SIMULATION1 SIMULATION vs. OPTIMIZATION In an optimization model, the values of the decision variables are outputs. The result of the model is a set of values for the decision variables that will maximize (or minimize) the value of the objective function. In a simulation model, the values of the decision variables are inputs. The model evaluates the objective function for a particular set of values. The result of the model is a measure of the quality of a suggested solution and the variability in various performance measures due to randomness in the 3 inputs. When should simulation be used? Simulation is one of the most frequently used tools of quantitative analysis today because: 1. Analytical models may be difficult or impossible to obtain, depending on complicating factors. 2. Analytical models typically predict only average or “steady-state” (long-run) behavior. 3. Simulation can be performed with a variety of software on a PC or workstation. The level of computing and mathematical skill required to design and run a simulator has been substantially reduced. 4 Steps in simulation 1. Problem formulation 2. Set objectives and overall project plan Phase 1 Problem Definition 3. Model conceptualization 4. Data Collection Phase 2 Model Building 5. Model Translation No No 6. Verified Yes 7. Validated Yes 8. Experimental Design 9. Model runs and analysis Yes No 11. Documentation, reporting and implementation No Phase 3 Experimentation Phase 4 Implementation 10. More runs 5 Model Verification and Validation Verification (efficiency) • Is the model correctly built/programmed? • Is it doing what it is intended to do? Validation (effectiveness) • Is the right model built? • Does the model adequately describe the reality you want to model? • Does the involved decision makers trust the model? Two of the most important and most challenging issues in performing a simulation study 6 running the model with one job type at a time • Reducing labor pool sizes to one worker Build the model in stages/modules and incrementally test each module • • • Uncouple interacting sub-processes and run them separately Test the model after each new feature that is added Simple animation is often a good first step to see if things are working as intended 7 .Model Verification Methods Find alternative ways of describing/evaluating the system and compare the results • Simplification enables testing of special cases with predictable outcomes • Removing variability to make the model deterministic • Removing multiple job types. 9 . the behavior of one or more factors is not known with certainty.Simulation and Random Variables MONTE CARLO METHOD: Simulation models are often used to analyze a decision under risk. Under risk. The behavior of the random variable can be described by a probability distribution. For example: demand for a product during the next month the return on an investment the number of trucks that will arrive to be unloaded The factor that is not known with certainty is called the random variable. T r u c k D o c k 3 T r u c k D o c k 2 T r u c k D o c k 1 Exit Entrance Truck waiting Truck waiting 10 . arrive at a warehouse to be unloaded.Design of Docking Facilities. In the following model. trucks of different sizes carrying different types of loads. The uncertainties are: When will a truck arrive? What kind and size of load will it be carrying? How long will it take to unload the trucks? Each uncertain quantity would be a random variable characterized by a probability distribution. The planners must address a variety of design questions: How many docks should be built? What type and quantity of material-handling equipment are required? How many workers are required over what 11 periods of time? . 12 . Management must balance the cost of acquiring and using the various resources against the cost of having trucks wait to be unloaded.The design of the unloading dock will affect its cost of construction and operation. Determination of Inventory Control Policies. the factory produces goods that are sent to the warehouses to satisfy customer demand. Simulation can be used to study inventory control models. The random variables are: daily demand at each warehouse and shipping times from factory to warehouse. Factory Warehouse 1 Demand Warehouse 2 Demand Warehouse 3 Demand In this model. 13 . Simulation can be used to study inventory control models. . Some of the operational questions are: When should a warehouse reorder from the factory and how much? How much stock should the factory maintain to satisfy the orders of the warehouses? The main costs are: Cost of holding the inventory Cost of shipping goods from a factory to a warehouse Cost of not being able to satisfy customer demand at the warehouse The objective is to find a stocking and ordering policy 14 that keeps the total cost low while meeting demand. Two broad categories of random variables: Discrete Can assume only certain specific values (e. integers) Continuous Can take on any fractional value (an infinite number of values) 15 Using a Random Number Generator in a Spreadsheet .. draw a random sample from a given probability distribution.Generating Random Variables To generate a random variable.g. The ability to generate discrete uniform random variables 2. The distribution (in the form of the cumulative distribution function) of the random variable to 16 be generated . The distribution of the discrete random variable to be generated To generate a continuous random variable. two things are needed: 1. two things are needed: 1. The ability to generate continuous uniform random variables on the interval 0 to 1 2.A GENERALIZED METHOD: To generate a discrete random variable with the RAND() function in a spreadsheet. Conversion of a Random No. to a Uniform Distribution 17 . 200 Random Numbers Generated Between 0 and 100 18 . 200 Uniform Discrete Random Numbers Generated Between 20 and 100 19 . 6 11 0.0 20 . D.3 10 0. Consider a random variable.8 12 0.9 13 1. the CDF for key values of D is: X F(x) 8 0. F(x) = Prob{D < x} Knowing the probability distribution for D. The CDF for D [called F(x)] is then defined as the probability that D takes on a value < x.The Cumulative Distribution Function (CDF).1 9 0. the demand. 8 0.2 1 0.4 0. To generate a discrete demand using the graph: Probability 1.Here is a graph of the CDF.2 u Step 1: Locate the particular value of U on this axis F(x) Step 2: Read the particular value of the random quantity. d. on this axis d x 0 7 8 9 10 11 12 13 14 21 .6 0. Suppose you want to model a discrete uniform distribution of demand where the values of 8 through 12 all have the same probability of occurring (uniform. that returns a random number between 0 and 1. To create a discrete uniform distribution. However. use the INT() function. =RAND(). The spreadsheet has a function. For example: In general. equally likely). if you want a discrete. this will result in a continuous uniform distribution. uniform distribution of integer values between x and y. use the formula: 22 INT(x + (y – x + 1)*RAND() ) . Its CDF is given by: F(x) = Prob{t>T} = e-lT Where 1/l is the mean of the random variable T. Therefore. we want to solve the following equation for w: u = e-lT The solution is: T = -1/l ln(u) where we can get “u” from Random number which represents cumulative distribution function 24 .Generating from the Exponential Distribution. The exponential distribution is often used to model the time between arrivals in a queuing model. 1000. . Consider drawing a random demand from a normal distribution with a mean (m) of 1000 and a standard deviation (s) of 100. dev. Excel has a built-in function that can do this: = NORMINV( RAND() . we can draw from a unit normal distribution. So. If Z is a unit normal random variable (normally distributed with a mean of 0 and a standard deviation of 1) then m + Zs is a normal random variable with mean m and standard deviation s. 100) Excel will automatically return a normally distributed 25 random number with mean 1000 and std.Generating from the Normal Distribution. The normal distribution plays an important role in many simulation and analytic models. 100. CAPITAL BUDGETING PROBLEM 26 . 000 75% of revenues Tax depreciation on the new equipment would be $10.A CAPITAL BUDGETING EXAMPLE: ADDING A NEW PRODUCT LINE Airbus Industry is considering adding a new jet airplane (model A3XX) to its product line. 27 . The following financial information is available: Startup Costs Sales Price Fixed Costs (per year) Variable Costs (per year) $150.000 $ 15. Salvage value of the equipment at the end of the 4 years is estimated to be 0.000 per year over the 4 year expected product life.000 $ 35. Airbus’ cost of capital is 10% and tax rate is 34%. assume that the demand for A3XXs is 10 units for each of the next 4 years: =C16 + C13 =NPV($D$3. For example. then a spreadsheet can be used to calculate the net present value (NPV).C17:F17)+B17 =-$B$2 28 .If demand is known. 9.THE MODEL WITH RANDOM DEMAND It is unlikely that demand will be the same every year. 12 . 9. 10. Now.. This model of demand is appropriate when there is a constant base level of demand that is subject to random fluctuations from year to year. 10. use the formula =INT(8 + 5*RAND() ) to sample from a discrete uniform distribution on the integers 8. or 12 units with each value being equally likely to occur. This is an example of a discrete uniform distribution. A more realistic model would be one in which demand each year is a sequence of random variables. Sampling Demand with a Spreadsheet: Assume initially that the demand in a year will be either 8. .g. Multiple trials can be performed by pressing the 29 recalculation key for the spreadsheet (e. 11. 11. F9). therefore. =INT(8+5*RAND() ) Hitting the F9 key would result in a different sample of demands.Using this formula results in random demands. The demands are random variables. and possibly a different NPV. . the NPV 30 is also a random variable. To run the simulation automatically and capture the resulting NPV in a separate spreadsheet. use the Data Table command. What is the probability that the NPV assumes a negative value (making the proposal to add the A3XX less attractive)? To answer these questions. 31 . What is the mean or expected value of the NPV? 2. a simulation model must be built.EVALUATING THE PROPOSAL Two questions need to be answered about the NPV distribution: 1. look 39 at the minimum and maximum NPVs. . Downside Risk and Upside Risk: To get a better idea about the range of possible NPVs that could occur.The resulting analysis gives the estimated mean NPV and standard deviation. the Frequency (column B) indicates the number of trials that fell into the bins (categories) defined by column A. The cumulative % column indicates the cumulative percentage of observations that fall into each category 41 or bin.In the resulting analysis. . The histogram gives a visual representation of the distribution of NPVs. Note that it is somewhat bell shaped. 42 . How much confidence do we have in the answers from the first trial? 2. 2. Would we be more confident if we ran more trials? 43 . the probability is <15%.100. the mean is $12.How Reliable is the Simulation? Now the two questions about the distribution can be answered: 1. What is the mean or expected value of the NPV? In this trial. What is the probability that the NPV assumes a negative value (making the proposal to add the A3XX less attractive)? In this trial. The next questions to ask are: 1. 96(standard deviation) In this case.96*$E$8/SQRT($E$16) =$E$4+1.521.679 and $14. . the upper and lower confidence limits are: =$E$4-1. Based on this trial. we have 95% confidence that the true mean NPV is 44 somewhere between $9.For a 95% confidence interval. the formula is: estimated mean + 1.96*$E$8/SQRT($E$16) So. the standard deviation is the standard error (the standard deviation divided by the square root of the number of trials). 11. 45 . Sampling Demand with a Spreadsheet: Assume initially that the demand in a year will be either 8. This model of demand is appropriate when there is a constant base level of demand that is subject to random fluctuations from year to year. 10. This is an example of a discrete uniform distribution. 9. A more realistic model would be one in which demand each year is a sequence of random variables.THE MODEL WITH RANDOM DEMAND It is unlikely that demand will be the same every year. or 12 units with each value being equally likely to occur. assume the mean demand will stay the same over the next four years. we allowed for random variation in mean demand (between 8 and 12 units) and discrete distribution. In addition. we started with equal mean demands of 10 for each period (year). with all values being equally likely. uniform distribution between 6 and 14.OTHER DISTRIBUTIONS OF DEMAND Originally. Now. we can explore the impact of different demand distributions on the NPV. 46 . somewhere between 6 and 14 units a year. This scenario can be modeled as a continuous. Then. MAINTAINANCE PROBLEM 47 . Corrective Maintenance versus Preventive Maintenance • The Heavy Duty Company has just purchased a large machine for a new production process. Therefore. • The breakdowns always occur on the fourth. and the two motors are rotated in use. or sixth day that the motor is in use. Cost of Replacement Cycle that Begins with Breakdown Replace a Motor Lost production during replacement Overhaul a motor Total $2. • The machine is powered by a motor that occasionally breaks down and requires a major overhaul.000 4.000 5. it takes fewer than three days to overhaul a motor. fifth.000 48 $11. so a replacement is always ready.000 . a second standby motor is kept. Fortunately. 7500 to 0.2499 5 6 7 or more 0.25 0. 3 4 Probability of a Breakdown 0 0.25 Corresponding Random Numbers 0.9999 49 .0000 to 0.2500 to 0. 2.7499 0.Probability Distribution of Breakdowns Day 1.5 0. 000 $33.000 $11.75 $11.000 $11.000 $11.6155 0.000 $110.000 Breakd own Cost Average Cost per Day $2.Computer Simulation of Corrective Maintenance Breakdown Random Time Since Last Cumulative Number Breakdown Day Cost Cumulative Cost Distribution of Time Between Breakdowns Number Prob Cumu.000 $308.7805 0.9091 0.000 $77.245 50 .000 $297.000 $11.000 $99.25 0 0.6920 0.000 $11.0026 0.000 $66.000 $11.000 $11. 0.4700 4 4 6 5 5 4 5 5 6 6 5 5 5 4 5 4 8 14 19 24 28 33 38 44 50 128 133 138 142 147 $11. Prob of Days no.6161 0.1343 0.6223 0.000 $319.000 $11.1523 0.9189 0.000 $11.000 $11.1493 0.000 $11.000 $55.25 0.6415 0.000 4 5 6 1 2 3 4 5 6 7 8 9 10 26 27 28 29 30 0.3599 0.5 0.000 $11.000 $11.000 $11.000 $11.000 $88.000 $22.000 $44.25 0.000 $330.000 $286.6869 0. Cost of Replacement Cycle that Begins without a Breakdown Replace a motor on overtime Lost production during replacement Overhaul a motor before a breakdown Total $3. • The goal is to provide maintenance early enough to prevent a breakdown.000 51 $6.000 0 3. • Scheduling the overhaul enables removing and replacing the motor at a convenient time when the machine is not in use. even if a breakdown has not occurred. so no production is lost.000 .Preventive Maintenance Options • Preventive maintenance would involve scheduling the motor to be removed (and replaced) for an overhaul at a certain time. 000 $11.0726 0.000 $183.4908 0.000 $6.000 $11.000 $6.000 $11.000 $126.000 $103.000 $217.5361 0.0593 0.000 $6.1657 0.1650 0.1035 0.000 $69.000 $11.000 $6.8207 0.6483 5 5 6 5 4 4 5 4 5 6 6 4 4 5 4 5 4 4 6 5 4 5 5 4 4 5 5 5 4 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 Replacement Replacement Replacement Replacement Breakdown Breakdown Replacement Breakdown Replacement Replacement Replacement Breakdown Breakdown Replacement Breakdown Replacement Breakdown Breakdown Replacement Replacement Breakdown Replacement Replacement Breakdown Breakdown Replacement Replacement Replacement Breakdown Replacement 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120 $6.4904 0.000 $81.000 $6.000 $11.000 $120.000 $35.000 $11.000 $205.6811 0.5 0.000 $109.000 $24.000 $6.4092 0.1032 0.000 $6.000 $11.25 5 0.000 $148.000 $6.9948 0.000 $11.75 6 $11.9297 0.3926 0.000 $6.000 $6.000 $137.2356 0.1084 0.000 $160.000 $12.000 $234.000 $11.000 $154.3365 0.000 $171.000 $11.000 $11.000 $6.000 $6.7241 0.3092 0.000 $211.000 $223.000 $6.000 $52.000 $6.Replace Motor on 4 Days Cycle Random Number Time Until Scheduled Time Event That Cumulative Breakdown Until ReplacementConcludes Cycle Day Cost Cumulative Cost Distribution of Time Between Breakdowns 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.5908 0.5408 0.6132 0.000 .000 $63.25 0.9918 0.000 4 days 52 Average Cost per Day $2.000 $6.000 $6.0718 0.000 $6.000 $177.25 Breakdown Cost Replacement Cost Replace After Number Cumulative of Days 0 4 0.000 $46.0267 0.5737 0.1388 0.000 $6.000 $75.000 $6.000 $18.000 $240.1035 0.000 $194.000 $11.000 Probability 0.000 $92. 000 $6.3543 0.4620 0.000 Probability 0.000 $11.000 $11.5844 0.000 $62.000 $11.000 $11.9815 0.000 $11.000 $263.1051 0.000 $179.000 $168.25 Number Cumulative of Days 0 4 0.000 $6.3945 0.6667 0.000 Replacement Cost $6.000 $6.000 Replace After 5 days Average Cost per Day $2.1892 0.000 $207.7489 0.8739 0.000 $252.000 $11.000 $6.000 $196.000 $235.5809 0.9448 0.000 $56.000 $11.5796 0.000 $22.000 $213.2253 0.9239 0.000 $6.000 $11.000 $11.000 $11.000 $11.000 $118.0949 0.000 $11.Replace Motor After 5 Days Cycle Random Number Time Until Breakdown Scheduled Time Until Replacement Event That Concludes Cycle Cumulative Day Cost Cumulative Cost Distribution of Time Between Breakdowns 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.000 $11.000 $6.000 $39.2480 0.4318 0.000 $96.0987 0.8282 0.000 $246.000 $6.000 $107.2204 0.75 6 Breakdown Cost $11.0583 0.000 $11.000 $6.000 $190.9055 0.000 $157.000 $6.000 $79.000 $33.000 $140.000 $11.5078 5 4 4 6 6 5 6 5 6 6 4 5 5 4 5 6 5 4 4 4 6 4 6 5 5 5 6 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 Breakdown Breakdown Breakdown Replacement Replacement Breakdown Replacement Breakdown Replacement Replacement Breakdown Breakdown Breakdown Breakdown Breakdown Replacement Breakdown Breakdown Breakdown Breakdown Replacement Breakdown Replacement Breakdown Breakdown Breakdown Replacement Breakdown Breakdown Breakdown 5 9 13 18 23 28 33 38 43 48 52 57 62 66 71 76 81 85 89 93 98 102 107 112 117 122 127 131 136 141 $11.000 $85.021 53 .000 $11.000 $146.000 $224.000 $11.5229 0.000 $73.25 0.5 0.000 $45.000 $11.7658 0.25 5 0.6972 0.7721 0.000 $129.8745 0.0157 0.000 $11.000 $11.000 $274.000 $285.000 $11. 54 . of times of occurrence 2 24 3 12 4 4 The management is of the opinion that proportion of stockouts should not exceed 5%.Inventory Problem The daily demand of an item has the following distribution: Demand per day (units) No. which follows the following distribution: Lead time (days) No. of days on which demand occurred 0 4 1 10 2 15 3 39 4 22 5 10 When an order is placed to replenish inventory there is a delivery time lag. Queuing Problem 55 . 56 . • The problem is to analyze the system by simulating the arrival and service of 20 customers.Single server queue • A small grocery store with just one checkout counter • Interarrival time uniform between 1-8 mins • Service time varies between 1-6 mins. Probability distribution given. 95 1.60 0.20 0.30 0.10 0.30 0.Given distributions Service Times (Minutes) 1 2 3 4 5 6 Probability 0.00 Interarrival Times (minutes) Minimum 1 Maximum 8 Uniform 57 .25 0.10 0.85 0.05 Cumulative Probability 0.10 0. 2/5 0.83 – Lq = the average number of people waiting in line – Wq = the average time spent waiting in line Lq = 0. – The average arrival rate = l = 5 customers per hour – The average service rate = m = 6 customers per hour • Using our knowledge of queuing theory we obtain – = the server utilization = 5/6 0.832/(1-0. • Empirical data gathering indicates that inter-arrival and service times are exponentially distributed.83) 4.Simulating of a M/M/1 Queue • Assume a small branch office of a local bank with only one teller.83 • How do we go about simulating this system? – How do the simulation results match the analytical ones? 58 .2 Wq = Lq/ l 4. NEWSPAPER BOY PROBLEM 59 . 7. and q – D 0 left over. 0) – X ~ normal (m = 135. same every day • Demand during a day: D = max (X. satisfy all demand.00 each • Each morning.03 each • If D > q.Newspaper Boy Problem • Newsvendor sells newspapers on the street – Buys for c = $0. from historical data – X rounds X to nearest integer • If D q.55 each.1). sells for r = $1. sells out (sells all q copies). sell for scrap at s = $0. s = 27. buys q copies – q is a fixed number. no scrap – But missed out on D – q > 0 sales • What should q be? 60 . Newspaper Boy Problem – Formulation • Choose q to maximize expected profit per day – q too small – sell out. applications – Much research on exact solution in certain cases – But easy to simulate.45 profit per paper – q too big – have left over. 0) – cq Sales revenue Scrap revenue Cost – W(q) is a random variable – profit varies from day to day • Maximize E(W(q)) over nonnegative integers q 61 . q) + s max (q – D. scrap at a loss of $0. extensions.52 per paper • Classic operations-research problem – Many versions. as a function of q: W(q) = r min (D. miss $0. variants. even in a spreadsheet • Profit in a day. 7.Steps: Newspaper boy Problem – Simulation • Set trial value of q.µ. compute profit for that day – Then repeat this for many days independently. 0) – So need to generate X ~ normal (m = 135. s = 27. average to estimate E(W(q)) • Also get confidence interval. 0) 62 . histogram of W(q) – Try for a range of values of q • Need to generate demand D = max (X. σ).1) •Demands = MAX(ROUND(NORMINV(RAND(). estimate of P(loss). generate demand D. 3. It is particularly well-suited for problems that are difficult or impossible to solve mathematically. 4. 2. It can serve as a mode for training decision makers by enabling them to observe the behavior of a system under different conditions.Advantage of Simulation 1. It allows an analyst or decision maker to experiment with system behavior in a controlled environment instead of in a real-life setting that has inherent risks. 63 . It enables a decision maker to compress time in order to evaluate the long-term effects of various alternatives. especially in cases in which a large number of trials are indicated. rather than optimal solutions. A certain amount of expertise is required in order to design a good simulation.Limitations of Simulation 1. Good simulations can be costly and time-consuming to develop properly. 2. 3. 4. 64 . Analytical techniques may be available that can provide better answers to problems. Probabilistic simulation results are approximations. they also can be time-consuming to run. and this may not be readily available. 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