Supplement to
Chapter 18
Simulation
Supplement Outline
Introduction and Simulation
Process, 2
Random Number and Random
Variate Generation, 4
Some Simulation Models, 13
Some Simulation Software and
Applications, 21
Key Terms, 31
Solved Problems, 31
Discussion and Review
Questions, 32
Internet Exercises, 33
Problems, 33
Mini-case: Valencia Kidney
Waiting Line, 38
Mini-case: Dofasco, 38
Learning Objectives
After completing this supplement,
you should be able to:
Explain what is meant by the
LO 1
term simulation, list some of
the reasons for simulation’s
popularity as a tool for
decision making, and
describe the steps in
simulation.
LO 2 Explain how random
numbers and random
variates are generated in
simulation.
LO 3 Model and solve typical
problems that require the use
of simulation—manually
and using Excel.
LO 4 Describe some simulation
software, be able to use
Arena for simple problems,
and describe some
applications of simulation.
2
PART EIGHT Waiting-Line Analysis
Introduction and Simulation Process
LO 1
simulation A descriptive
technique that enables a
decision maker to evaluate the
behaviour of a model under
various conditions.
Simulation, as used in decision making, is a descriptive technique in which a model of
a process is developed and then experiments are conducted on the model to evaluate its
behaviour under various conditions. Simulation is not an optimizing technique. It does not
produce a solution per se. Instead, simulation enables decision makers to test their solutions
on a model that reasonably duplicates a real process. Simulation models enable decision
makers to experiment with decision alternatives using what-if approach.
Simulation has applications across a broad spectrum of operations management problems. In some instances, the simulations models are quite modest, while others are rather
complex. Their usefulness in all cases depends on the degree to which decision makers
are able to successfully answer their what-if questions.
A list of operations management topics would reveal that many have simulation applications. For instance, simulation is often helpful in process design, facility layout, capacity
planning, line balancing, testing alternative inventory policies, scheduling, waiting lines,
and project management.
Generally, analysts use the simulation approach because the assumptions required by
an optimizing technique are not reasonably satisfied in a given situation. Waiting line
problems are a good example. Although waiting line problems are pervasive, the rather
restrictive assumptions of arrival and service distributions in many cases are simply not
met. Very often, analysts will then turn to simulation as a reasonable alternative for obtaining descriptive information about the system in question.
Other reasons for the popularity of simulation include:
1. Many situations are too complex to permit development of a mathematical solution.
2. Simulation models are fairly simple to use and understand.
3. Simulation enables the decision maker to conduct experiments on a model that will
help in understanding the process behaviour while avoiding the risks of conducting
tests in real life.
4. Extensive computer software packages make it relatively easy to model fairly sophisticated processes.
Simulation Process
Certain basic steps are used for all simulation models:
1. Define the manufacturing or service process and set simulation objective(s).
2. Develop the simulation model.
3. Test the model to be sure that it is working as intended (called model verification) and
reflects the system being studied (called model validation).
4. Develop one or more experiments (conditions under which the model’s behaviour
will be examined).
5. Run the simulation and evaluate the results.
6. Repeat steps 4 and 5 until you are satisfied with the results.
The first step in simulation is to clearly define the manufacturing or service process to
be modelled. A clear statement of the simulation objective(s) can provide not only guidance for model development but also the basis for evaluation of the success or failure of
a simulation study. In general, the goal is to determine how a process will behave under
certain conditions. The more specific a manager is about what he or she is looking for, the
better the chances that the simulation model will be designed to accomplish that. Toward
that end, the manager must decide on the scope and level of detail of the simulation. This
determines the necessary degree of complexity of the model and the information requirements of the study.
As part of system definition, the type of simulation should be determined: terminating
(or transient) and non-terminating (or steady state). A terminating state or event might
SUPPLEMENT TO CHAPTER 18 Simulation
be when a particular number of jobs have been completed. A terminating point in time
might be the closing of shop at the end of a business day. In a terminating simulation,
performance during successive time intervals during the simulation is measured. A nonterminating simulation means that the simulation could theoretically go on indefinitely with
no statistical change in behaviour of the system. The modeller must determine a suitable
length of time to run the model. An example of a non-terminating simulation is a model
of a manufacturing operation in which work always continues exactly as it left off at the
end of previous shift.
The next step is simulation model development. Typically, this involves deciding on the
structure of the model and using a computer to carry out the simulation. Data gathering is
a significant part of model development. The amount and type of data needed are a direct
function of the scope and level of detail of the simulation. The data are needed for both
model development and validation.
Once a model is developed, it must be verified, i.e., debugged to ensure that it works
correctly. This is much easier to do if a model is built in stages and with minimal detail.
To help debug the model, most simulation software provide a trace capability in the form
of audit trail, screen messages, or graphic animation. A “walk-through” of the model
input is always advisable. Model validation’s main purpose is to determine if the model
adequately depicts real process performance. An analyst usually accomplishes this by
comparing the results of simulation runs with known performance of the process under the
same circumstances. If such a comparison cannot be made because, for example, real-life
data are difficult or impossible to obtain, an alternative is to employ a test of reasonableness (called face validation), in which the judgments and opinions of individuals familiar
with the process or similar processes are relied on for confirmation that the results are
plausible and acceptable. Still another aspect of validation is careful consideration of the
assumptions of the model and the values of parameters used in testing the model. Again,
the judgments and opinions of those familiar with the real-life process and those who must
use the results are essential. Finally, note that model development and model verification/
validation go hand in hand: model deficiencies uncovered during verification/validation
prompt model revisions, which lead to the need for further verification/validation efforts
and perhaps further revisions.
The fourth step in simulation is developing experiments (conditions under which the
model’s behaviour will be examined). Experiments are the essence of a simulation; they
help answer the what-if questions posed in simulation studies. For example, should we use
one server or two servers, or use order quantity of 5 units, 6 or 7.
The fifth step is to run the simulation model. If a simulation model is deterministic and
all parameters are known and constant, only a single run will be needed for each what-if
question. But if the model is probabilistic, with parameters subject to random variability,
multiple runs will be needed to obtain confidence in the results. In this chapter supplement,
probabilistic simulation (also called the Monte Carlo method or simulation) is the focal
point of the discussion, and comments are limited to it.
Probabilistic simulation is essentially a form of random sampling, with each run representing one observation (for non-terminating simulation) and a sequence of observations
(for terminating simulation). Consequently, statistical theory can be used to determine
appropriate sample sizes. In effect, the larger the degree of variability inherent in simulation
results, the greater the number of simulation runs needed to achieve a reasonable level of
confidence in the results as true indicators of model behaviour. Another variance reduction
technique is to use common random numbers, i.e., use the same stream of random numbers in various experiments. For example, in queueing, if we are comparing two different
configurations of tellers in a bank, we would want the (random) time of arrival of the nth
customer to be generated using the same random number for both configurations.
In probabilistic simulation, each random component of the process under study has a
probability distribution. Random samples taken from probability distributions are analogous
to observations made on the process itself. Random sampling is accomplished by the use
of random numbers and random variates.
3
Monte Carlo method or
simulation Probabilistic
simulation technique, used
when a process has one or
more random component(s).
PART EIGHT Waiting-Line Analysis
4
Random Number and Random Variate
Generation
LO 2
random number An integer,
usually 2 digits, chosen
randomly from a range of
integers such as [00, 99]
random variate A random
value for a distribution such as
Poisson or Normal
Random number generation is the process of choosing a number in a given range where every
number in the range has equal probability of being picked. For example, a two-digit random
number in the range [00 to 99] is any of the 100 possible 2-digit numbers (note that 0 to 9
are considered as 00 to 09, i.e., two digits), and each has an equal chance of being picked.
In this chapter supplement, we will distinguish between random numbers that are integers,
usually 2 digits, from a range such as [00, 99], and random values (called random variates)
from distributions such as Normal and Poisson. The reason for this distinction is that we
will use both random numbers and variates below, with different roles. The objective is to
generate random variates, and we will use random numbers to do so when we are simulating
manually. Computer software use random numbers too but these are hidden from the user.
For example, in Excel’s Data Analysis module, there is the “Random Number Generation”
program that directly generates random variates from common probability distributions
such as Poisson and Normal. Note that Excel calls random variates random numbers. We
will first consider how we can obtain random numbers (e.g., two-digit integers).
Obtaining Random Numbers
The random numbers used in probabilistic simulation come from one of two sources: (a) a
table of random numbers like Table 18S-1, or (b) computer software such as Excel. In either
case, the method of generation is usually a recursive equation such as Zi = (a Zi-1 ) mod m,
i.e., the next random number Zi is the remainder after dividing a times the current random
number Zi-1 by m. This process results in random numbers between 0 and m – 1. The derived
random number can then be transformed to obtain a random number in any desired range,
e.g., [00 to 99]. The problem with the recursive equation method is that the resulting sequence
of numbers repeats itself after a certain number of numbers (called a cycle) are generated.
Therefore, the parameters a and m should be chosen carefully. In the ProModel simulation
software, a = 630,360,016 and m = 231 – 1, which results in a cycle of m. Because the resulting
numbers are not truly random, they are sometimes referred to as pseudo random numbers.
Fortunately, we don’t have to generate random numbers and will only use them. In
manual simulation we will use Table 18S-1 (or Table 18S-2 for Normal variates), and in
computer simulation we will bypass random numbers and use Excel to generate random
variates directly. If in Excel it is necessary to first generate random numbers, this can be
done by using the function @rand, which will generate a real number greater or equal to
0 and less than 1. To generate a random (integer) number between 00 and 99, we can use
the formula “= round(rand()*100-.5,0)”.
The numbers in Table 18S-1 have two digits for convenience, but the digits can be used singly,
in pairs, or in whatever grouping a given problem calls for. Each digit ranges from 0 to 9.
Table 18S-1
Random numbers
1
2
3
4
5
6
7
8
9
10
11
12
1
18
20
84
29
91
73
64
33
15
67
54
07
2
25
19
05
64
26
41
20
09
88
40
73
34
3
73
57
80
35
04
52
81
48
57
61
29
35
4
12
48
37
09
17
63
94
08
28
78
51
23
5
54
92
27
61
58
39
25
16
10
46
87
17
6
96
40
65
75
16
49
03
82
38
33
51
20
7
23
55
93
83
02
19
67
89
80
44
99
72
8
31
96
81
65
60
93
75
64
26
90
18
59
9
45
49
70
10
13
79
32
17
98
63
30
05
10
01
78
32
17
24
54
52
44
28
50
27
68
11
41
62
57
31
90
18
24
15
43
85
31
97
12
22
07
38
72
69
66
14
85
36
71
41
58
SUPPLEMENT TO CHAPTER 18 Simulation
5
For any size grouping of digits (e.g., two-digit numbers), every possible outcome (e.g.,
34, 89, 00) has the same probability of appearing. This implies that there are no discernible
patterns in sequences of numbers to enable one to predict numbers further in the sequence.
Therefore, the numbers can be read across rows and up or down columns.
When using the table, it is important to avoid always starting in the same spot. For our
purposes, the starting point will be specified so that everyone obtains the same results.
Generating Random Variates
To generate most random variates, the inverse transformation method is used:
1. Construct the Cumulative probability distribution F (from which a random variate is
desired).
2. Generate a random real number u greater or equal to 0 and less than 1.
3. Determine the value x such that F(x) = u.
Below, we will illustrate the inverse transformation method for some common probability
distributions. In each case we will first perform the random variate generation manually,
and then use Excel.
Generating Discrete Random Variates
The manager of a machine shop is concerned about machine breakdowns. He has made a
decision to simulate breakdowns for a 10-day period. Historical data on breakdowns over
the last 100 days are given in the following table:
Number of Breakdowns
Frequency
Example S-1
0
0............
10
1............
30
2............
25
2
3............
20
3
4............
10
4
5............
5
5
1
100
Simulate breakdowns for a 10-day period. Read two-digit random numbers from
Table 18S-1, starting at the top of column 1 and reading down.
a. Develop cumulative probabilities for breakdowns:
Solution
i. Convert frequencies into relative frequencies by dividing each frequency by the
sum of the frequencies. Thus, 10 becomes 10/100 = .10, 30 becomes 30/100 = .30,
and so on.
ii. Develop cumulative relative frequencies (i.e., cumulative probabilities) by ­successive
summing. The results are shown in the following table:
Number of
Breakdowns
Frequency
Relative
Frequency
Cumulative
Probability
0
10
.10
.10
1
30
.30
.40
2
25
.25
.65
3
20
.20
.85
4
10
.10
.95
5
5
.05
1.00
100
1.00
6
PART EIGHT Waiting-Line Analysis
b. Assign random-number intervals to correspond to the cumulative probabilities for breakdowns. (Note: Use two-digit numbers because the cumulative probabilities are given
to two decimal places.) You want a 10-percent probability of obtaining the event “0
breakdowns” in our simulation. Therefore, you must designate 10 percent of the possible
random numbers as corresponding to that event. There are 100 two-digit numbers, so
we can assign the 10 numbers 01 to 10 to that event.
Similarly, assign the numbers 11 to 40 to “one breakdown,” 41 to 65 to “two breakdowns,” 66 to 85 to “three breakdowns,” 86 to 95 to “4 breakdowns” and 96 to 00 to
“five breakdowns.”
Note that “00” is assigned as if it was “100,” not “0.” This makes the interval maximums correlate well to the cumulative Probabilities.
Number of
Breakdowns
Frequency
Relative
Frequency
Cumulative
Probability
Corresponding
Random Numbers
0
10
.10
.10
01 to 10
1
30
.30
.40
11 to 40
2
25
.25
.65
41 to 65
3
20
.20
.85
66 to 85
4
10
.10
.95
86 to 95
5
5
.05
1.00
96 to 00
100
1.00
c. Obtain the random numbers from Table 18S-1, column 1, as specified in the problem:
18 25 73 12 54 96 23 31 45 01
d. Convert the random numbers into numbers of breakdowns on each day, starting from
day 1:
18 falls in the interval 11 to 40 and corresponds, therefore, to one breakdown on
day 1.
25 falls in the interval 11 to 40 and corresponds to one breakdown on day 2.
73 corresponds to three breakdowns on day 3.
12 corresponds to one breakdown on day 4.
54 corresponds to two breakdowns on day 5.
96 corresponds to five breakdowns on day 6.
23 corresponds to one breakdown on day 7.
31 corresponds to one breakdown on day 8.
45 corresponds to two breakdowns on day 9.
01 corresponds to no breakdowns on day 10.
The following table summarizes these results:
Day
Random
Number
Simulated Number
of Breakdowns
1
18
1
2
25
1
3
73
3
4
12
1
5
54
2
6
96
5
7
23
1
8
31
1
9
45
2
10
01
0
17
SUPPLEMENT TO CHAPTER 18 Simulation
The mean number of breakdowns for this 10-day simulation is 17/10 = 1.7 breakdowns per
day. Compare this to the expected number of breakdowns based on the historical data:
0(.10) + 1(.30) + 2(.25) + 3(.20) + 4(.10) + 5(.05) = 2.05 per day.
The two (1.7 and 2.05) are close but because of random variability, not equal. As the length
of simulation increases, however, the sample mean should approach the population mean.
Using Excel to Generate Discrete Random Variates While the inverse transformtion method
can be used to generate discrete random variates in Excel using rand() and VLookup functions, there is a more direct way. Excel’s Data Analysis module contains the Random Number
Generation program that can directly generate Discrete random variates, as well as some other
variates presented below. The Data Analysis module should be in the top right-hand corner
of the Data folder (see the screenshot below). If it is not, you can add it in Excel 2010 as follows:, click File, then click Options, next click Add-ins, then click Analysis Toolpak on the
top, next click Go in the bottom, next tick the “Analysis Toolpak,” and finally click OK.
To use the Random Number Generation program in Excel, first we need to enter the
data (the numbers of breakdowns and the probability of each value) in the worksheet. Then
we click on Data Analysis; in the small window that opens we find and click on Random
Number Generation, and finally click OK. If we want 10 Discrete random variates from
the breakdown distribution (in cells A4 to B9), arranged in column D starting with cell
D1, we fill the Random Number Generation window as follows:
7
8
PART EIGHT Waiting-Line Analysis
Finally, clicking OK will result in the following:
The problem with using Random Number Generation program is that it does not recalculate the numbers automatically when the recalculate key F9 is pushed. The recalculation
is useful when we want to quickly replicate the simulation. Also, it is useful when Data
Tables (explained later) are used to rerun the simulation for various values of an input
variable. Therefore, we need to briefly look at the generation of a Discrete random variate
using rand() and Vlookup.
First we need to enter the values and the probabilities, and compute the cumulative
probabilities:
In the above spreadsheet, the formula in cell C4 is “= B4,” the formula in cell C5
is “= B5 + C4,” and the formula in cell C6 is “= B6 + C5,” etc. Now, we will add the
Minimum of Interval in column D as follows: In cell D4 we enter 0, in cell D5 we
enter “= C4,” and copy it down. In effect, we are just shifting the cumul. Prob. values
one row down:
SUPPLEMENT TO CHAPTER 18 Simulation
9
Now, we just copy column A to column E:
Suppose we wish to generate 10 random values from the breakdown distribution and put
them in cells G1 to G10. All we have to do is to enter “= VLOOKUP(RAND(),D$4:E$9,2)”
in cell G1 and copy it down. What this formula does is that first rand() will generate a real
number greater or equal to 0 and less than 1, then VLookup will take the random number to the
range D4 to E9, and will find the Interval where the random number fits (in the left column of
the range) and will look up the corresponding value in the right column of the range; 2 in the
VLOOKUP formula means that it should look up the second column in the given range:
Generating Poisson Random Variates Generation of a Poisson random variate requires
the mean of the distribution. Given the mean, we can obtain the cumulative probabilities
for the Poisson distribution from Appendix B, Table C. Then, the inverse transformation
method can be used as in the Discrete case above. However, we should obtain three-digit
random numbers from Table 18S-1 because the Poisson cumulative probabilities are given
in three digits. Example S-2 illustrates this.
The average number of lost-time accidents at a large plant has been determined from historical records to be two per day. Moreover, it has been determined that this accident rate can
be well approximated by a Poisson distribution. Simulate five days of accident experience
for the plant. Read random numbers from columns 1 and 2 of Table 18S-1.
First obtain the cumulative Poisson probability distribution from Appendix B, Table C for
a mean of 2.0, and determine the intervals:
x
Cumulative
Probability
Random Number
Intervals
001 to 135
0............
.135
1............
.406
136 to 406
2............
.677
407 to 677
3............
.857
678 to 857
4............
.947
858 to 947
5............
.983
948 to 983
6............
.995
984 to 995
7............
.999
996 to 999
8............
1.000
000
Example S-2
Solution
10
PART EIGHT Waiting-Line Analysis
Next obtain three-digit random numbers from Table 18S-1. Reading from column 1 and
2 as instructed, you find 182, 251, 735, 124, and 549.
Finally, convert the above random numbers into number of lost-time accidents using the
established set of intervals above. Because 182 falls in the second interval, it corresponds
to one accident on day 1. The second random number (251) falls in the same interval,
indicating one accident on day 2. The number 735 falls between 678 and 857, which corresponds to three accidents on day 3; 124 corresponds to 0 accidents on day 4; and 549
corresponds to two accidents on day 5.
Using Excel to Generate Poisson Random Variates In the Random Number Generation win-
dow, choose Poisson as the Distribution and enter the mean (2) for Lambda. For example,
the following will generate 5 Poisson random variates with mean = 2 in cells A1 to A5:
Generating Normal Random Variates There are a number of ways to generate a Normal
random variate of a given mean and standard deviation, but perhaps the simplest is to
use Table 18S-2, a table of standard Normal random variates, i.e., a Normal distribution
with mean 0 and standard deviation 1.00. Like all such tables, the numbers are arranged
randomly, so that when they are read in any sequence they exhibit randomness. Numbers
obtained from Table 18S-2 can be converted to the desired variates by multiplying them
by the standard deviation and adding this amount to the mean. That is:
Normal
Simulated value = Mean + Standard
random number × Standard deviation
(18S-1)
In effect, the standard Normal random variate is a z value, which indicates how far a
­particular value is above or below the distribution mean.
Table 18S-2
Standard Normal random
variates
1
2
3
4
5
6
7
8
9
10
1
1.46
−0.09
−0.59
0.19
−0.52
−1.82
0.53
−1.12
1.36
−0.44
2
−1.05
0.56
−0.67
−0.16
1.39
−1.21
0.45
−0.62
−0.95
0.27
3
0.15
−0.02
0.41
−0.09
−0.61
−0.18
−0.63
−1.20
0.27
−0.50
4
0.81
1.87
0.51
0.33
−0.32
1.19
2.18
−2.17
1.10
0.70
5
0.74
−0.44
1.53
−1.76
0.01
0.47
0.07
0.22
−0.59
−1.03
6
−0.39
0.35
−0.37
−0.52
−1.14
0.27
−1.78
0.43
1.15
−0.31
7
0.45
0.23
0.26
−0.31
−0.19
−0.03
−0.92
0.38
−0.04
0.16
8
2.40
0.38
−0.15
−1.04
−0.76
1.12
−0.37
−0.71
−1.11
0.25
0.59
−0.70
−0.04
0.12
1.60
0.34
−0.05
−0.26
0.41
0.80
−0.06
0.83
−1.60
−0.28
0.28
−0.15
0.73
−0.13
−0.75
−1.49
9
10
SUPPLEMENT TO CHAPTER 18 Simulation
The time required to perform a certain task can be described by a Normal distribution that
has a mean of 30 minutes and a standard deviation of 4 minutes. Simulate times for three
jobs using the first three values in column 1 of Table 18S-2.
The first three values are: 1.46, −1.05, and 0.15. The simulated values are:
For 1.46:
For −1.05:
For 0.15:
11
Example S-3
Solution
30 + 1.46(4) = 35.84 minutes.
30 − 1.05(4) = 25.80 minutes.
30 + 0.15(4) = 30.60 minutes.
Using Excel to Generate Normal Random Variates In the Random Number Generation
window, choose Normal as the Distribution and enter its mean and standard deviation.
For example, the following will generate 3 Normal random variates with mean = 30 and
standard deviation = 4 in cells A1 to A3:
Alternatively, we can use the formula “= NORMINV(rand(),mean,stddev).” For
example, to generate a random value from the above Normal distribution in a cell, we
enter “= NORMINV(rand(),30,4)” in that cell.
Generating Continuous Uniform Random Variates The Continuous Uniform distribution
U[a, b] has equally probable values anywhere between a and b, a < b.
Simulated value = a + (b − a)(random real number u greater or
equal to 0 and less than 1)
(18S-2)
We can convert the random numbers of Table 18S-1 to u by simply placing a decimal
point to the left of the number. For example, 77 becomes 0.77.
Example S-4 illustrates the generation of Continuous Uniform variates.
Job times vary uniformly between 10 and 15 minutes. Use Table 18S-1 to simulate job
times for four jobs. Read numbers from column 9, going down.
Example S-4
12
PART EIGHT Waiting-Line Analysis
Solution
a = 10 minutes, b = 15 minutes, b − a = 5 minutes
a. Obtain the random numbers: 15, 88, 57, and 28.
b. Convert to Continuous Uniform variates:
Simulated Value
(minutes)
Random Number
Computation
15 . . . . . . . . . . . .
10 + 5(.15)
=
10.75
88 . . . . . . . . . . . .
10 + 5(.88)
=
14.40
57 . . . . . . . . . . . .
10 + 5(.57)
=
12.85
28 . . . . . . . . . . . .
10 + 5(.28)
=
11.40
Using Excel to Generate Continuous Uniform Variates In the Random Number Generation
program, choose Uniform as the Distribution and enter the extreme values a and b after
“Between” and “and,” respectively. For example, the following will generate four Continuous Uniform random variates between 10 and 15 in cells A1 to A4:
Alternatively, we can use the formula “= a + rand()*(b - a).” For example, to generate a random value from the above Continuous Uniform distribution in a cell, we
enter “= 10 + rand()*5” in that cell.
Generating Exponential Random Variates An Exponential distribution is portrayed in
Figure 18S-1. The probability is fairly high that the random variable will assume a value
close to zero, and it decreases as the value of the random variable increases. The probability that the Exponential random variable will take on a value greater than some specified
value T, given mean of 1/λ, is:
P(t > T) = e−λT
FIGURE 18S-1
(18S-3)
f(t)
An Exponential distribution
P(t > T) = u
0
T
t
SUPPLEMENT TO CHAPTER 18 Simulation
13
To simulate an Exponential value manually, we obtain a real random number u between
0 and 1, set this equal to the probability P(t > T ), and solve Formula 18S-3 for T. The result
is a random variate from the Exponential distribution with mean of 1/λ. This concept is
illustrated in Figure 18S-1.
We can obtain an expression for T by taking the natural logarithm of both sides of the
equation. Thus, with P(t > T ) = u, we have
ln(u) = ln(e−λT)
The natural logarithm of a power of e is equal to the power itself, so
ln(u) = − λT
Then
T =−
1
ln(u )
λ
(18S-4)
As for the Continuous Uniform case above, a random number can be obtained from
Table 18S-1 and converted to u simply by placing a decimal point to the left of it. This is
demonstrated in the following example.
Times between breakdowns of a piece of equipment can be described by an Exponential
distribution with mean of five hours. Simulate the time between two pairs of breakdowns.
Read two-digit random numbers from column 3 of Table 18S-1.
The mean, 1/λ, is 5 hours. The random numbers are 84 and 05. Using formula 18S-4, the
simulated times are:
Example S-5
Solution
For 84: T = –5[ln(.84)] = −5[−0.1744] = 0.872 hours.
For 05: T = –5[ln(.05)] = −5[−2.9957] = 14.979 hours.
Note that the smaller the value of the random number, the larger the simulated value of T.
Using Excel to Generate Exponential Random Variates In Excel use the formula
“= −5*ln(rand()).”
Some Simulation Models
LO 3
Real problems involve many components or steps, with process flows that may be complicated. In some cases, it is helpful to construct a process flowchart (similar to a process flow
diagram). If the model is complicated, one needs to use a specialized simulation software
such as Arena. Otherwise, Excel can be used. Below, we illustrate some simple models of
operations decision making using Excel.
An Inventory Control Model
While we solved most inventory control problems analytically in Chapter 12, we made
some simplifying assumptions. For example, we calculated EOQ and ROP separately. If
a more exact solution is required, simulation should be used. Simulation is also used to
illustrate the behaviour of the system over time so that the decision maker can gain insight/
confidence in it.
The manager of a truck dealership wants to acquire some insight into how a proposed
policy for reordering trucks from the manufacturer might result in shortage. Under the
new policy, two trucks are to be ordered whenever the number of trucks on hand at the
Example S-6
14
PART EIGHT Waiting-Line Analysis
end of the day plus number of trucks on order is two or fewer. Assume that purchase lead
time is only two full days. According to the dealer’s records, the probability distribution
for daily demand (i.e., its sales) is:
Solution
Demand, x
P(x)
0.........
.50
1.........
.40
2.........
.10
a. Draw a flowchart that describes a 10-day simulation. Assume a beginning inventory of
four trucks and that all shortage will be back-ordered.
b. Manually simulate the inventory system for 10 days. Use two-digit random numbers from
Table 18S-1, column 11, reading down. Estimate the probability of shortage.
c. Enter the model for this problem in Excel and run the simulation for 1,000 days. Estimate the probability of shortage.
a.
Generate
demand, x
Update
inventory
I=I+2
No
Update inventory
I=I–x
Is
arrival
day = Day?
Yes
I + on order _< 2?
I = Amount of
End-of-Day
inventory
Yes
Reorder 2
No
Start
I=4
Day = 0
No
Update day
Day = Day + 1
Day = 10?
Yes
Stop
b. Specify random number intervals for demand:
Cumulative
x
P(x)
0
.50
P(x)
.50
Ranges
01–50
1
.40
.90
51–90
2
.10
1.00
91–00
2. Obtain random numbers, convert to demand, update inventory accordingly, and reorder
when necessary:
Day
Beginning
Inventory
Random
Number
Demand,
x
1........
2........
Ending Inventory + on order
4
54
1
3
3
73
1
2 (reorder 2; arrival day = 5)
3........
2
29
0
2 + 2
4........
2
51
1
1 + 2
5........
3 (Day 2 order arrived)
87
1
2 (reorder 2; arrival day = 8)
6........
2
51
1
1 + 2
7........
1
99
2
–1 + 2 (reorder 2; arrival day = 10)
8........
1 (Day 5 order arrived)
18
0
1 + 2
9........
1
30
0
1 + 2
10 . . . . . . .
3 (Day 7 order arrived)
27
0
3
SUPPLEMENT TO CHAPTER 18 Simulation
The only negative Ending Inventory is in Day 7. Therefore, estimate for probability
of shortage = 1/10 = .10. However, this is probably not very accurate. If we simulate
another 10 days, we may get a different answer, e.g., no shortage. Therefore, we must
simulate longer (e.g., 1,000 days) and possibly run many replications (e.g., 10) and
average their results.
c. The top part of the worksheet is shown below. The reorder decision components have
been explicitly defined here. Therefore, there are separate columns for “Quantity On
Order,” “Order?,” and “Day of Arrival.” On the other hand, we have used the Random
Number Generator program of Excel to generate the demand in Column D (so we did
not need to generate random numbers first).
The formula for “Units Recei’d” in cell B17 is “= IF(A17 = H14,$C$9,0),” i.e., if
the day number equals the “Day of Arrival” two full days before, then a shipment is
received. In this case, H14 will be evaluated as 0 by Excel. The formula for “Beginning Stock Level” in cell C17 is “= B17 + E16,” i.e., it equals “Units Recei’d” plus
“Ending Stock” the previous day. The formula for “Ending Stock” in cell E17 is
“= C17 – D17,” i.e., the “beginning Stock Level” minus Demand. The formula for
“Quantity On Order” in cell F17 is “= IF(G16 = “Yes”,F16 + C$9 – B17,F16 –
B17),” i.e., if Order? on previous day is “Yes”, then Quantity On Order = Quantity
On Order the previous day + order quantity – any amount received today; else Quantity On Order = Quantity On Order the previous day – any amount received today.
The formula for “Order?” in cell G17 is “= IF(E17 + F17 > $C$10, “No”, “Yes”),”
i.e., if “Ending Stock” plus “Quantity On Order” are larger than Reorder Point, then
“Order?” = “No”, else “Order?” = “Yes”. The formula for “Day of Arrival” in cell
H17 is “= IF(G17 = “Yes”,A17 + C$11 + 1,“ ”),” i.e., if “Order?” is “Yes” then “Day
of Arrival” = today’s “Day No.” + lead time + 1, else it is blank. The “+1” is because
the lead time is 2 full days and we are ordering in the end of the current day and will
receive it the morning of day of arrival.
Finally, the formula in cell E12 is “= COUNTIF(E17:E1016, “<0”)” which counts
the negative “Ending Stocks” for 1,000 days. In this case, there are 47 days when shortage occurred. Therefore, our estimate for probability of shortage is 47/1,000 = 4.7%.
Note that even a run of 1,000 days is not long enough if an estimate within ±1% is
required. This is because when the simulation is replicated, the number of short days
varies between 30 and 60 days. To increase accuracy, we should increase the run length
and the number of replications.
15
16
PART EIGHT Waiting-Line Analysis
A Waiting Line Model
While we presented many queueing models in Chapter 18, we made some simplifying
assumptions. For example, for most models we assumed that the distributions of interarrival and service times are Exponential, and that only steady state results are desired.
Also, some performance measures are difficult to obtain analytically, e.g., maximum wait
time and maximum number of customers in the queue. In these cases, simulation should
be used. Simulation will also illustrate the behaviour of the system over time so that the
decision maker can gain insight/confidence in it.
Example S-7
Solution
The time between mechanics’ requests for tools in a large plant is Normally distributed
with a mean of 10 minutes and a standard deviation of 1 minute. The time to fill requests
is also Normal with a mean of 9 minutes per request and a standard deviation of 1 minute.
Mechanics’ waiting time represents a cost of $2 per minute, and servers’ time represents
a cost of $1 per minute.
1. Manually simulate nine mechanic requests and their service times, and determine the
mechanics’ waiting time, assuming one server. Would it be economical to add another
server? Explain. Use Table 18S-2, column 8 for requests and column 9 for service.
2. Enter the model for this problem in Excel and simulate 1,000 arrivals. Does your answer
to part a change?
1. i. Obtain standard Normal random variates from Table 18S-2 and convert to times
[see columns (a) and (b) in the following table for requests for tool (i.e., arrivals)
and columns (f) and (g) for service]. For example, the first Time Between Arrivals = 10 - 1.12(1) = 8.88 and the first Service Time = 9 + 1.36(1) = 10.36.
CUSTOMER ARRIVALS
(a)
Random
Number
(b)
(c)
Time
Between
Arrivals
Arrival
Time
SERVICE
(d)
(e – c) Service
Wait
Start
Time
Time
(f)
(g)
(h)
Random
Number
Service
Time
(e + g)
Service
End Time
−1.12
8.88
8.88
.00
8.88
1.36
10.36
19.24
−.62
9.38
18.26
.98
19.24
−.95
8.05
27.29
−1.20
8.80
27.06
.23
27.29
.27
9.27
36.56
−2.17
7.83
34.89
1.67
36.56
1.10
10.10
46.66
.22
10.22
45.11
1.55
46.66
−.59
8.41
55.07
.43
10.43
55.54
.00
55.54
1.15
10.15
65.69
.38
10.38
65.92
.00
65.92
−.04
8.96
74.88
−.71
9.29
75.21
.00
75.21
−1.11
7.89
83.10
−.26
9.74
84.95
.00
84.95
.41
9.41
94.36
4.43
(e)
82.60
ii. Determine Arrival Times [column (c)] by adding to previous Arrival Time to the
Time Between Arrivals in column (b).
iii. Use Arrival Times for Service Start Times unless service is still in progress
on a previous request. In that case, determine how long the arrival must wait
(e − c). Column (h) values are the sum of Service Start Time and Service Time
[column (g)].
iv. Total waiting time is 4.43 minutes (see the table).
SUPPLEMENT TO CHAPTER 18 Simulation
v. The total cost is:
Waiting cost (including service time):
(4.43 + 82.60) minutes at $2 per minute = $174.06
Server cost:
94.36 minutes at $1 per minutes = 94.36
$268.42
vi. Usually, a second simulation with two servers would be needed (e.g., for 18 arrivals). However, in this case it is apparent that a second server would increase server
cost by about $94 but could not eliminate more than approximately $8.86 of waiting
cost. Hence, the second server would not be justified.
2. The top part of the worksheet is shown below. The column headings are identical to the
manual simulation table above, except for the “Mechanic No.” column (to keep track
of numbers of arrivals). Also, we have used the Random Number Generator program of
Excel for “Time Between Arrivals” and “Service Time,” so we did not need to generate
random numbers first.
The formula for “Arrival Time” in cell C9 is “= C8 + B9”, i.e., the previous Arrival
Time plus the current “Time Between Arrivals.” The formula for “Wait Time” in cell
D9 is “= MAX(G8,C9)-C9”, i.e., the maximum of “Service End Time” of the previous
mechanic and the “Arrival Time” of current mechanic minus the “Arrival Time” of
current mechanic. The formula for “Service Start Time” in cell E9 is “= C9 + D9”, i.e.,
the “Arrival time” plus the “Wait Time.” The formula for “Service End Time” in cell G9
is “= E9 + F9”, i.e., “Service Start Time” plus “Service Time.” The formula in cell D3 is
“= SUM(D9:D1008)”, i.e., sum of the “Wait Time” of 1,000 mechanics, and the formula
in cell F3 is “= SUM(F9:F1008)”, i.e., the sum of “Service Time” of 1,000 mechanics.
The answer to the question of the need for the second server does not change because
the average wait time is almost the same as in part 1: 401/1000 = .401 minutes per
mechanic ≈ 4.43/9 = .492 minutes per mechanic from part 1.
It is interesting to compare the maximum wait times of part 1 and 2. In part 1, the maximum wait time is 1.67 minutes, whereas in part 2 the maximum wait time is approximately
5 minutes (not shown). This is expected because a longer simulation run allows for more
extreme cases to occur.
17
18
PART EIGHT Waiting-Line Analysis
“Optimization” in Simulation
Simulation is usually used to compare two or more configurations of the system, based
on some measure of performance. Choosing the best configuration is sometimes called
optimization. We have to run the simulation for each configuration, and choose the configuration with the best measure. However, to make the right decision, the simulation results
should be accurate enough. For this reason, we need to have long runs, and therefore need
to use the computer. Excel has a program called Data Table that will run the simulation
for different values of a decision variable. In the following example, we will use Excel
and Data Table to determine the “optimal” order quantity of the single period inventory
control model of Chapter 12.
Example S-8
Solution
Suppose you are responsible for ordering soft drinks for an event such as a sports game.1 One
canister can serve 100 drinks. Demand is Exponentially distributed with mean of five canisters. Suppose that unmet demand costs $40 per canister (or $.40 per drink) for lost profit,
and returning excess soft drink to the bottler costs $10 per canister (or $.10 per drink).
Use Excel to determine the order quantity that will minimize the expected total cost. Run
2,000 replications of the event for each order quantity between 5 and 10.
In the following worksheet, the formula for Demand in cell B13 is “= -C$6*LN(RAND()),”
i.e., negative of the mean of the Exponential distribution times the natural logarithm of a
(real) random number greater or equal to 0 and less than 1. The formula for Shortage Cost
in cell C13 is “= IF(B13 > C$9,B13 – C$9,0)*C$3”, i.e., if Demand is larger than Order
Quantity, shortage = Demand – Order Quantity, else shortage = 0, and multiply the shortage
by the unit shortage cost. The formula for Excess Cost in cell D13 is = IF(C$9 > B13,C$9 –
B13,0)*C$4”, i.e., if Order Quantity is larger than Demand, excess = Order Quantity –
Demand, else excess = 0, and multiply the excess by the unit excess cost. The formula for
Total Cost in cell E13 is “= C13 + D13,” i.e., the sum of the Shortage and Excess Costs.
The formula for Avg. Total Cost in cell F2 is “= AVERAGE(E13:E2012),” i.e., the average
of Total Cost of 2,000 games.
The values in cells F3 to F8 are the Avg. Total Cost for Order Quantities 5 to 10,
respectively. These are determined by Excel’s Data Table program. Data Table was used as
follows: select cells E2 to F8, then click on Data folder, next click What-if Analysis, then
click Data Table; in the small window enter $C$9 (the location of the Order Quantity) in
front of Column input cell:, and click OK:
It can be seen from cells F2 to F8 that Order Quantity = 8 canisters has the lowest Avg.
Total Cost. However, even 2,000 replications may not be enough for an accurate decision.
If the whole experiment is repeated (press the F9 key to recalculate), in some instances
other Order Quantities, in particular 7 and 9, will have a lower Avg. Total Cost than 8. To
get a more accurate result, one should increase the number of games.
1
A. F. Seila et al, Applied Simulation Modeling, Australia: Thomson Learning, 2003, p. 49.
SUPPLEMENT TO CHAPTER 18 Simulation
19
A Maintenance Model
While we solved some maintenance problems analytically in the Supplement to Chapter 15,
we made some simplifying assumptions. For example, we calculated the preventive maintenance interval approximately, and we assumed that individual units will not fail again
before group replacement time. If a more exact solution is required, simulation should be
used. Simulation is also used to illustrate the behaviour of the system over time so that the
decision maker can gain insight/confidence in it.
A milling machine has two different bearings that fail.2 The distribution of the life of
each bearing is identical and is shown below. When a bearing fails, the machine stops,
a mechanic is called, and he installs a new bearing. Each bearing costs $32. It takes an
average of 27 minutes to change the bearing. Downtime for the machine costs $10 per
minute. The engineering staff has proposed a new policy: replace both bearings whenever one fails. This will take the mechanic an average of 37 minutes. Simulate 1,000
failure/replacement cycles for each bearing for the current and for both bearings for the
proposed replacement policy and determine the lower-cost policy.
2
Based on J. Banks et al, Discrete-event System Simulation, 5th ed, 2010, New Jersey: Prentice-Hall,
pp. 65–67.
Example S-9
20
PART EIGHT Waiting-Line Analysis
Bearing Life (Hours)
Solution
Prob
1,000
.08
1,100
.15
1,200
.26
1,300
.19
1,400
.13
1,500
.09
1,600
.06
1,700
.03
1,800
.01
The individual replacement policy is modelled in Excel as shown in the worksheet below.
Each cycle is the length of life of a new bearing. These are created using the Discrete
Distribution of the Random Number Generation program of Excel. The formula in cell
H13 is “= C4*2000”, i.e., the cost of replacing 2,000 bearings. The formula in cell H14 is
“= C5*2,000,” i.e., the cost of downtime of 2,000 individual failures The formula in cell
H15 is “=H13 + H14”. The formula in cell H16 is “= SUM(B13:B1012),” i.e., the total
lives of Bearing 1s, and the formula in cell H17 is “= SUM(C13:C1012),” i.e., the total
lives of Bearing 2s. The formula in cell H18 is “=H16+H17”. Finally, the formula in cell
H19 is “= H15/H18*1000,” i.e., the total cost divided by total life hours times 1,000, or
expected cost per 1,000 hours of work.
The group replacement policy is modelled in Excel as shown in the worksheet below.
It is very similar to the individual replacement policy worksheet above, but for each cycle
the minimum length of life of the two bearings (1 and 2) is computed (called First Failure)
and used for both bearings, because both are replaced at that time. Therefore, the formula
in cell D13 is “= MIN(B13:C13)”. The formula in cell I14 is “= C5*1000”, i.e., the total
downtime cost of joint replacement (note only 1,000 downtimes). The formulas in cells
SUPPLEMENT TO CHAPTER 18 Simulation
I16 and I17 are = SUM(D13:D1012),” i.e., the total lives of the first failures. The other
formulas are the same as the individual policy.
Because the cost per 1,000 hours $184.02 < $234.69, group replacement will be cheaper.
We ran each policy many times to make sure that the above results are not flukes.
Some Simulation Software and
Applications
LO 4
There are many simulation software available such as SIMSCRIPT III, GPSS/H, Extend,
Witness, Micro Saint Sharp, SimProcess, ProModel, Arena, and @RISK. These have
some features in common. They provide model development, random variate generation,
collection and tabulation of simulation results, time keeping, and animation. For a list
of simulation software, see http://lionhrtpub.com/orms/surveys/Simulation/Simulation1.html.
The following briefly describes some applications of some simulation software.
SIMPROCESS®. SIMPROCESS® is a CACI Software Company product. We will
briefly describe the use of SIMPROCESS® to model the operations of an emergency department. The objectives were to determine the right level of staffing (administrative clerks,
technicians, nurses, doctors) and other resources (emergency rooms, wheelchairs, etc.),
and work flow design, given the expected number and mix of patients arriving. Patients are
triaged: Level 1 patients are serious cases, whereas levels 2 and 3 can wait. Registration is
performed before or after treatment depending on the state of the patient.
Modelling involves defining patients as “entities” that are “generated” in the “entrance
door” and “ambulance” according to historical data. Then, they go through a series of operations (as modelled below). The time spent in an operation is generally a random variable
that is entered in the “Delay Properties” of the operation (by clicking on the icon). Also,
21
PART EIGHT Waiting-Line Analysis
22
the resources used in each operation are entered (these affect the times). The analyst needs
to specify the level of resources to use and run the simulation for a certain length of time
to see its effects on entities (e.g., percentage of time spent in an operation) and resources
(e.g., percentage of time idle, costs). Important data can also be represented graphically
(called dashboard monitors). The simulation is run several times for each different level
of resources and the best level of resources is chosen. For more details, see http://www.
simprocess.net/solutions/er_model.html.
Emergency room
Not Level 1
Level 2 or 3
Registration
Triage
Sign In
Level 1
Entrance Door
Ambulance
Not Level-1
Transfer
to room
Level 1
Always Level 1
Level-1
Release
and
Admission
Treatment
Source: http://www.simprocess.net/solutions/models_html/EmergencyRoom.
Extend. Extend is a product of Imagine That! Software company. Modelling in Extend
is similar to SIMPROCESS®. Just click on the library menu and select a type of operation (Generator, Queue Activity, Plotter, etc.) to include in the model. Then, connect it
to another operation by left-clicking in the middle of the right edge of the icon and dragging the line to the input activity. For further information, see http://www.extendsim.
com/downloads/manuals/support_manuals_dl.html. An Extend model for bank staffing
is shown below.
Source: Bank Line Tutorial in Extend Demo Software, downloaded from http://www.extendsim.com/prods_
demo.html. Screen capture courtesy of Imagine That Inc. of San Jose, CA.
SUPPLEMENT TO CHAPTER 18 Simulation
STA.1
LOADING
STATION
STA. 2
STA. 11
STA. 12
STA.39
OFF-LINE
WASH
STA.36
UNLOADING
STATION
23
STA. 16
REPAIR BAY #1
REPAIR BAY #3
STA. 17
STA. 16
STA. 19
REPAIR BAY #2
STA. 20
STA. 37
STA. 36
STA. 31
STA. 26
STA. 25
STA. 24
STA. 22
STA. 21
FIGURE 18S-2
WITNESS. WITNESS is a product of the Lanner Group, based in the UK (http://www.
lanner.com). The following describes an application of WITNESS.Visteon is a subsidiary
of Ford Motor Company. In the mid-1990s, the Sterling Heights plant (in Michigan) was
the only source of Ford’s axles. With the introduction of new cars, trucks, and SUVs, Ford
needed to substantially increase production of the front axle line (see Figure 18S-2).
The front-axle line used a closed-loop conveyor. At Station 1 (the loading station), parts
and kits are loaded onto pallets. Pallets progress through subsequent stations to reach the
unloading station, where the fully assembled axles are unloaded from the pallets. The finished axle assemblies leave the system while the empty pallets are moved to the loading
station to receive new parts.
Visteon formed a project team that used WITNESS to model and simulate the production
of the front axle line. Most stations used a machine with varying operation times (a range
of 8 seconds to 33 seconds per axle). In addition, there were machine failures (a range of
11 minutes to 11,000 minutes for mean time between failures) and repair (a range of 1
minute to 5 minutes for mean time to repair). Also, a small percentage of axles had to be
torn down for repair at various stages during the process.
Although there were more pallets circulating in the front axle line than stations, too
few pallets resulted in stations “starving,” but too many pallets resulted in “blocking.”
So, the team wanted to identify the optimal number of pallets. Also, the team wanted to
know the bottleneck station and if a given station-time reduction would help increase the
line capacity. As a bottleneck was relieved and capacity of the line increased, another station would become the bottleneck. After each improvement, simulation was run again to
investigate the best way to increase the capacity next. At the start of the study, capacity of
the line was 57 axles per hour. By the end of project 1.5 years later, capacity was increased
to 75 axles per hour (a 30 percent increase).
Simulation was also used to show that the line capacity could not easily be increased much
more. Therefore, Ford decided to set up another front axle line to meet the excess demand.
ProModel. ProModel is a product of the ProModel corporation of Utah (http://www.
promodel.com). The following is a simple application of ProModel. Gasification converts
coal into carbon monoxide and hydrogen, which can then be converted into chemicals or
put in other uses. Eastman Chemicals used ProModel to determine how many gasifiers
to use in a gasification plant in order to produce the required output.3 It is also possible
to use a gasifier as standby. For example, when 4 gasifiers are used, if the reliability
3
http://www.promodel.com/solutions/manufacturing/Eastman-Chemical-web.pdf.
Visteon’s front axle line
Source: G. Pfeil, et al.,
“Visteon’s Sterling Plant
Uses Simulation-based
Decision Support in
Training, Operations, and
Planning,” Interfaces 30(1),
January–February 2000,
pp. 115–133.
24
PART EIGHT Waiting-Line Analysis
(­ percentage uptime) of each gasifier is 90 percent, then the output will be approximately
92 percent of planned output, whereas if there is an additional standby gasifier, the output
will be about 98 percent of planned output.
Source: http://www.promodel.com/solutions/manufacturing/Eastman-Chemical-web.pdf.
Arena. Arena is a product of Rockwell Automation (http://www.arenasimulation.com).
An application is as follows. Dofasco used Arena to model its primary steelmaking process.
The steps of this production line are (1) iron making and de-sulfurizing, (2) steelmaking, (3)
ladle metallurgy facility, (4) vacuum degassing (if pure steel is needed), and (5) continuous
casting (see the following figure, from left to right). Dofasco managers wanted to know if
relatively small expenditures in some critical parts of the process would result in productivity improvements, or whether major expenditures would be needed. After construction
of the model in Arena, which used 87 different steps/activities, the model was validated
against previous year’s actual performance (e.g., the quantity of slabs made, number of
ladles fed into the continuous caster, etc.). The results showed that only simultaneously
increasing the uptimes of major steps of the process would increase productivity. In other
words, there were no unique bottlenecks. Later, the model was also used to evaluate the
major changes to the process.
oxygen
Off Gas
oxygen
Source: http://www.informs-sim.org/wsc06papers/255.pdf
SUPPLEMENT TO CHAPTER 18 Simulation
How to Use Arena
Download a free trial version of Arena from http://www.arenasimulation.com (you have to
register first). Unzip the file and install it. Arena is relatively simple to use. After starting
Arena, the initial screen looks like the following:
The trial version has access to only Basic Processes (see the left pane). We will model a
single server queueing model, Example 2 of Chapter 18, in Arena. The time between arrivals and service times both have Exponential distributions with mean of 4 and 3 minutes,
respectively. We wish to determine the average and maximum wait time (in queue), and
average and maximum numbers in the queue. Click on Create icon in the left pane and
drag it to the main window:
25
26
PART EIGHT Waiting-Line Analysis
This is the arrival process. In the bottom pane, we need to change the Value into 4
and the Units into minutes. Note that the distribution of arrivals is already Expo, i.e.,
Exponential:
Now, with Create 1 selected we click on Process icon in the left pane and drag it to the
main pane and drop it to the right of Create 1:
SUPPLEMENT TO CHAPTER 18 Simulation
In the bottom pane for Process 2, we need to change the Action into Seize Delay Release
(or else no queue will be developed), Delay Type into Expression (because Exponential is
not one of the distributions we can choose directly here), Units into Minutes, and Expression into EXPO(3) (we need to type in the 3):
Now double-click on Process 2 because we need to assign a resource to it (to seize
the arrivals):
In the Process window above, click Add. . .:
27
28
PART EIGHT Waiting-Line Analysis
In the Resources window above, click OK:
Resource 1 will appear in the middle (see above). Now, click OK again (the Process
window will close). Then, in the main window, with Process 2 selected click on Dispose
icon in the left pane, and drag it and drop it to the right of Process 2. Then click on Run
(on the top) and next click Setup:
SUPPLEMENT TO CHAPTER 18 Simulation
Run Setup window will open:
In the Run Setup window, change the Replication Length from Infinite to 24 (because
we want to run it for 24 hours), and change the Base Time Units to Minutes. Then, click
OK (Run Setup window will close). In the main window, click Go under the Run (in the
top). Simulation starts. A simple animation shows that a paper (representing a customer)
comes out of Create 1, possibly queues above Process 2, and exits in Dispose 2. The numbers
under Create and Dispose are the number of entities generated and disposed, respectively.
The number under Process is the number of customers in the queue and the paper icons
above it represent the queue:
29
30
PART EIGHT Waiting-Line Analysis
At the end of simulation, the following window will open. Click Yes:
The Report will Preview. Move to the second page:
These are the results of simulation: Average wait time is 7.5226, maximum wait time
is 34.022 minutes, average number in the queue is 1.8841, and maximum number in the
queue is 9. Finally, we need to end the simulation run (click on End under Run at the top)
and exit Arena (File → Exit) after saving the model.
@RISK. @RISK is an Excel add-in software that performs risk analysis using Monte
Carlo (probabilistic) simulation. @RISK is a product of Palisade Corporation (http://www.
palisade.com/risk). BC Hydro used it to evaluate the uncertainties surrounding its energy
conservation program.4 BC Hydro plans to meet most of its future electricity needs through
demand management, i.e., energy conservation by its customers. Around 60 projects were
considered, such as compact fluorescent light promotions, subsidies for energy-efficient
appliances, variable-speed motor promotions for home furnaces, and promotional activity
4
http://www.palisade.com/cases/bchydro.asp.
SUPPLEMENT TO CHAPTER 18 Simulation
31
aimed at motivating customers to use less energy. The uncertainties were analyzed on a
case-by-case basis, and a probability distribution for the aggregate conservation-savings
forecast was developed. BC Hydro used estimates from experts in various industries to
provide input to the @RISK model. The displays assisted the experts to visualize how their
predictions influenced the final outcome of the decision-making process. Also, interrelationships among key uncertainties were explored.
Monte Carlo method, 3
random numbers, 4
random variates, 4
simulation, 2
The number of customers who arrive during each hour at a car repair shop can be described by
a Poisson distribution. Assuming that the average number of customers per hour is three during
the first four hours of a day, simulate customer arrivals for the first four hours. Read three-digit
random numbers from Table 18S-1, columns 4 and 5, going down.
d. Obtain cumulative probabilities of the Poisson distribution from Appendix B, Table C, for the
mean specified (3). Then, determine the random number intervals.
x
Cumulative
Probability
Random Number
Intervals
0..........
.050
001 to 050
1..........
.199
051 to 199
2..........
.423
200 to 423
3..........
.647
424 to 647
4..........
.815
648 to 815
5..........
.916
816 to 916
6..........
.966
917 to 966
7..........
.988
967 to 988
8..........
.996
989 to 996
9..........
.999
10 . . . . . . . . .
1.000
Key Terms
Solved Problems
Problem 1
Solution
997 to 999
000
e. Obtain three-digit random numbers 299, 642, 350, and 091.
f. Convert the random numbers to simulated values. Note where each number falls in the randomnumber interval list. For instance, 299 falls between 200 and 423. Interpret this to mean that
two customers arrive in the first hour. Similarly, 642 is interpreted to mean that three customers
arrive in the second hour, 350 implies two customers in the third hour, and 091 implies one
customer in the fourth hour.
g. To summarize, the number of customers per hour for the four-hour simulation is:
Hour
Number of
Arrivals
1..........
2
2..........
3
3..........
2
4..........
1
Jobs arrive at a workstation at fixed intervals of one hour. Processing time is approximately Normal
and has a mean of 56 minutes per job and a standard deviation of 4 minutes per job. Using the
fifth row of the table of standard Normal random variates (Table 18S-2), simulate the processing
times for four jobs and determine the amount of operator idle time and job waiting time. Assume
that the first job arrives at time = 0.
Problem 2
PART EIGHT Waiting-Line Analysis
32
Solution
a. Obtain the random numbers from the table: 0.74, −0.44, 1.53, and −1.76.
b. Convert the random numbers to simulated processing times:
Random
Number
Computation
Simulated Time
0.74 . . . . . . . . . .
56 + 4(0.74)
=
58.96
−0.44 . . . . . . . . .
56 + 4(−0.44)
=
54.24
1.53 . . . . . . . . . .
56 + 4(1.53)
=
62.12
−1.76 . . . . . . . . .
56 + 4(−1.76)
=
48.96
Note that three of the times are less than the inter-arrival times for the jobs (i.e., 1 hour), meaning that the operator will be idle after those three jobs. One time exceeds the one-hour interval,
so the next job must wait, and possibly the job following it if the waiting plus processing time
exceeds 60 minutes.
c. Calculate waiting and idle times:
Job
Number
Arrives
at
Processing Time,
t (minutes)
60 − t
Operator Idle
(minutes)
t − 60
Next Job Waits
(minutes)
1
0
58.96
1.04
—
2
60
54.24
5.76
—
2.12
3
120
62.12
—
4
180
48.96
8.92*
15.72
— 2.12
*60 − 2.12 − 48.96 = 8.92.
Discussion and
Review Questions
1. What is simulation (as used in operations management)? (LO1)
2. What are some of the primary reasons for the widespread use of simulation in practice?
(LO1)
3. What are some of the ways managers can use simulation? (LO1)
4. What is the difference between random numbers and random variates? (LO2)
5. How can Excel be used for generating random variates? (LO2)
6. How can random values from each of the following distributions be generated using random
number tables? (LO2)
a. Discrete
b. Normal
c. Continuous Uniform
d. Poisson
e. Exponential
7. Respond to the following comment: “I ran the simulation several times, and each run gave me
a different result. Therefore, the technique does not seem to be useful. I need one answer!”
(LO1–3)
8. Name three popular simulation software programs. (LO4)
SUPPLEMENT TO CHAPTER 18 Simulation
1. Visit http://www.promodel.com/solutions/logistics/, watch the video, and briefly summarize how
ProModel’s simulation helped the Salt Lake 2002 Winter Olympics. (LO4)
33
Internet Exercises
2. Visit http://www.lanner.com/en/case-studies.cfm, pick a case study, and summarize how
­WITNESS is being used. (LO4)
3. Visit http://www.palisade.com/cases/ctt.asp?caseNav=byProduct, pick a case study, and summarize how @RISK is being used. (LO4)
4. Google “IIE/RA contest problems,” pick a simulation contest problem, and determine what
topic it is on. Note: This is an annual contest sponsored and managed by Rockwell International (Arena). (LO3)
5. Visit http://arenasimulation.com/Tools_Resources_Video_Library.aspx, click on “What is
Simulation?,” register with Arena, watch the video, and summarize what, why, and when of
simulation. (LO1)
6. Visit http://arenasimulation.com/Tools_Resources_Video_Library.aspx, click on “Emergency
Department Simulation Demo,” register with Arena if necessary, watch the video, and summarize how Arena can help with the operations of an emergency department. (LO4)
7. Visit http://arenasimulation.com/Solutions_Solutions.aspx, click on an application area, find
a case study, and summarize how Arena is being used. (LO4)
1. The number of jobs received by a small machine shop is to be simulated for an eight-day
period. The shop manager has collected the following data for the number of jobs received
daily: (LO2)
Number of Jobs
2 or less
Frequency
0
3
10
4
50
5
80
6
40
7
16
8
4
9 or more
0
200
Use the third column of Table 18S-1 and read two-digit numbers, going down. Determine
the average number of jobs per day for the eight-day simulation period.
2. Jack sells insurance on a part-time basis. His records on the number of policies sold per week
over a 50-week period are: (LO2)
Number Sold per Week
Frequency
0
8
1
15
2
17
3
7
4
3
50
Simulate three five-week periods. Use Table 18S-1, column 6 for the first simulation, column
7 for the second, and column 8 for the third. In each case, read two-digit numbers, beginning
Problems
34
PART EIGHT Waiting-Line Analysis
at the bottom of the column and going up. For each simulation, determine the percentage of
weeks during which two or more policies are sold.
3. After a careful study of requests for a special tool at a large tool crib, an analyst concluded
that demand for the tool can be adequately described by Poisson distribution with mean of two
requests per hour. Simulate demand for a 12-hour period for this tool crib using Table 18S-1.
Read three-digit numbers from columns 5 and 6 combined, starting at the top and reading
down (e.g., 917, 264, 045). (LO2)
4. The number of lost-time accidents per month at a forestry company can be described using a
Poisson distribution that has mean of four accidents. Using the last two columns of Table 18S-1
(e.g., 540, 733, 293), simulate the number of accidents for a 12-month period. (LO2)
5. The time a surgeon spends performing a particular surgery can be modelled using a Normal
distribution that has mean of 20 minutes and standard deviation of 2 minutes. Using the table
of standard Normal random variates (Table 18S-2), simulate the times the surgeon might spend
on the next seven surgeries. Use column 4 of the table; start at the bottom of the column and
read up. (LO2)
6. The time between job arrivals at a workstation tends to be Normally distributed with a mean
of 15 minutes and a standard deviation of 1 minute. Job processing time is also Normally
distributed with a mean of 14 minutes and a standard deviation of 2 minutes. (LO3)
a. Using Table 18S-2, simulate the arrival and processing of five jobs. Use column 4 of the
table for job inter-arrival times and column 3 for processing times. Start each column at
row 4 and go down. Find the total time jobs wait for processing.
b. The company is considering the use of a new equipment that would result in processing
times that is Normal with mean of 13 minutes and standard deviation of 1 minute. Job
waiting represents a cost of $3 per minute, and the new equipment would represent an
additional cost of $.50 per minute. Would the equipment be cost justified? (Note: Use the
same arrival times and the same random numbers for processing times.)
7. Daily usage of sugar in a small bakery can be described by a Continuous Uniform distribution
with endpoints of 30 kg and 50 kg. Simulate daily usage for a 10-day period. Read four-digit
numbers from Table 18S-1, columns 5 and 6, going up from the bottom. (LO2)
8. Weekly usage of a specific spare part in a maintenance storeroom can be described by Poisson distribution with mean of 2.8 parts per week. Lead time to replenish the spare part is two
weeks (constant). Simulate the total usage of the part during lead time 6 times (don’t overlap
the lead times), and then determine the frequency of lead time demand (i.e., what percentage
of times was the demand equal to 2, 3, 4, etc.?). Read three-digit numbers from Table 18S-1,
columns 8 and 9, going down. (LO2 & 3)
9. (Excel exercise.) Repeat Problem 8 for 100 lead-time periods. Hint: To make a frequency
distribution, use the Histogram tool in Data Analysis. (LO2 & 3)
10. 10. A machine shop breaks an average of .6 unit of a tool per day. The number of breakages
(i.e., demand) can be described by Poisson distribution. The number of days required to obtain
replacement is almost constant (3 days). Because ordering is at the end of a day and receiving
is at the beginning of another day, the order will arrive 4 days after the order date. Four units
are ordered whenever the stock on hand and on order is two or less. Initially we have three
units on hand and none on order. All shortage is back-ordered. (LO3)
a. Draw a flowchart to describe this process.
b. Simulate this problem for a 12-day period. Read three-digit numbers from Table 18S-1,
columns 5 and 6, going down (e.g., 917, 264), for tool breakage. What is the probability
of shortage?
11. (Excel exercise.) Repeat Problem 10 for 100 days. What is the probability of shortage?
(LO3)
12. The time between arrivals of a part to a workstation varies Uniformly between 10 and 20
minutes. Process time is Normal with mean of 15 minutes and standard deviation of 2 minutes. (LO3)
SUPPLEMENT TO CHAPTER 18 Simulation
a. Simulate waiting times and processing for nine parts. Read three-digit numbers going down
columns 9 and 10 of Table 18S-1 for inter-arrivals (e.g., 156, 884, 576). Use column 8 of
Table 18S-2 for processing time. Calculate total waiting time.
b. If management can reduce the range of inter-arrival times to between 13 and 17 minutes,
what would the impact be on part waiting times? (Use the same service times and the
same random numbers for inter-arrival times from part a.) Round inter-arrival times to
two decimal places.
13. At a call centre for a cell phone company, customer service associates are employed to respond
to customer calls and complaints.5 During a non-peak period, on average 6 customers call per
hour (Poisson), i.e., the inter-arrival times are Exponential with mean of 10 minutes. The length
of calls has a triangular distribution with minimum of 2 minutes, maximum of 10 minutes,
and a most likely value of 5 minutes. Use Arena to model this problem and simulate for 200
hours to answer the following question: How many associates should the company employ if
the policy is both of the following: (LO4)
a. Average wait time in the phone queue should not be larger than 2 minutes.
b. Maximum number of calls waiting should be no more than 5.
14. A product requires that a hole be drilled into a metal block and that a cylindrical shaft be
inserted into it.6 The probability distribution of shaft radius is Normal with mean of 1 inch and
standard deviation of .001 inch. Similarly, the probability distribution of hole radius is Normal
with a mean of 1.003 inches and standard deviation of .001 inch. The clearance between a
hole and a shaft is the difference in their radii. The objective is to determine how frequently
interference (i.e., negative clearance) will occur. Simulate 1,000 pairs. (LO3)
15. An analyst found that the length of telephone conversations in an office could be described by
an Exponential distribution with a mean of four minutes. Reading two-digit random numbers
from Table 18S-1, column 6, simulate the length of five calls and calculate the simulated average time. Why is the simulated average different from the mean of four minutes? (LO2)
16. The length of time between calls for service of a certain piece of equipment can be described
by an Exponential distribution with a mean of 40 minutes. Service time can be described by
a Normal distribution with mean of eight minutes and standard deviation of two minutes.
Simulate the queuing system for four breakdowns. For breakdowns, read two-digit numbers
from Table 18S-1, column 7; for service times, read numbers from Table 18S-2, column 8.
Calculate the total wait time. (LO3)
* 17. A simple project has only two activities: A and B. A is an immediate predecessor of B. A has
a triangular distribution with minimum a = 10 days, mode m = 20 days, and maximum b = 50
days. B also has a triangular distribution with minimum = 15 days, mode = 25 days, and maximum = 60. Simulate 100 replications of the project in Excel to determine the distribution of
the project completion time. What is a reasonable point estimate for the project duration?
Hint: Excel does not provide random triangular distribution variates. We need to generate
a (real) random number U greater or equal to 0 and less than 1; if U ≤ (m – a)/(b – a), then
random variate = a + U ( b − a )( m − a ) ; else random variate = b − ( 1 − U ) ( b − a )( b − m ) .
See e.g., “Generating Triangular-distributed random variates” section in http://en.wikipedia.
org/wiki/Triangular_distribution. (LO3)
18. A service operation consists of three steps. The first step can be described by a Continuous
Uniform distribution that ranges between five and nine minutes. The second step can be
described by a Normal distribution with a mean of seven minutes and a standard deviation of
one minute, and the third step can be described by an Exponential distribution with a mean
of five minutes. Simulate three services using two-digit numbers from Table 18S-1, row 4 for
step 1; Table 18S-2, row 6 for step 2; and two-digit numbers from column 4 of Table 18S-1
for step 3. Determine the simulated time for each of the three services. (LO3)
5
C. Harrell et al, Simulation Using ProModel, 2000, Boston: McGraw-Hill, pp. 397–399.
Based on F. S. Hillier and G. J. Lieberman, Operations Research, 2nd ed, 1974, San Francisco: Holden-Day,
p. 653.
6
35
36
PART EIGHT Waiting-Line Analysis
19. A project consists of five major activities, as illustrated in the diagram below. Activity times
are Normally distributed with means and standard deviations as shown in the following table.
Note that there are two paths through the project: A-B-D-End and A-C-E-End. Project duration
is defined as the larger sum of times along the two paths. Simulate 10 times for each activity.
Use column 1 of Table 18S-2 for activity A, column 2 for activity B, column 3 for activity
C, column 4 for activity D, and column 5 for activity E. Determine the project duration for
each of the 10 sets, and prepare a frequency distribution of project duration. Use categories
of 25 to less than 30, 30 to less than 35, 35 to less than 40, 40 to less than 45, and 45 or more.
Determine the proportion of time that a simulated duration of less than 40 days occurred. How
might this information be used? (LO3)
B
D
A
End
C
E
Mean (days)
Standard
Deviation (days)
A
10
2
B
12
2
C
15
3
D
14
2
E
8
1
Activity
* 20. A drug manufacturer has recently accepted an order from its best customer for 950 ounces
of a new drug and wants to plan its production schedule to meet the promised delivery date.7
The order has to be produced in 1 or more batches. There are three sources of uncertainty: (a)
time required to produce a batch (Normal with mean of 8 days and standard deviation of 1
day rounded to the nearest integer), (b) the yield of a batch, in ounces (triangular distribution
[600, 1000, 1100]), and (c) the probability that a batch passes final inspection (prob = .99). If
the batch fails inspection, it will be discarded. Use simulation and Excel to decide how many
days prior to the due date the production should begin. (LO3)
*21. A newsvendor must decide how many papers to buy from the newspaper publisher each
weekday.8 Each weekday she will order the same quantity. A newspaper costs her $.55
and sells for $1. Daily demand during each weekday is Normal with mean of 136 and
standard deviation of 27 newspapers. Any unsold newspapers will be sold to a recycler
for $.03 each. A shortage results in loss of profit. Simulate demand for 1,000 days using
Excel and determine the number of newspapers the newsvendor should buy each ­morning
7
Based on W. L. Winston and S. C. Albright, Practical Management Science, 2nd ed, 2001, Pacific Grove,
California: Duxbury, pp. 626–630.
8
W. D. Kelton et al, Simulation with Arena, 4th ed, 2007, Boston: McGraw-Hill, p. 38.
SUPPLEMENT TO CHAPTER 18 Simulation
to minimize expected total shortage and excess cost. Assume that she can buy only in
multiples of 10. (LO3)
* 22. The inter-arrival times to a single server queue are Exponentially distributed with a mean of
1.6 minutes.9 Service times have a Continuous Uniform distribution between .27 and 2.29
minutes. Download a trial version of Arena and use it to perform 100 hours of simulation to
determine the maximum customer wait time in the queue. (LO3)
* 23. A technical support centre is staffed by two people, Able and Baker.10 Able is more experienced than Baker, and hence provides faster service. If both are idle, Able takes the call. If
both are busy, the caller waits. The distributions of inquiry inter-arrival times during peak
periods, Able’s service times, and Baker’s service times are all Exponential with means of
2.25, 3.35, and 4.25 minutes, respectively. Model this problem in Excel, run the simulation
for 1,000 calls, and determine the average wait time (in the queue). Should another person be
assigned to the centre? (LO3)
11
**24. An appliance retail chain keeps inventories of its most popular fridges in its warehouse.
Consider one such model and size. The company uses the fixed interval inventory model for
replenishing the fridge from the manufacturer. The order interval used is 1 week (5 workdays). The distribution of daily demand is Normal with mean of 2 and standard deviation
of 1 (but non-negative) and lead time from the manufacturer is Exponential with mean of
1.5 workdays (but minimum = 1 day and maximum = 4 days; 1 day LT = next-day arrival).
Currently the order up to level used for this fridge is 13 units. Model this problem in Excel,
run the simulation for 1,000 days, and determine the average daily number of fridges
short. (LO3)
* 25. A multi-user mainframe computer includes two disk drives that are prone to failure.12 If a
disk drive fails, users lose their files, which need to be restored. Restoration is achieved by
copying copies of the files, held on magnetic tapes, onto the disk backup. This restoration
is inconvenient and so a new operation policy is being considered. At the moment, the disk
drives are repaired and restored as and when they fail. The proposal is to introduce a joint
repair system. The life of a disk drive is Normally distributed with a mean of 4 months and
a standard deviation of 1 month. Under the current policy, it costs $50 to repair and restore
a failed disk drive. The joint repair and restore policy would operate as follows: when either
unit fails, the failed unit is repaired and restored, and the other unit is cleaned to bring it to
a state equivalent to having been just repaired. This will cost $75 in total. Is the new joint
repair and restore policy cost effective? Use Excel to simulate 200 individual failures and
100 joint repairs. (LO3)
* 26. An airport hotel has 100 rooms.13 On any given night the hotel takes up to 105 reservations because of the possibility of no-shows. Past records indicate that the number of daily
reservations is uniformly distributed over the integer range [96, 105]. The no-shows have a
Normal distribution with mean of 2.5 and standard deviation of 1, but rounded to the nearest integer. Develop and run a simulation model in Excel for 100 days to find the expected
number of rooms used per night and the percentage of nights when more than 100 rooms
are claimed. (LO3)
9
W. D. Kelton et al, Simulation with Arena, 4th ed, 2007, Boston: McGraw-Hill, pp. 43–44.
Based on J. Banks et al, Discrete-event System Simulation, 5th ed, 2010, New Jersey: Prentice-Hall, pp. 51–52.
11
J. Banks et al, Discrete-event System Simulation, 5th ed, 2010, New Jersey: Prentice-Hall, pp. 61–62.
12
M. Pidd, Computer Simulation in Management Science, 1984, Chichester, UK: John Wiley, pp. 22–24.
13
W. L. Winston, Operations Research, 2nd ed, 1991, Boston: PWS-Kent Publishing, p. 1159.
10
37
PART EIGHT Waiting-Line Analysis
38
MINI-CASE
Valencia Kidney Waiting Line
S
ome staff researchers at the Valencia Regional Health
Authority of Spain are wondering what will happen to the
waiting line for kidney transplants in their region. They believe
that the waiting line is likely to decrease. During the 1997
to 1999 period (1,095 days), the (new) demand for a kidney
transplant was 531, with the number of (new) patients per day
distributed approximately according to Poisson distribution.
Hence, the average number of “arrivals” to the queuing system
is estimated to be 531/1,095 = 0.485 patients per day. There
were 241 double donations and 82 single donations during
the same time period. Hence, the average number of kidneys
donated per day was (241 × 2 + 82 × 1)/1,095 = .515. This can
be used to estimate the average number of patients receiving a
kidney each day (there was no instance of a patient receiving
two kidneys as transplants). The number of kidney donations
per day can also be approximated by Poisson distribution. There
were 446 people in the queue as of December 31, 1999. What
is the probability that the waiting line will decrease after 365
days? Hint: (a) First in Excel compute the cumulative Poisson
distribution with means of .485 and .515 using = POISSON.
DIST(x, mean, cumulative). For example, to get the cumulative
probability that X takes the values up to and including 2 use
POISSON.DIST(2, .485, TRUE) for the arrival distribution, (b)
Use the cumulative Poisson distributions of part a to simulate
365 days of arrivals and services, keeping track of the length
of the queue (initially it is 446 people long), (c) use the Data
Table to obtain 100 values for the end-of-the-year number of
people in the queue, and (d) count the numbers that are smaller
than 446.
Source: Based on J. J. Abellan, et al., “Predicting the Behaviour
of the Renal Transplant Waiting List in the Pais Valencia (Spain)
Using Simulation Modelling,” Proceedings of the 2004 Winter
Simulation Conference, pp. 1969–974. http://www.informs-sim.org/
wsc04papers/263.pdf
MINI-CASE
Dofasco
The superintendent of blast furnaces at Dofasco has to decide
whether or not to recommend that an engineering modification
costing $600,000 be made to the lance desulfurizing plant. The
lance desulfurizing plant is part of the process for converting
iron into steel. Molten iron from the blast furnaces is transported
in special rail cars, called torpedoes, to the lance desulfurizing
plant where sulfur content of torpedoes is reduced. After desulfurizing, a locomotive is called to haul the torpedoes to the
melt shop where the iron is converted to steel.
The desulfurizing plant consists of two side-by-side ports,
each capable of desulfurizing a torpedo. However, the lance used
at each port tends to break down, and repairs take up to a week.
Management is concerned that lance breakdowns could cause a
bottleneck, which would hold up a large number of torpodeos,
causing the blast furnaces to shut down. There were 23 torpedoes
circulating between the blast furnaces, desulfurizing plant, and
melt shop. The proposed modification would install a backup
system, which would solve the problem.
There are four blast furnaces. Each needs approximately
two torpedoes for each discharge. The discharge time for each
blast furnace is Normally distributed with a mean of 200 minutes and standard deviation of 20 minutes. The processing
time of each desulfurizing port is approximately Normally
distributed with a mean of 15 minutes and standard deviation
of 3 minutes. The wait time for a locomotive to haul a torpedo
out of a port is approximately Exponential with a mean of
8 minutes.
Model this problem in Excel, assuming that a lance has broken down for a week (i.e., just use one server). What is the implication for the number of torpedoes stuck in the lance queue?
Management won’t like to see more than five to six torpedoes
in the lance line. Hint: The distribution of sum of two or more
independent Normal distributions tends to the Continuous Uniform distribution. Therefore, one can assume that inter-arrival
times are Uniform.
Source: P. C. Bell, Management Science/Operations
Research, 1999, Cincinnati: South-Western College Publishing,
pp. 203–206.