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BU.520.601 Decision Models Simulation Summer 2013 Simulation BU.520.601 1 Simulation Simulation Process • Many definitions. • It is the process of studying the behavior of a real system using a computer-based model that replicates the behavior of that system. • Used in situations involving probabilistic elements (e.g. random arrivals, service times and process yields) • Used in situations where the complexity or the size of the problem makes it difficult to use optimizing models. • Useful in both service or manufacturing systems. Simulation BU.520.601 2 Simulation characteristics In a simulation model we have “transactions” (customers, cars, ..) and “events” (arrivals, receiving service, departure,..). • When probability distributions are associated with events, we use a method called random deviate generation to get numbers from the probability distribution to simulate events. • Timing of event may or may not be important in a simulation. For simulation of a warehouse operation, if inventory is charged on items at the end of the month, we do not need to know precise timing of withdrawal of items. We only need to know how many items were withdrawn during the month. For simulation of toll booths, we need timings of two types of events – when each car arrives and how long it takes to pay. • When time is involved, simulation may be done by changing simulation clock in fixed increment or by changing clock from one event to the next (this is the preferred method). Simulation BU.520.601 3 Simulation characteristics..cont Simulation is not an optimization tool, rather we try to establish values of performance measures to arrive at better decision making. Here is an example. Suppose we would like improve customer satisfaction at a bank drive-in facility. We study arrival patterns, service times etc., and simulate the operation with one drive-in window, with two and may be with three drive-in windows. Obviously three windows will be most satisfactory from the customer point of view. But then we take into account the cost (initial and operating) and other factors to make the final decision. We will use EXCEL for some simple simulation exercises. In EXCEL we will use a function RAND() quite frequently. This function is volatile (it recalculates all the time). You should disable automatic calculations (switch to manual). Press F9 key and all values are recalculated. Simulation BU.520.601 4 Ex. 1 Simulate the tossing of a coin. Model construction: No simulation clock is involved • • • • • Each transaction will be the toss of a coin. We will generate 500 transactions (an arbitrary decision). We will assume that the coin is “fair”. The random variable X takes two values (0, 1 for T and H) with equal probabilities. To generate of a transaction, we need a very simple formula. Generate a random number (RN). If RN < 0.5, it’s a head; otherwise it’s a tail. In Excel, we will use the following formula in 500 cells =IF(RAND()<0.5,"H","T“). Performance measure: We will compute the probability of tails based on our simulation to see if it is close to 0.5. Simulation BU.520.601 5 Ex. 1.. Excel: Simulate the tossing of a coin. Simulation BU.520.601 6 Generating random deviates (variates) • For every probability distribution, as the variable X goes the minimum value to the maximum value, the cumulative probability increases from 0 to 1. • Random number generators produce numbers between 0 to 1 (uniform distribution). Thus for any value of a random number, there is one matching point on the cumulative distribution of X. We match the value and generate X. • RAND() generates a random number (say RN, 0 RN < 1). function automatically. Discrete uniform distribution For the discrete uniform distribution , when X varies between integers A and B, Excel has a special function = RANDBETWEEN(A, B). Simulation A A+1 BU.520.601 … B-1 B X 7 Empirical distribution Demand: X 300 400 500 600 Pr(X) 0.3 0.4 0.2 0.1 F(X) 1.0 0.3 0.7 0.9 1.0 The logic is simple. We match RN with cumulative probability. If RN < 0.3, X = 300. If 0.3 RN < 0.7, X = 400. If 0.7 RN < 0.9, X = 500. If 0.90 RN < 1.0, X = 600 F(X) Suppose we pick RN = 0.632. 0.632 0.0 We need cumulative probabilities. 300 400 500 600 X We get corresponding X value (= 400). Simulation BU.520.601 8 Using LOOKUP What is the value of X if RN = 0.632? 1.0 F(X) 0.632 0.0 300 400 500 600 F(X) 0.3 0.7 0.9 1.0 Demand: X 300 400 500 600 Pr(X) 0.3 0.4 0.2 0.1 Suppose we use HLOOKUP to find demand corresponding to F(X) = 0.632 We need to use the function without exact match. But since 0.3 < 0.632 < 0.7, Excel will match the value equal to 300. X To avoid this, we need to replace F(X) with some variable G(X), in which F(X) values are shifted to the right. G(X) 0.0 0.3 0.7 0.9 Demand: X 300 400 500 600 Simulation BU.520.601 9 Ex. 2 Ships arrive in the night at Arrival distribution a facility with two docks. X 0 1 2 3 4 5 If a dock is available in the morning, Pr(X) 0.30 0.30 0.20 0.10 0.05 0.05 it is assigned to a waiting ship for the whole day and the ship leaves in the evening. If a dock cannot be assigned, there is a fee of $10,000 per day per ship. Simulate the operation and estimate the annual fee. Model construction: We will start with a flow chart Population Dock 1 Arrival Queue Departure Dock 2 Our model is simpler because both docks take 1 day to process. Simulation BU.520.601 10 Ex. 2.. Dock simulation Population Arrival • • • • Queue Docks: Service 1 or 2 Departure Every day, we will generate new arrivals with HLOOKUP. Assume ship arrive between midnight and 6 a.m. These ships will be added to the queue. We will assign up to 2 ships from the queue (assumed FIFO – First In First Out) and calculate remaining ships waiting. These waiting ships will incur fee for that day. We will simulate the operation of a year and calculate the fee. We can replicate the experiment many times. Performance measure: We will compute the annual fee. Simulation BU.520.601 11 Ex. 2… Excel: Dock simulation Run for 365 days ? Simulation generated Simulation BU.520.601 12 Excel: Dock simulation Ex. 2…. Frequency of ships waiting 250 No of days 200 150 100 50 0 0 Simulation 1 2 3 4 5 No of ships BU.520.601 6 7 >7 13 Ex. 2….. Excel: Dock simulation Arrivals are generated with HLOOKUP. Simulation BU.520.601 14 Random Deviates: Continuous distributions Suppose X has continuous probability distribution (range 100 to 500) and we can find the cumulative distribution F(X). Every F(X) varies between 0 and 1. 1.0 F(X) 0.52 0.0 100 264 X 500 We can use random numbers (RN) to generate X values because RN also vary between 0 and 1 and there is a one to one correspondence. Suppose we pick RN = 0.52. We can find corresponding X value (say 264). Simulation BU.520.601 15 Random Deviates…… Triangular distribution Uniform distribution A A X B Uniform: = A + RAND()*(B – A) X B C Triangular: Let p = (B - A) / (C – A) =IF(RAND() ≤ p, X, Y) where X = A + SQRT(RAND() * (C – A) * (B – A)), Y = C - SQRT((1-RAND()) * (C – A) * (C – B)) Normal distribution = NORMINV(RAND(),Mean, Std. dev.) Log-Normal = LOGINV(RAND(),Mean, Std. dev.) Exponential distribution = (-Mean)*LN(RAND()) Simulation BU.520.601 16 Ex. 3 Retirement Planning You are 30 years old, and would like to invest 3000 dollars at the end of each year from now till you reach 60. Assume interest paid to be N(12, 2) meaning normally distributed with mean = 12% and std. dev. = 2%; interest is paid at the end of year. You would like to estimate probability of reaching the target of one million dollars at the age 60. You would like to know chances of achieving the target if you increase the annual amount invested. Simulation Age 30 Investment value X30 = 3000 31 32 X31 = 3000 + X30 + interest on X30 X32 = 3000 + X31 + interest on X31 BU.520.601 17 Ex. 3.. =B12*(1+NORMINV(RAND(), Mean_R,STDV_R))+ Annual_contr Simulation BU.520.601 18 Ex. 3… Retirement Planning sensitivity analysis Effect of changing contribution on the probability of achieving the desired outcome. Simulation BU.520.601 19 Ex. 4 An IPO is to be launched with the opening price expected to be from the distribution shown. Opening Stock Price 10 11 12 13 14 15 X Pr(X) 0.10 0.20 0.30 0.20 0.10 0.10 For the next five years, the stock price is expected to increase by an amount given by a lognormal distribution with mean of 1.5% and standard deviation of 0.5% provided the company does not fail. The probability of failure is 40%, 30%, 20%, 20% and 10% during the next five years. Estimate the following using simulation: (a)Price of the stock at the end of the 5 year period assuming the company has not failed. (b)Probability of survival at the end of 5 years. Theoretical answer to part (b) is (1 – 0.4) * (1 – 0.3) * (1 – 0.2) * (1 – 0.2) * (1 – 0.1) = 0.24192 Simulation BU.520.601 20 Ex. 4.. Simulation IPO Launching BU.520.601 21 Ex. 4… IPO Launching How? Count failures through Y1, then Y2 – Y1, etc. Fail in Year 4: =IF(K3="Y","Y",IF(RAND()<F$15,"Y","N")) 22 Simulation BU.520.601 Ex. 4…. IPO Launching Stock Price in Year 4: =IF(L4="N",Q4*(1+0.01*LOGINV(RAND(),Log_mean, Log_Stdv)),0) 23 Simulation BU.520.601 Time Based Event Oriented Simulation We will consider the following: Transactions (customers) enter the system in a single line and are processed at a single facility (server) on a FIFO basis also called “First Come First Served” (FCFS). Population Arrival Queue Service Departure We will consider several different situations. First, the dock example appeared to have same flow chart but it was somewhat different. Docks were open only during day time (say from 7). This means ships arrive in the night could be considered as arriving at 7 and using the docks for fixed amount of time. Each row generated new arrivals for a new day. Simulation BU.520.601 24 A simple example Population Arrival Queue Service Departure A machine take exactly 5 minutes to process a job. Our work load is only 10 jobs per hour. So we don’t need simulation, we can simply schedule a job every 6 minutes. M1 5 min M1 Job 1 Job 2 5 Job 3 11 17 There is an idle time of 1 minute after every job. Machine utilization is (5/6)* 100 = 83.3%. There will be no queue. Simulation BU.520.601 25 Example 5 M1 5 min Population Arrival Queue Service Departure We are now going to consider different arrival and service time distributions. Case Arrival distribution Service time distribution A Uniform (discrete) 1 to 11 min Fixed ( 5 minutes) B Uniform (continuous) 1 to 11 minutes Fixed ( 5 minutes) C D Triangular (1, 5, 9) minutes Poisson: 10 arrivals/hour (time Exponential (Average service time 5 between arrival 6 minutes – minutes) exponential) Note that the average time between arrival is 6 minutes and average service time is 5 minutes. Simulation BU.520.601 26 Arrival Flow chart Arrival SS=1 When an arrival event happens, the following is checked at that point in time. SS? SS=0 Enter service Set SS = 1 Set next Dep. Time Join Q Add 1 to Q Server Status (SS) Determine next arrival time SS = 0 Idle Q SS = 1 Busy Dep. Time Departure Time Simulation BU.520.601 Queue 27 Departure Flow chart When a departure event happens, the following is checked at that point in time. Departure Q not empty Q. Status Remove first Tr. From Q , start service & set Dep. Time Shorten Q by 1 Q Queue SS = 0 Server Status: Idle Tr. Transaction Simulation BU.520.601 Q Empty Set SS = 0 28 Example 5.. Single server system simulation – 4 cases Case Arrival distribution Service time distribution A Uniform (discrete) 1 to 11 min Fixed ( 5 minutes) B Poisson: 10 arrivals/hour Exponential (Average time 5 min.) Model Parameters: Each model may have many parameters. Examples are: arrival and service rates, capacities, etc. Statistics of interest /performance measures: Statistics on performance measures can be useful in validating the model and for decision making. Some examples. What’s the average waiting time? Maximum waiting time? What is the maximum queue length? What is the server utilization? How many people had to wait in queue before using the server? How many people waited more than X minutes? Simulation BU.520.601 29 Example 5… Single server system simulation When the simulation starts at time zero, the system is empty. Value of a performance measure such as server utilization keeps on changing as time progresses. After some time, when the process reaches a “steady state”, value of the performance measure comes close to the expected value of that measure. In all four cases since average time between arrivals is 6 minutes and the average service time (when the server is busy) is 5 minutes. The server utilization will stabilize at about 83% (=100*75/90). Expected Parameter value Parameter value 0 t Time For better estimate of the performance measure values, we generally chop off initial observations (up to period t). For our example, we will start collecting data from observation 201 (to 1200). Simulation BU.520.601 30 Example 5A Arrival Service time Uniform (discrete) 1 to 11 min Fixed ( 5 minutes) How did we get numbers in the table below? Simulation BU.520.601 31 Example 5A.. Simulation Arrival Service time Uniform (discrete) 1 to 11 min Fixed ( 5 minutes) BU.520.601 32 Example 5B Arrival Service time Poisson: 10 / hour Exponential: 5 min When the number of arrivals is Poisson (with 10 units / hr.), the time between arrivals is exponential (with average time = 6 minutes). Simulation BU.520.601 33 Example 5D.. Arrival Service time Poisson: 10 / hour Exponential: 5 min The graph here shows that it is not easy to determine when steady state may be reached (how many observations to chop off), nor do we know how many total observations to collect in a simulation run or how many time to replicate. Simulation BU.520.601 34 Example 5: Comparison Case Arrival distribution Service time distribution A Uniform (discrete) 1 to 11 min Fixed ( 5 minutes) D Poisson: 10 arrivals/hour Exponential (Average time 5 min.) A Simulation B BU.520.601 35 Simulation: general comments One can use visual basic macros within EXCEL. Risk solver also includes some simulation capability. Many specialized simulation languages have been developed. These can handle even more complex situations. Examples: AutoMod, Arena, GASP, GPSS, Promodel, SIMSCRIPT, Simula. Many simulation software packages also come with “animation” capability (there are even “free” ones). This can make a tremendous impact in visualizing the operations. One note of caution. Impressive visual display may give some false impressions even though data used in the simulation or the simulation logic is faulty. Simulation BU.520.601 36 Advantages 1. 2. 3. 4. 5. 6. 7. Simulation Disadvantages Flexibility. Can handle large and complex systems. Can answer “what-if” questions. Does not interfere with the real system. Allows study of interaction among variables. “Time compression” is possible. Handles complications that other methods can’t. Simulation 1. 4. Can be expensive and time consuming. Does not generate optimal solutions. Managers must choose solutions they want to try (“what-if” scenarios). Each model is unique. BU.520.601 37 2. 3. ATM Simulation 1 vs. 2 Average number of arrival per hour: 40 (Poison) Average service time 75 sec. (Exponential) Simulation BU.520.601 38