bu.520.601 bu.520.601 decision models simulation1 simulation summer 2013
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BU.520.601Simulation 1
BU.520.601 Decision Models
Simulation
Summer 2013
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• 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.
SimulationSimulation Simulation ProcessSimulation Process
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Simulation characteristicsSimulation characteristicsIn 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 We only need to know how many items were withdrawn during the month.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).
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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.
Simulation characteristics..cont
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.
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Ex. 1Ex. 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.
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Ex. 1..Ex. 1.. Excel: Simulate the tossing of a coin.
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Generating random deviates (variates)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.
For the discrete uniform distribution , when X varies between integers A and B, Excel has a special function
= RANDBETWEEN(A, B).
Discrete uniform distributionDiscrete uniform distribution
A A+1 … B-1 B X
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We get corresponding X value (= 400).
300 400 500 600 X
1.0
0.0
F(X)
Suppose we pick RN = 0.632.
0.632
Empirical distributionEmpirical distribution
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
Demand: X 300 400 500 600
Pr(X) 0.3 0.4 0.2 0.1
F(X) 0.3 0.7 0.9 1.0
We need cumulative probabilities.
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F(X) 0.3 0.7 0.9 1.0Using LOOKUPUsing LOOKUP
What is the value of X if RN = 0.632?
300 400 500 600 X
1.0
0.0
F(X)
0.632
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.
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
Demand: X 300 400 500 600
Pr(X) 0.3 0.4 0.2 0.1
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Ex. 2Ex. 2 Ships arrive in the night at a facility with two docks.
Model construction: We will start with a flow chart
If a dock is available in the morning, 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.
Arrival distributionX 0 1 2 3 4 5
Pr(X) 0.30 0.30 0.20 0.10 0.05 0.05
Population
QueueArrival
Dock 1
DepartureDock 2
Our model is simpler because both docks take 1 day to process.
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Population
QueueArrival Docks:
Service 1 or 2
Departure
Ex. 2..Ex. 2..
• 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.
Dock simulation
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Ex. 2…Ex. 2…
Run for 365 days
Simulation generated
?
Excel: Dock simulation
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Ex. 2….Ex. 2…. Excel: Dock simulation
0 1 2 3 4 5 6 7 >70
50
100
150
200
250
Frequency of ships waiting
No of ships
No
of d
ays
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Ex. 2…..Ex. 2…..
Arrivals are generated with HLOOKUP.
Excel: Dock simulation
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Random Deviates: Continuous distributionsRandom 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.
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.
We can find corresponding X value (say 264).
0.52
Suppose we pick RN = 0.52.
264
F(X)
100 X 500
1.0
0.0
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A X B
Uniform distribution
Random Deviates……Random Deviates……
Uniform:= A + RAND()*(B – A)
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())
A X B C
Triangular distribution
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Ex. 3Ex. 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.
Age Investment value30 X30 = 3000
31 X31 = 3000 + X30 + interest on X30
32 X32 = 3000 + X31 + interest on X31
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Ex. 3..Ex. 3..
=B12*(1+NORMINV(RAND(), Mean_R,STDV_R))+ Annual_contr
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Ex. 3…Ex. 3… Retirement Planning sensitivity analysis
Effect of changing contribution on the probability of achieving the desired outcome.
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Ex. 4Ex. 4 An IPO is to be launched with the opening price expected to be from the distribution shown.
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.
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.
Opening Stock PriceX 10 11 12 13 14 15
Pr(X) 0.10 0.20 0.30 0.20 0.10 0.10
Theoretical answer to part (b) is
(1 – 0.4) * (1 – 0.3) * (1 – 0.2) * (1 – 0.2) * (1 – 0.1) = 0.24192
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Ex. 4..Ex. 4.. IPO Launching
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Ex. 4…Ex. 4…IPO Launching
Fail in Year 4: =IF(K3="Y","Y",IF(RAND()<F$15,"Y","N"))
How?Count failures through Y1, then Y2 – Y1, etc.
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Ex. 4….Ex. 4…. IPO Launching
Stock Price in Year 4: =IF(L4="N",Q4*(1+0.01*LOGINV(RAND(),Log_mean, Log_Stdv)),0)
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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
QueueArrival 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.
Time Based Event Oriented SimulationTime Based Event Oriented Simulation
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Job 2 11
A simple exampleA simple example
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.
Population
QueueArrival Service Departure
M1 5 min
M1 Job 1 5
Job 3 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.
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Example 5Example 5Population
QueueArrival Service Departure
We are now going to consider different arrival and service time distributions.
Case Arrival distribution Service time distributionA Uniform (discrete) 1 to 11 min Fixed ( 5 minutes)
B Uniform (continuous) 1 to 11 minutes
Fixed ( 5 minutes)
C Triangular (1, 5, 9) minutes
D Poisson: 10 arrivals/hour (time between arrival 6 minutes – exponential)
Exponential (Average service time 5 minutes)
Note that the average time between arrival is 6 minutes and average service time is 5 minutes.
M1 5 min
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Arrival
SS?
Join QAdd 1 to Q
Determine next arrival time
SS=1
Enter serviceSet SS = 1
Set next Dep. Time
SS=0
Q Queue
Dep. Time Departure Time
Arrival Flow chartArrival Flow chart
Server Status (SS)
SS = 0 Idle
SS = 1 Busy
When an arrival event happens, the following is checked at that point in time.
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Departure
Q. Status
Remove first Tr. From Q , start service & set Dep. Time
Shorten Q by 1
Q not empty
Set SS = 0
Q Empty
Q Queue
SS = 0 Server Status: Idle
Tr. Transaction
Departure Flow chartDeparture Flow chart
When a departure event happens, the following is checked at that point in time.
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Single server system simulation – 4 cases
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?
Example 5..Example 5..
Case Arrival distribution Service time distributionA Uniform (discrete) 1 to 11 min Fixed ( 5 minutes)
B Poisson: 10 arrivals/hour Exponential (Average time 5 min.)
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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).
Parameter value
0 Time
Expected Parameter value
t
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).
Example 5…Example 5…
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Example 5AExample 5A Arrival Service timeUniform (discrete) 1 to 11 min Fixed ( 5 minutes)
How did we get numbers in the table below?
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Example 5A..Example 5A.. Arrival Service timeUniform (discrete) 1 to 11 min Fixed ( 5 minutes)
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Example 5BExample 5B Arrival Service timePoisson: 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).
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Example 5D..Example 5D.. Arrival Service timePoisson: 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.
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Case Arrival distribution Service time distributionA Uniform (discrete) 1 to 11 min Fixed ( 5 minutes)D Poisson: 10 arrivals/hour Exponential (Average time 5 min.)
Example 5: ComparisonExample 5: Comparison
A B
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Simulation: general commentsSimulation: 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.
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Simulation Advantages Disadvantages 1. Flexibility.2. Can handle large and
complex systems.3. Can answer “what-if”
questions.4. Does not interfere with the
real system.5. Allows study of interaction
among variables.6. “Time compression” is
possible.7. Handles complications that
other methods can’t.
1. Can be expensive and time consuming.
2. Does not generate optimal solutions.
3. Managers must choose solutions they want to try (“what-if” scenarios).
4. Each model is unique.
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ATM Simulation 1 vs. 2ATM Simulation 1 vs. 2
Average number of arrival per hour: 40 (Poison)Average service time 75 sec. (Exponential)