an order. The following ordering policy has been suggested. evel becomes less than or equal to 2,000 units (reorder level), an order Policy: Wh inventor Dr Wey the are sem current inventory balance and the specified maximum equal to the difference betw ori replenishment level is equal to 4,000 units is plas zed. Scanned with CamScanneri) beginning it is 3,000 units, (ii) no Simulate the policy for a week's period assuming that the (i) beginning inventory is 3,000 » ) it inventory level back orders are permitted, (ii) each order is placed at the beginning of is ae pean an el is less than or equal to the reorder level, and (iv) the replenishment orders of the week. i bet Solution Using weekly demand and lead time distributions, assign an appropriate ee a random numbers to represent value (range) of variables as shown in Tables 19.9 and 19.10, respectively. Table 19.9 Probabilities and Random Number Interval for Weekly Demand Wookly demand Probability Cumulative probability Random number (in thousand) interval ee eee 0 020 020 00-19 1 a0 060 20-59 2 030 0.90 60 —89 3 0.10 1.00 90-99 Table 19.10 Probabilities and Random Number Interval for Lead Time Lead time Probability Cumulative probability Random number (weeks) - interval 2 030 030 00 - 29 3 040 07 30-69 4 030 1,00 70-99 a a I The simulation experiment conducted for 10 weeks period is shown in Table 19.11. The simulation process begins with an inventory level of 3,000 units. The following four steps occur in the simulation process. |. Begin each simulation week by checking whether any order has just arrived. If it has, increase the beginning (current) stock (inventory) by the quantity received. 2. Generate a weekly demand from the demand probability distribution in Table 19.9 by selection of a random number. This random number is recorded in column’, The demand simulated ig recorded in column 5. ‘The random number 31 generates a demand of 1,000 units when it is subtracted from the ial inventory level value of 3,000 units, yields an ending inventory of 2,000 units at the end of the first week. 3. Compute the ending inventory every week and record it-in column 7, Ending inventory ~ Beginning inventory ~ Demand = 3,000 — 1,000 = 2,000 {fon hand inventory is not sufficient to meet the week's demand, then record the number of units short in column 6, se nut Determine whether the week's ending inventory has reached the reorder level. If it has, and if there is no outstanding (back orders), then place an order Since ending inventory of 2,000 units is equal to the reurder level, therefore, an order for 4,000 — 2,000 = 2,000 units is placed, Scanned with CamScannerjead time for the new order is si , The lead | : is simulated by first choosi oo. $F ial hs random nanber icone i choosing 8 random mur and resoing it in getripution in 72 fe 19.10. lead time (column 9) by using the lead time be Fie eg me So de 2 al te a 5 . since there he ‘summing these cost yields a total inventory cost (column 10) for no lea a # ny shortage e step-by-ste] is ‘The same step-by-step process is repeated for the remaining 10 weeks of the simulation experiment. Analysis: of Inventory Cost 1,000 total unit Average ending inventory = Nett ut 00 unis per week. _ Zorders ‘Average number of orders placed = 02 order per week. 10 weeks 7,000 ‘Average number of lost sales © = 7 ggg = 7 units per week. Total average inventory cost = Ordering cost + Holding cost + Shortage cost = (Cost of placing one order) x (Number of orders placed per Week) + (Cost of holding one unit for one week) * (Average ending inventory) + (Cost per lost sale) * (Average number of lost sales per week) 100 16, 70 5 4 4 = 1041647 = Rs. 18.6 10 10 10 Table 19.11 Inventory Simulation Experiments Maximum inventory level = 4,000 units Reorder level = 2,000 units Ending Quantity Random Lead Total cast (TC) Week innii Demand leek Order Beginning Random iy Random ome Cot Gt Ge = TCS) inventory order receipt inventory number 10 3,000 31 1,000 2,000 a 2» 2 0 4 =S4 200 wo «| 2.0000 ‘a 2 8M 7008 33-1000 1000)? ay u 4 2p “apo as 20a a a 5 0 a hoo 1000) 6 6 0 73,000 2,000) ° 7 9 0 6 1,000, 1,000) to 0 os 3000, 2,000) ‘ 9 gp apog 5000, 0 A i S308 Scanned with CamScanner19.6 SIMULATION OF QUEUING PROBLEMS inute appointments. Some of the patients take 1 work to be done. The following summary tally needed to complete the work: Example 19.6 A dentist schedules all his patients for 30 mi more or less than 30 minutes depending on the type of dental shows the various categories of work, their probabilities and time actu it sbubility Category of Time required Prol service (minutes) of category (minutes) YC Filling 45 040 Crown o 0.15 Cleaning 15 0.15, Extraction 45 0.10 Checkup 15 020 Checkup Simulate the dentist's clinic for four hours and determine the average waiting time for the patlents as well as the idleness of the doctor. Assume that all the patients show up at the clinic at exactly their scheduled arrival time starting at 8.00 am. Use the following random numbers for handling the above problem: 40 82 1 34 25 66 17 79 [CA, Nov. 1990] Solution ‘The cumulative probability distribution and random number interval for service time are shown in Table 19.12. Table 19.12 Catagory Service time required Probability Cumulative Random number of service (minutes) probability interval Filling 45 040 0.40 00-39 (Crown 60 Os O55 40-54 Cleaning 15 ous 070 55-69 Extraction 45 0.10 080 10-79 Checkup 15 020 1.00 80-99 The various parameters of a queuing system such as arrival pattern of customers, service time, waiting time in the context of the given problem are shown in Tables 19.13 to 19.15. Table 19.13 Arrival Pattern and Nature of Service Pationt Scheduled Random Category of Service ume number arrtvat number service (minutes) 1 8.00 40 Crown @ 2 830 2 Checkup 15 3 9.00 u Filling 4s 4 930 34 Filling 4s 5 10.00 25 Filling 45 6 1030 6 Cleaning 1s 1 11.00 7 Filling 45 8 1130 9 Extraction 4S “ ton Scanned with CamScannerTime Event (Patient number) 8.00 ‘L arrive 830 arrive eon 1 departs; 3 arrive 915 2 depart 930 arrive iy 3 depart; 5 arrive 1030 6 arrive 1045.4 depart 11.00 7 arrive 11305 depart; 8 arrive 11.45 6 depart 12.00 End ‘The dentist was not ‘idle during the entire simulated period: The follows. Patient number (ime to exit) 160) 10) 2(15) 3(43) 3G0) 4(43) 41s) 5 (45) 5(30) 615) 7(48) 70)" Table 19.14 Ce Waiting (Patient number) wel wnt o 2 4 .e waiting times for the patients were a5 Table 19.15 Computation of Average Waiting Time Patient. Arrival time _ Service sturls at The average waiting time = Example, 19.7 A firm has @ probability distributions: Inter-arrival time Probability (minutes) 0.10 i 035 20 030 8 025 20 0.10 ee eran = 800 830 9.0 930 10.00 10.30 11.00 11.30 2780/8 — 35 minutes. 8.00 9.00 9.15 10.00 10.45 11,30 1145 1230 single, channel service stat Waiting time 0 30 15 30 45 60 45 60. (minutes) 280 on with the. following, arrival and service time Service time Probability (minutes) 5 0.08 10 0.14 15 O18 20 024 25 022 30 014 Scanned with CamScannerSolution The cumulative probability distributions and random number interval for inter-arrival time and service time are shown in Table 19.19. Table 19.19 Arrival time ‘Cumulative Random Service time Cumulative Random Minutes Probability Probability number Minutes Probability Probability mnumber fiaterual + interval 2 os outs oo-14 1 0.10 0.10 00 - 09 4 0.23 038 15-37 3 02 032 10-31 6 035 073 38-72 5 035 0.67 32-66 g 0.7 090 B-89" 7 023 090 67-89 10 0.10 1.00 90-99 9 010 1.00 90-99 “The simulation work cheet developed for the given problem is shown in Table 19.20. ‘Table 19.20 Random Inter- Arrival Service Random Service Service Waiting time > Line umber arrival time time starts mumber” time ends” Auendunt’ Customer length a (min) — (min) (min) (2), (min) (nin) (min) (min) % 10 910 9.10 1 1 917 10 = 4 2 912 9.17 8 5 922 - 5 1 n 6 918 922 4 3 925 - 4 1 10 # 920. 925 3 5 930 - 5 1 21 4 924 930 a 5 935 - 6 1 al 8 932-035 2 5 9.40 = 3 1 a7 8 9.40 9.40 7 1 941 = = S 0 10 950 9.50 4 5 955 9 = - 38 6 956 956 6 s 10.01 1 - - Total 56 4l 0 B 5 ( Average queue length = 5/9 = 0.56 = | customer (approx.) Il) Average wai me of customer before service = 23/9 = 2.56 minutes. i) Average service idle time = 20/9 = 2.22 minutes. (iv) Average service time = 41/9 = 4.56 minutes (v) Time a customer spends in the system = (4.56 + 2.96) = 7.12 minutes. (vi) Percentage of service idle time = 20/(20 + 41) = 0.33. 19.7. SIMULATION OF INVESTMENT PROBLEMS Example 19,9 The Investment Corporation wants to study the investment projects based on three factors: market demand in units; price per unit minus cost per unit and investment required. ‘These-factors are felt Scanned with CamScannerjndependent of each : be inden ch other. In analysing a new consumer product, the Corporation estimates the faowing probability distributions: An it de 7, Fae _ ar Price minus cost per unit Investment required = ability Rs Probability Rs Probability 204 00s 25.000 aT 3.00 0.10 17,50,000 025 50000 ron 5.00 020 20,00,000 0.50 35000 7.00 040 25,00,000 025 35 030 9.00 020 ‘ 40,000 020 10.00 0.10 45,000 0.10 0,000 005 ; Using simulation process, repeat the trial 10 times, compute the return on investment for each trial taking these three factors into account. What is the most likely return? Solution ‘The return’ per annuny can be computed by the following expression Return (R) = (Price - Cos) x Number of units demanded Investment ity distribution corresponding to each of the three factors, an ened to represent each of the three factors as shown in Tables Developing a cumulative probabil appropriate set of random numbers is ass 19.21, 19.22. and 19.23 Table 19.21 Annual demand Probability Cumulative probability Random number Annual demant’ 20,000 0.05 00s 00-14 25,000 010 os 05-14 30,000 020 035 15-34 35,000 030 065 35-64 ‘au000 20 08s 65-84 45,000 oo 0.95 85-94 50,000 0s 1.00 95-99 ‘Table 19.22 Cumulative Random number Price minus Probability aan cost per unit probability alse? 0.10 0.10 00-09 io 020 030 10-19 a 040 0.70 20-69 7.00 ea ea aa on 0.10 uo 9099 10. . Scanned with CamScannerTable 19.23 Cumulative Random number Investment Probubility required probability 17,50,000 02s 025 00-24 20,00,000 050 075 25-74 25,00,000 025 1.00 5-99 “The simulated return (R) is also calculated by using. ‘The simulation worksheet is prepared for 10 trials lation are shown in Table 19.24. the formula for R as stated before, The results of simul Table 19.24 Tia Random Simulated Random Simulated Random Simulated Simulated return number for demand —mumber for profit mumber for investment (26): Demand * demand — ("000) profit (price— investment (000) Profit per unit cost) per unit amare 100 1 2” 30 19 5.00 18 1750 857 2 37 35 a 3.00 él 2000 32 3 0 35 % 10.00 16 1750 20.00 4 7 30 oe 3.00 1 2000, 450 5 a 8 37 7.00 B 2000 1225 6 2 30 B 5.00 8 2000 750 1 2 30 9 5.00 41 2000 7.50 8 3B 35 ce 9.00 Pa 1750 19,00 9 6 38 8 7.00 19 1750 14.00 0 30 30 4l 7.00 ” 2500 8.40 ‘As shown in Table 19.24, the highest likely return is 20 per cent which corresponds to annual demand of 34,000 units yielding a profit of Rs 10 per unit and investment required is Rs 17,50,000. 19.8 SIMULATION OF MAINTENANCE PROBLEMS Example 19.10 A plant has a large number of similar machines. The machine breakdowns or failures are random and independent. ‘The shift in-charge of the plant collected the data about the various machines breakdown times and the repair time required on hourly basis, and the record for the past 100 observations as shown below was: Time between recorded — Probability Repair time Probability machine breakdowns (hours) required (hours) os 00s 1 028 i as 2 0.52 Ol 3 020 2 033 25 021 3 0.19 Scanned with CamScanner{00 Simulation was For each hour that one i gg Mby WaY Of lOst production A OM Ae to being or wi : fe nA repairman Is pala 2 ey alli to be repaired, the plant loses (@) Simulate this maintenance system for 15 breakdo a ; akdowns (6) How many repairmen should the plant hire for repair work solution The random numbers codi 19.25 and 19.26. ing for the hourly breakdowns and the repair times are shown in Tables Tabl fable 19.25 Random Number Coding for Breakdowns Sa Time between eR eT. Se Pes Salata: RY aaa, ae mete 05 0.05 005 00-04 i 0.06 oll os 05-10 a 0.16 027 11-26 = - 0.33 0.60 27-59 e 25, 021 O81) 60 - 80 wi ORL e. 0.19 = 1.00 81-99 Table 19.26 Randdin Number Coding for Repairs Repair time Probability Cumulative’ Random number required (hours) probability range : 1 028 028 00-27 2 052 080° 28279 3 apy 020. 1.00 80-99 : t een sie eS at The simultion worksheet i shown In Table 19.27. 1 assumed that the ist day hepin a midnight (00.00 hours) and also.the repairman begins work at 00.00 hours. The first breakdown occurred at 2.30 A.M, and the second occurred after.3 hours at clock time of 5.30 A.M. = [dle time cost + Repairman’s wage + (Repair time + Waiting time) * Hourly rate + Total hours * Hourly wages = 57.30 70 + 38:30 x 20 = Rs 4,777 Total current maintenance cost Maimenance Cost with Additional Repairman UF the plant hi re repairmen, then no machin plant hires two more repainmen, HCP TT rar, be only the repairing time of 36.00 ico 436 » 10 + (3830 * 2) * 20= Rs 4,052 Total cost = , airmen would ony inerease the ftal maintenance cos Hence, . ’ This shows’ that hiring ‘more than two TEP" ° tional repairman. ‘he plant may hire one addit e will wait for its repair. Thus, total idle time would Scanned with CamScanner‘Table 19.27 Simulation Worksheet Breakdown Random Time Time of Repair Random — Repair Repair Total idle Waiting ‘number number for between break- work mumber for time . work — time ime ‘break break-down begins repair required. ends at (hours) (hours) downs downs atime @ 2 m0 9 6 7 @ @ 09 1 61 25, 02.30 02.30 87 3 05.30 3.00 a: 85 3 0530 0.3039 2° 0730 © 7.00 S 3 16 15, 07.00 07.30 28 2 09.30 2.30 0.30 4 46 2 09.00. 09.30 7 3 1230 330 © 030 5 88 3 12.00 12.30 69 2 14.30 2.30 0.30 6 0g 1 1300. 14308 3 1730 4301.30 1 2 3 16001730 $2 2 1930 3.30 1.30 8 56 2 18.00 19.30 52s 2 21,30 3.30 130 . 9 2 13 19.30 21.30 1s 1 22.30 3.00 2.00 10 49 2 2130 72.30 85. 3 01.30 4.00 1.00 ub 44 2 23.30 01.30 4t 2 03.3¢ 4.00 2.00 2 33 2 0130 03.30 82. 3 06.30 5.00 2.00 b 7 25, 04.00 06,30 Biss 3 09.30 5.30 230 i“ 87 3 a7m = 09.30. 99 3 12.30 5.30 2.30 15 54. 2 09.00 12.30 23 2 1430 5303.30 3830 36.00 57.30 21.30 19.9 SIMULATION OF PERT PROBLEMS Example 19,11 A project consists of eight activities A to H, The completion time for each activity is a le. The data concerning probability distribution along with completion times for each activity Activity Immediate Time (day)/Probabi predecessor(s) A = - = oe ke B - - ~ - 05 - os - c A Me a MOTT OE Bc - = = = 9 te E A - = - 02 =~ = 08 F DE : 698 OAs mic ove iat OE halen G E a ieee ee u F 04 re ee (a) Draw the network diagram. and identify the critical path using the expected activity times. () Simulate the project to determine the activity times, Determine the criti completion time ty ‘ermine the critical path and project expected (©) Repeat the simulation four times and state estimated duration of the project in each of the tials. Scanned with CamScannersotution (2) The network diagram based’ on the precedence relationships is shown in Fig. 19.2. Tl expected completion time of each activity is obtained by using the formula: Expected time ~ . (Activity time x Probability) - = 40.2 +6 * 0.447 «0.4 = 6 days (activity A) The critical path of the project is: 1-2-3456 7, with expected completion time of 23.6 day The random number coding for each of the activities expected time is shown in Table 19.28. E)=6 B= 146 Ly=146 WO Bs 236 : 4228 4 & Ey=0° zo ; b3=9.3 1y=93 ; Fig. 19.2. Network Diagram Table 19.28. Random Number Coding for Activity Times Activity —. Time Probability. Cumulative Random number probability range A 4 ' 020 020 00-19 6 0.40 0.60 20-59 7 040 1.00 60-99 B 6 050 050 00 - 49 8 050 1.00 50-99 c 3 0.70 0.70 00 - 69 4 030 1.00 70-99 D5 0.90 090.,, 00-89 ° 8. O10 ay evr l.00, 90-99 E 5 020 020 00-19 9 080 ry 9 100,., 20-99 F 4 0.60 . 0.60 00-59 _ 5 040 1.00 60-99 040 "040 00-39 z 3 0.40 : 0.80 40-79 6 020 1.00 80-99 0.40 “0039 2 H 7 11.00 40-99 a Scanned with CamScannerThe simulation worksheet for four simulation runs.is shown in Table 19.29. For each run the project time is obtai Total time = Larger of s A) B and C + Larger of times for activities D and E + Larger of times for‘activities F and G + Time for activity H. Using the data given i Table 19.29, we have the simulation results shown in Table 19.30. Table 19.29 Simulation Worksheet Run . Activity times (days) A » cB Ci (VE. EF G A RNo. Time R.No. Tine R No: Tine No, TunveR-No. Time R. No. Time R. No. Time R. No. Time 12 6°17 6 68: 3 65. 5S 84 9 BS SH BD 2 92 7 35 6 6l 3 09 $s 43 9 95 5 06 3 87 7 3 02 4 22 6 57 3 St 5 58 9 24 4 82 6 03 2 4 47 6 19 6 36 3 27 Ss 59 9 AG . 4 13 3 719 7 5 93 7 37 6 66, 3 85 's 52 9 05 4 30 3 62 7 Total 30 30" 1S s AS 2 2 3s Average 6 6 3 5 9 44 42 5 Table 19.30 Simulation Kesults Simulation © Activity time’ Project diivation * Longést (critical) path run (days) ~ 1 6494642 1B __1-2-3-4-5-6-7 1=3-4-5-6-7 2 ¢ (7494547 mR 1-3-4-6 7 3 6494642 2B 1-2-3-4-5-6-7 4 6494447 % 1-2-3-4-6-7 ‘ 1-3-4-6-7 ’ TF9Fat7 a , 1-3-4-6-7 127 Here it may be ndfed that simulated mean project completion time, 25.4 days is almost two days longer than the 23.6 days completion time indicated using expected values alone 19.10 ROLE OF COMPUTERS IN SIMULATION The role of computers i simulation is vital, They are used to generate random mumbers, simulate te given Broblem with varying values of variables in few minutes and help the decision-maker to prepare reports which enable him to make decisions quickly as well as draw valid conclusions Computer languages available to help the simulation process can be divided into two categories: 19,10.1 General Purpose Programming Languages general purpose a jaming languages include FORTRAN, ai COBOL, PL/I, Pascale Tous Scanned with CamScannerthese languages for sinwlaton Process an extensive: programming experience {s required. As can be seen, even ina simple qi ig problem, many tedious details are involved in a simulation model. 19.102 Special Purpose Simulation Languages special simulation languages have tew advantages such-as: (i) They rediice programme preparation time and cost with features specially designed for simulation model. Such features generally include a master sequencing routine to automatically maintain an event sequence. and to keep track of simulated time sub- routines to handle arrivals and departures in a queuing syste, (ii) They have the capability to readily generate different types of random variates automatic generation of certain-types of statistical tables, and vious other features. (ii) They require litle or no prior programming knowledge for use. Major special purpose simulation languages are ‘ * © GPSS (General Purpose System Simulation) Usually, it does not requite programme writing, The system model is constructed via ‘block diagrams using block commands. The third version of this language, ie. GPSS III consists of two parts. The first part is an assembly programme which converts the system descriptors into input for the second part that performs the simulation. This langage was’ developed by IBM in early i960s. (i) SIMSCRIPT This language neither depends on any predefined coding forms nor on any intermediate language such as FORTRAN for its impleméntation, This language was developed by RAND Corporation in early 1960s. : Gi) DYNAMO It is a computer programme which’ is capable of taking Input in the foun of a set of equations describing the system. These equations are evaluated continuously for each time jiterval to understand the behaviour of the system. This language was developed at MIT in 1959 and ie best suited for econometric modelling of industrial complexes, urban, social and world systems planning. ‘The choice of a simulation package depends mainly on the specific purpose, the availability of simulation languages on a particular computer, the training and experience in simulation modeling and programming, and the availability of experienced programmers. 19.11 APPLICATIONS OF SIMULATION There is a wide range of applications ‘of computer-based simulation models because it is an approach rather ‘han an application of specific techniques. © / : ‘The major use of computer-based ‘Monte-Carlo simulation model has been in the solution of complex Nevin, 3 : , ‘ASamber of job shop simulation programmes Have beeh developed invaving detente tes for the individual operations of a given order. Due to different processing times for similar operations and different order operations sequences, itis difficult 0 predict the waiting time for a particular job at any given Work centre, For better scheduling, orders must be scheduled with a provision of waiting at the various Work centres they will pass through. Simulation can help in estimating accurately such waiting times. A good deal of work has been done in the development of inventory simulation models such as determination of optimal reorder level and lot size under conditions of probabilistic demand and lead time, optimal review period and ordering policy for, contin review inventory models, A number of network simulation ‘models have also been developed. For example, with a randomly Selected activity times the critical path can De evaluated, Repeating this process many times, the probability distribution of project completion time can be obtained.as.well as the probability that each given activity ‘Son the critical path. 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