Operations Management Help
klikki27
queuing_problems2.pdf
1. Modification of Problem 9.3 from the book:
Jean Clark is the manager of the Midtown Saveway Grocery Store. She now needs to replenish her
supply of strawberries. Her regular supplier can provide as many cases as she wants. However, because
these strawberries already are very ripe, she will need to sell them tomorrow and then discard any that
remain unsold. Jean estimates that she will be able to sell 10, 11, 12, or 13 cases tomorrow. She can
purchase the strawberries for $3 per case and sell them for $8 per case. Jean now needs to decide how
many cases to purchase.
Jean has checked the store’s records on daily sales of strawberries. On this basis, she estimates that the
prior probabilities are 0.2, 0.4, 0.3, and 0.1 for being able to sell 10, 11, 12, and 13 cases of strawberries
tomorrow, respectively.
a) Develop a decision analysis formulation of this problem by identifying the decision
alternatives, the states of nature, and the payoff table. (Build a table similar to the
Table 9.3 in the textbook or the table on Slide 9 of Lecture Notes 11 – Decision
Analysis).
State of Nature
Alternative Sell 10 cases Sell 11 cases Sell 12 cases Sell 13 cases
Buy 10 cases $50 $50 $50 $50
Buy 11 cases $47 $55 $55 $55
Buy 12 cases $44 $52 $60 $60
Buy 13 cases $41 $49 $57 $65
Prior Probability 0.2 0.4 0.3 0.1
b) If Jean is dubious about the accuracy of these prior probabilities and so chooses to
ignore them and use the maximax criterion, how many cases of strawberries should
she purchase? Show how you reach to your answer using the table you have in part
a).
Max(Buy 10) = $50,
Max(Buy 11) = $55,
Max(Buy 12) = $60,
Max(Buy 13) = $65.
Maximax = $65 with buying 13 cases.
State of Nature
Alternative Sell 10 cases Sell 11 cases Sell 12 cases Sell 13 cases
Buy 10 cases $50 $50 $50 $50
Buy 11 cases $47 $55 $55 $55
Buy 12 cases $44 $52 $60 $60
Buy 13 cases $41 $49 $57 $65
c) How many cases should be purchased if she uses the maximin criterion? Show how
you reach to your answer using the table you have in part a).
Min(Buy 10) = $50,
Min(Buy 11) = $47,
Min(Buy 12) = $44,
Min(Buy 13) = $41.
Maximin = $50 with buying 10 cases.
State of Nature
Alternative Sell 10 cases Sell 11 cases Sell 12 cases Sell 13 cases
Buy 10 cases $50 $50 $50 $50
Buy 11 cases $47 $55 $55 $55
Buy 12 cases $44 $52 $60 $60
Buy 13 cases $41 $49 $57 $65
d) How many cases should be purchased if she uses the maximum likelihood criterion?
Show how you reach to your answer using the table you have in part a).
The most likely state of nature is to sell 11 cases. Under this state, she should buy 11
cases with a payoff of $55.
State of Nature
Alternative Sell 10 cases Sell 11 cases Sell 12 cases Sell 13 cases
Buy 10 cases $50 $50 $50 $50
Buy 11 cases $47 $55 $55 $55
Buy 12 cases $44 $52 $60 $60
Buy 13 cases $41 $49 $57 $65
Prior Probability 0.2 0.4 0.3 0.1
e) How many cases should be purchased according to Baye’s decision rule? Show your
calculations using the table you have in part a).
Jean should buy 12 cases. The maximum expected payoff is $53.60.
2. Modification of 9.7 from the book:
Consider a decision analysis problem whose payoffs (in units of thousands of dollars) are given
by the following payoff table:
a) Which alternative should be chosen under the maximax criterion?
Max(A1) = 80, Max(A2) = 50, Max(A3) = 60.
Maximax = $80 thousand when choosing alternative A1
b) Which alternative should be chosen under the maximin criterion?
1 2 3 4 5 6
7 8 9 10
11 12
A B C D E F G H
Purchase Price $3 Selling Price $8
Payoff Table Sell Sell Sell Sell Expected
Alternative 10 11 12 13 Payoff
Buy 10 $50 $50 $50 $50 $50.00 Buy 11 $47 $55 $55 $55 $53.40 Buy 12 $44 $52 $60 $60 $53.60
Buy 13 $41 $49 $57 $65 $51.40
Prior Probabil ity 0.2 0.4 0.3 0.1
State of Nature
State of Nature
Alternative S1 S2
A1 80 25
A2 30 50
A3 60 40
Prior Probability 0.4 0.6
State of Nature
Alternative S1 S2
A1 80 25
A2 30 50
A3 60 40
=0.2*50+0.4*50+0.3*50+0.1*50
=0.2*47+0.4*55+0.3*55+0.1*55
=0.2*44+0.4*52+0.3*60+0.1*60
=0.2*41+0.4*49+0.3*57+0.1*65
Min(A1) = 25, Min(A2) = 30, Min(A3) = 40.
Maximin = $40 thousand when choosing alternative A3.
c) Which alternative should be chosen under the maximum likelihood criterion?
S2 is the most likely outcome. For this state, the maximum payoff of $50 thousand
occurs with alternative A2.
d) Which alternative should be chosen under the Baye’s decision rule?
Alternative A3 has the highest expected payoff of $48 thousand.
e) Draw a decision tree and apply Baye’s decision rule using the decision tree.
1 2 3
4 5 6
7 8
A B C D E
Payoff Table ($thousand) Expected Payoff
Alternative S1 S2 ($thousand)
A1 80 25 47 A2 30 50 42 A3 60 40 48
Prior Probabil ity 0.4 0.6
State of Nature
0.4
State 1
80
Alternative 1 80 80
0 47 0.6
State 2
25
25 25
0.4
State 1
30
Alternative 2 30 30
3
48 0 42 0.6
State 2
50
50 50
0.4
State 1
60
Alternative 3 60 60
0 48 0.6 State 2
40 40 40
3. Modification of 9.11 from the book:
You are given the following payoff table (in units of thousands of dollars) for a decision analysis
problem:
a) According to Baye’s decision rule, which alternative should be chosen?
Alternative A2 has the highest expected payoff of $1,000.
b) Find the expected value of perfect information.
With perfect information, choose A1 for when the state is S1, A2 when the state is S2,
and A3 when the state is S3.
EP(with perfect information) = (0.2)(4) + (0.5)(2) + (0.3)(1) = $2,100
EVPI = EP(with perfect information) – EP (without more information)
= $2,100 – $1,000 = $1,100.
c) Check your answer in part b by recalculating it with the help of a decision tree.
(Hint: In case of perfect information, your decision tree will start with an event
node!)
1 2 3
4 5 6
7 8
A B C D E F
Payoff Table ($thousands) Expected State of Nature Payoff
Alternative S1 S2 S3 ($thousands)
A1 4 0 0 0.8 A2 0 2 0 1 A3 3 0 1 0.9
Prior Probabil ity 0.2 0.5 0.3
State of Nature
Alternative S1 S2 S3
A1 4 0 0
A2 0 2 0
A3 3 0 1
Prior Probability 0.2 0.5 0.3
EVPI = EP(with perfect information) – EP (without more information)
= $2,100 – $1,000 = $1,100.
d) You are given the opportunity to spend $1,000 to obtain more information about
which state of nature is likely to occur. Given your answer to part b, might it be
worthwhile to spend this money?
Since the information will cost $1,000 and the value is no more than $1,100, it
might be worthwhile to spend the money.
Alternative 1 4
4 4
0.2 State 1 Alternative 2
1 0 0 4 0 0
Alternative 3 3
3 3
Alternative 1 0
0 0
0.5 State 2 Alternative 2
2 2 2.1 0 2 2 2
Alternative 3 0
0 0
Alternative 1 0
0 0
0.3 State 3 Alternative 2
3 0 0 1 0 0
Alternative 3 1
1 1
4. Modification of 9.28 from the book:
The comptroller of the Macrosoft Corporation has $100 million of excess funds to invest. She
has been instructed to invest the entire amount for one year in either stocks or bonds (but not
both) and then to reinvest the entire fund in either stocks or bonds (but not both) for one more
year. The objective is to maximize the expected monetary value of the fund at the end of the
second year.
The annual rates of return on these investments depend on the economic environment, as
shown in the following table:
Rate of Return
Economic Environment Stocks Bonds
Growth 20% 5%
Recession -10% 10%
Depression -50% 20%
The probabilities of growth, recession, and depression for the first year are 0.7, 0.3, and 0,
respectively. If growth occurs in the first year, these probabilities remain the same for the
second year. However, if a recession occurs in the first year, these probabilities change to 0.2,
0.7, and 0.1, respectively, for the second year.
Construct a decision tree for this problem and then analyze the decision tree to identify the
optimal policy based on Baye’s decision rule.
The comptroller should invest in stocks the first year. If there is growth during the first year
then she should invest in stocks again the second year. If there is a recession during the first
year then she should invest in bonds for the second year. The expected payoff is $122.94
million.
0.7 Growth
144 Stocks Y2 24 144
0 133.2 0.3 Recession
0.7 108 Growth -12 108
1 20 133.2 0.7
Growth 126
Bonds Y2 6 126
0 127.8 0.3 Recession
132 12 132
Stocks Y1 0.2 Growth
100 122.94 108 18 108
0.7 Stocks Y2 Recession
81 0 82.8 -9 81
0.1 Depression
0.3 45 Recession -45 45
2 -10 99 0.2
Growth 94.5
4.5 94.5
0.7 Bonds Y2 Recession
99 0 99 9 99
0.1 1 Depression
122.94 108 18 108
0.7 Growth
126 Stocks Y2 21 126
0 116.55 0.3 Recession
0.7 94.5 Growth -10.5 94.5
1 5 116.55 0.7
Growth 110.25
Bonds Y2 5.25 110.25
0 111.825 0.3 Recession
115.5 10.5 115.5
Bonds Y1 0.2 Growth
100 117.885 132 22 132
0.7 Stocks Y2 Recession
99 0 101.2 -11 99
0.1 Depression
0.3 55 Recession -55 55
2 10 121 0.2
Growth 115.5
5.5 115.5
0.7 Bonds Y2 Recession
121 0 121 11 121
0.1 Depression
132 22 132
5. FINAL QUESTION OF FALL 2012
Suppose that Dr. Konur operates a local grocery store in Rolla, Mo named Dino-cery. Dino-cery’s best product is
the authentic baklavas that Dr. Konur ships from the Turkish restraint in St. Louis at the beginning of each week.
He ships and sells two types of baklavas: walnut and pistachio. Dr. Konur observed the following weekly demand
for trays of baklavas sold:
For instance, 1-tray of walnut and 2-trays of pistachio baklavas will be sold in a week with probability of 0.3.
Each tray of walnut baklava costs Dr. Konur $100 and sold for $250. Each tray of pistachio baklava costs Dr.
Konur $125 and sold for $300. Each unsold tray of baklava is perished so it is wasted. There is no cost associated
with an unsatisfied customer. Dr. Konur now wants to decide on how many trays of each baklava type he should
buy at the beginning of each week. He is only considering buying 1 or 2 trays of each baklava type.
a) Express Dr. Konur’s baklava purchase problem as a decision analysis problem. Formulate a decision analysis
problem by constructing the payoff table with the information given above. That is, describe your
alternatives, states of natures, and calculate the payoff (profit) for each alternative and state of nature pair,
and note the prior probabilities for each state of nature.
b) Based on maximax criterion, how many trays of each baklava type Dr. Konur should buy?
c) Based on minimax criterion, how many trays of each baklava type Dr. Konur should buy?
d) Based on maximum likelihood criterion, how many trays of each baklava type Dr. Konur should buy?
e) Based on Baye’s decision rule, how many trays of each baklava type Dr. Konur should buy?
f) Dr. Konur can drink a Turkish coffee and have his fortune read from the coffee cup by a fortune teller. He
will specifically learn how many trays of each baklava type he will sell at the beginning of the week from the
fortune teller. The fortune tellers asks $150 to do one reading. Would Dr. Konur pay $150 to the fortune
teller? Explain why.
Walnut Pistacio Walnut Pistacio Walnut Pistacio Walnut Pistacio
1 1 1 2 2 1 2 2
probability 0.2 0.3 0.4 0.1
Note that what you sell is the minimum of what you buy and the demand
So, your profit from a specific baklava type will be
Minimum(number of trays you buy, demand for the type)*selling price - number of trays you buy*cost
And, your total profit will be the some of the profits of each type
a)
Walnut Pistachio Walnut Pistachio Walnut Pistachio Walnut Pistachio b) c) d) e)
Walnut Pistachio 1 1 1 2 2 1 2 2 Maximax Maximin Maximum likely Baye's Rule
1 1 325 325 325 325
1 2 500 200 200 320
2 1 475 225 475 350
2 2 650 100 350 345
(2,2) (1,1) (2,1) (2,1)
Bonus:
< 150
So, do not do the fortune telling
Alternative
PAYOFF TABLE
0.2 0.3 0.4 0.1
325 500 475 650
Prior Probability
Maximum of Each Alternative
225 225 475 475
100 400
325 325
Expected Payoff without Perfect Information
350
Expected Value of Perfect Information
120
350 650
200 500 200 500
Demand 1 Demand 2 Demand 3
alternative selected:
Expected Payoff with Perfect Information
470
Demand 4
325 325
6. 11.9 from the book:
Explain why the utilization factor ρ for the server in a single-server queueing system must equal
to 1-P0, where P0 is the probability of having 0 customers in the system.
The utilization factor represents the fraction of time that the server is busy. The server is busy
except when there are zero people in the system. P0 is the probability of having 0 customers in
the system. Hence, = 1 – P0.
7. Modification of 11.10 from the book:
The Friendly Neighbor Grocery Store has a single checkout stand with a full-time cashier.
Customers arrive randomly at the stand at a mean rate of 30 per hour (Note: when customers
are told to arrive randomly with a given rate, the textbook assumes that the interarrival time is
exponentially distributed). The service time distribution is exponential, with a mean of 1.5
minutes. This situation has resulted in occasional long lines and complaints from customers.
Therefore, because there is no room for a second checkout stand, the manager is considering
the alternative of hiring another person to help the cashier by bagging the groceries. This help
would reduce the expected time required to process a customer to 1 minute, but the
distribution still would be exponential.
The manager would like to have the percentage of time that there are more than two customers
at the checkout stand down below 25 percent. She also would like to have no more than 5
percent of the customers needing to wait at least five minutes before beginning service, or at
least seven minutes before finishing service.
a) Use the formulas for the M/M/1 model to calculate L, W, WQ, LQ, P0, P1, and P2 for
the current mode of operation. What is the probability of having more than two
customers at the checkout stand? That is, do not use the Excel template for part a.
L = /(–) = 30/(40–30) = 3 customers
W = 1/(- = 1/(40–30) = 0.1 hours
Wq = /[(-)] = 30/[40(40–30)] = 0.075 hours
Lq = Wq = 30(0.075) = 2.25 customers
P0 = 1– = 1–0.75 = 0.25
P1 = (1–) = (1–0.75)(0.75) = 0.188
P2 = (1–) 2 = (1–0.75)(0.75)2 = 0.141
There is a 1–P0–P1–P2 = 1–0.25–0.188–0.141 = 42% chance of having more than 2
customers at the checkout stand.
b) Use the Excel template for this model to check your answers in part a. Also, find the
probability that the waiting time before beginning service exceeds five minutes, and
the probability that the waiting time before finishing service exceeds seven minutes.
c) Repeat part a for the alternative being considered by the manager.
L = /(–) = 30/(60–30) = 1 customer
W = 1/(- = 1/(60–30) = 0.033 hours
Wq = /[(-)] = 30/[60(60–30)] = 0.017 hours
Lq = Wq = 30(0.017) = 0.5 customers
P0 = 1– = 1–0.75 = 0.5
P1 = (1–) = (1–0.75)(0.75) = 0.25
P2 = (1–) 2 = (1–0.75)(0.75)2 = 0.125
There is a 1–P0–P1–P2 = 1–0.5–0.25–0.125 = 12.5% chance of having more than 2
customers at the checkout stand.
d) Repeat part b for this alternative.
e) Which approach should the manager use to satisfy her criteria as closely as
possible?
The manager should adopt the new approach of adding another person to bag the groceries.
Template for the M/M/s Queueing Model
Data Results
1 = 30 (mean arrival rate) L = 3
0 = 40 (mean service rate) Lq = 2.25
0 s = 1 (# servers)
0 W = 0.1
0 Pr(W > t) = 0.311403 Wq = 0.075
0 when t = 0.116667
0 = 0.75
0 Prob(Wq > t) = 0.325949
0 when t = 0.083333 n Pn
0 0 0.25
0 1 0.1875
Template for the M/M/s Queueing Model
Data Results
1 = 30 (mean arrival rate) L = 1
0 = 60 (mean service rate) Lq = 0.5
0 s = 1 (# servers)
0 W = 0.033333333
0 Pr(W > t) = 0.030197 Wq = 0.016666667
0 when t = 0.116667
0 = 0.5
0 Prob(Wq > t) = 0.041042
0 when t = 0.083333 n Pn
0 0 0.5
0 1 0.25
8. Modification of 11.12 from the book:
Jerry Jansen, materials handling at the Casper Edison Corporation’s new factory, needs to decide
whether to purchase a small tractor-trailer train or a heavy-duty forklift for transporting heavy
goods between certain producing centers in the factory. Calls for the materials-handling unit to
move a load would come essentially at random at a mean rate of four per hour, i.e., they are
exponentially distributed with rate four per hour. The total time required to move a load has an
exponential distribution, where the expected time would be 12 minutes for the tractor-trailer
and 9 minutes for the forklift truck. The total equivalent uniform hourly cost (capital recovery
cost plus operating cost) would be $50 for the tractor-trailer train and $150 for the forklift truck.
The estimated cost of idle goods (waiting to be moved or in transit) because of increase process
inventory is $20 per load per hour.
Jerry also has established certain criteria that he would like the materials-handling unit to satisfy
in order to keep production flowing on schedule as much as possible. He would like to average
no more than half an hour for completing the move of a load after receiving the call requesting
the move. He also would like the time for completing the move to be no more than one hour 80
percent of the time. Finally, he would like to have no more than three load waiting to start their
move at least 80 percent of the time.
a) Obtain the various measures of performance if the tractor-trailer train were to be
chosen. Evaluate how well these measures meet the above criteria.
The train does not meet any of the criteria. The average time is more than half-an-
hour (W = 1 hour), it is no more than an hour less than 80% of the time (Pr(W > 1) =
36.8%), and there are three loads or fewer less than 80% of the time
(P0+P1+P2+P3+P4 = 67.2%).
b) Repeat part a if the forklift truck were to be chosen
The forklift truck meets all the criteria. The average time is less than half-an-hour (W
= 0.375 hours), it is no more than an hour more than 80% of the time (Pr(W > 1) =
6.9%), and there are three loads or fewer more than 80% of the time
(P0+P1+P2+P3+P4 = 92.2%).
c) Compare the two alternatives in terms of their expected total cost per hour
(including the cost of idle goods).
Tractor-trailer train: L($20)+$50 = (4)($20) + $50 = $130/hour
Forklift truck: L($20)+$150 = (1.5)($20) + $150 = $180/hour
d) Which alternative do you think Jerry should choose? Discuss why.
While the forklift truck has higher overall costs, it does a better job of meeting the
additional criteria.
9. Modification of 11.23 from the book:
The Security & Trust Bank employs four tellers to serve its customers. Customers arrive
randomly at a mean rate of two per minute, i.e., interarrival time is exponentially distributed
with rate of 2 customers per minute. However, business is growing and management projects
that the mean arrival rate will be three per minute a year from now. The transaction time
between the teller and customer has an exponential distribution with a mean of one minute.
Management has established the following guidelines for a satisfactory level of service to
customers. The average number of customers waiting in line to begin service should not exceed
one. At least 95 percent of the time, the number of customers waiting in lines should not exceed
five. For at least 95 percent of the customers, the time spent in line waiting to begin service
should not exceed five minutes.
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
B C D E G H I
Data Results
= 4 (mean arrival rate) L = 1.5
= 6.66666667 (mean service rate) Lq = 0.9
s = 1 (# servers)
W = 0.37500
Pr(W > t) = 0.06948345 Wq = 0.22500
when t = 1
= 0.6
Prob(Wq > t) = 0.04169007
when t = 1 n Pn Cumulative
0 0.4 0.4
1 0.24 0.64
2 0.144 0.784
3 0.0864 0.8704
4 0.05184 0.92224
a) Use the M/M/s model to determine how well these guidelines are currently being
satisfied.
All the guidelines are currently being met. The average number in line is 0.17, 99.7%
of the time there are 5 or fewer in line, and 0.000789% of customers wait more than
5 minutes.
b) Evaluate how well the guidelines will be satisfied a year from now if no change is
made in the number of tellers.
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
B C D E G H I
Data Results
= 2 (mean arrival rate) L = 2.173913043
= 1 (mean service rate) Lq = 0.173913043
s = 4 (# servers)
W = 1.086956522
Pr(W > t) = 0.00790187 Wq = 0.086956522
when t = 5
= 0.5
Prob(Wq > t) = 7.8956E-06
when t = 5 n Pn cumulative
0 0.130434783 0.130434783
1 0.260869565 0.391304348
2 0.260869565 0.652173913
3 0.173913043 0.826086957
4 0.086956522 0.913043478
5 0.043478261 0.956521739
6 0.02173913 0.97826087
7 0.010869565 0.989130435
8 0.005434783 0.994565217
9 0.002717391 0.997282609
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
B C D E G H I
Data Results
= 3 (mean arrival rate) L = 4.528301887
= 1 (mean service rate) Lq = 1.528301887
s = 4 (# servers)
W = 1.509433962
Pr(W > t) = 0.02390064 Wq = 0.509433962
when t = 5
= 0.75
Prob(Wq > t) = 0.00343254
when t = 5 n Pn cumulative
0 0.037735849 0.037735849
1 0.113207547 0.150943396
2 0.169811321 0.320754717
3 0.169811321 0.490566038
4 0.127358491 0.617924528
5 0.095518868 0.713443396
6 0.071639151 0.785082547
7 0.053729363 0.83881191
8 0.040297022 0.879108933
9 0.030222767 0.9093317
The first two guidelines will not be satisfied in one year but the third will be. The
average number in line is 1.53, 90.9% of the time there are 5 or fewer in line, and
0.34% of customers wait more than 5 minutes.
c) Determine how many tellers will be needed a year from now to completely satisfy
these guidelines.
10. Deli Store (FINAL QUESTION OF FALL 2012)
Suppose that you operate your own deli store in Rolla, MO that serves its customers delicious
sandwiches. The customers give their sandwich orders by calling your store. You observed that the time
between two consecutive order calls is exponentially distributed with mean inter-arrival time equal to
15 minutes. Once a customer gives a sandwich order via phone, he/she comes to your store to pick
his/her sandwich. There is one employee who is preparing any sandwich orders that you received on
phone. You observed that preparation time for any sandwich is exponentially distributed with average
preparation time equal to 10 minutes. Answer the following questions based on the above information.
a) Describe the above system as a queueing system by defining the customers and the arrival
rate, servers and the service rate, and express the queueing model using Kendall’s notation.
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4
5
6
7
8
9
10
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12
13
14
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18
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20
21
22
B C D E G H I
Data Results
= 3 (mean arrival rate) L = 3.354227405
= 1 (mean service rate) Lq = 0.354227405
s = 5 (# servers)
W = 1.118075802
Pr(W > t) = 0.0083184 Wq = 0.118075802
when t = 5
= 0.6
Prob(Wq > t) = 1.0721E-05
when t = 5 n Pn cumulative
0 0.04664723 0.04664723
1 0.139941691 0.186588921
2 0.209912536 0.396501458
3 0.209912536 0.606413994
4 0.157434402 0.763848397
5 0.094460641 0.858309038
6 0.056676385 0.914985423
7 0.034005831 0.948991254
8 0.020403499 0.969394752
9 0.012242099 0.981636851
b) Calculate L using the formula for L for M/M/1 model (that is, do not use excel template).
Then using little’s law and the relationship between measure of performance, calculate LQ,
W, WQ (that is, do not use excel template and do not use the specific formulas given for
each measure of performance under M/M/1). Only use the little’s law and the relationships
between performance measures to calculate LQ, W, WQ after you calculate L using the
formula under M/M/1.
c) A sandwich is in preparation from the time you received its order via phone until the time
your employee finishes its preparation. As the owner of the store, you think that there
This sysmte is a queueing system
Customers The customer calls for sandwiches
60/15=4 4 customers per hour is the arrival rate
Inter-arrivals are noted to be exponential
Server(s) The employee preparing the sandwiches (single server as there is one employee preparing the sandwiches)
60/10=6 6 customers are served per hour (5 sandwiches are prepared per hour)
The service time is noted to be exponential
Using Kendall's notation, the above queueing systems can be denoted as
M/M/1 because we have exponential interarrivals, exponential service time, 1 server, and no queue capacity.
= 4
= 6
In this part, we are asked to determine various measures of performance
using the explicity formulas derived for M/M/1 systems.
Particularly, we are first asked to determine L, the expected number of people in the system
= 4 from part a
= 6 from part a
Thus, L=4/(6-4) = 2 customers in the system on average
From Little's Law, we know that
L = W
Thus, W=2/4 = 0.5 hour is the expected time a customer waits until he/she get the order (or 30 minutes)
this might be interpreted as the expected time after a order call received until the sandwich is ready
From the relationship between L and Lq, we have
Thus, L q = 2-4/6=1.33 customer orders are in the line waiting on average
From Little's Law, we know that
L q = W q Thus, Wq = 1.33/4=0.333 hours is the expected time a customer order waits until the sandwich ordered starts to be prepared
(or 18 minutes)
=
=
1 L
L = Lq + /
should be less than 3 sandwiches in preparation at least 70% of the time. Determine
whether this criterion is being satisfied currently using the formulas for M/M/1. That is, do
not use the excel template for your calculation.
d) Since sandwiches are standard, each sandwich costs you $3. Suppose you sell sandwiches
for $10 each. Hence, your hourly revenue is the number of orders that you get in one hour
multiplied by 10$. You pay $4 per hour to your employee. What is the current hourly profit
of your store?
e) Suppose that you are considering the following promotion to increase the number of your
customers. If the sandwich is not ready to pick up in 30 minutes after the order call, you sell
the sandwich for its half price, i.e., for $5 instead of $10. You expect that, with this
promotion, more customers will call your store. In particular, you expect that the time
between two consecutive calls for sandwiches will be exponentially distributed with mean
inter-arrival time equal to 12 minutes. Calculate the hourly profit of your store with this
promotion. Recall that each sandwich costs you $3 and you pay $4 per hour to your
employee. Is it more profitable to offer this promotion compared to the case without the
As the owner of the store, you think that there should be less than 3 sandwiches in preparation at least 70% of the time.
So you want to have P0+P1+P2>=0.7
Using the formulas, we have the following probability calculations
P0= 1 / = 1 - 4/6 = 0.333
P1= rho(1-rho)=4/6(1-4/6)= 0.222222
P2= rho^2(1-rho)=16/25(1-4/5)= 0.148148
P0+P1+P2= 0.703 >0.7
thus, this criterion is currently satisfied
So, we sell 4 sandwiches per hour on average, each costs $3 each brings $10
revenues 40 per hour
sandwich cost 12 per hour
labor 4 per hour
profit 24 per hour
promotion? The case without the promotions has the hourly profit that you found in part e).
You can use excel templates provided.
Now, with this promotion, our arrival rate will change, it will be
= 5
= 6 service rate is not changing
We still have M/M/1
Template for the M/M/s Queueing Model
Data Results
1 = 5 (mean arrival rate) L = 5
0 = 6 (mean service rate) Lq = 4.166667
0 s = 1 (# servers)
0 W = 1
0 Pr(W > t) = 0.606531 Wq = 0.833333
0 when t = 0.5
0 = 0.833333
0 Prob(Wq > t) = 0.306566
0 when t = 1 n Pn
0 0 0.166667
0 1 0.138889
0 2 0.115741
0 3 0.096451
11. Dino-Mec (FINAL QUESTION OF Spring 2013)
Dr. Konur is an assistant professor in the Engineering Management and Systems Engineering department at the
Missouri University of Science and Technology. Aside from his work related to the university, he runs a car-
mechanic shop in Rolla, MO called Dino-Mec of Rolla. He works as a mechanic there repairing cars and he is the
only mechanic of the shop. Using his knowledge in statistics, Dr. Konur observed that cars are coming to Dino-
Mec for repair at a rate of 10 per month and the interarrival time is exponentially distributed. Dr. Konur can
repair cars at a rate of 12 per month. The cars are repaired in a first-come first-served manner.
a) Express Dr. Konur’s Dino-Mec shop as a queueing system by defining customers, arrival rate and expected
interarrival time, server or servers, service rate and expected service time. Use Kendall’s notation to define
the Dino-Mec queueing system.
Now, let's calculate the cost of this system
labor cost $4 per hour it did not change
What is changing is our revenues
Now we have 5 sales in an hour
However while some of these sales are $10 per sandwich
some of these sales are $5 because the sandwich was not ready in 30 minutes
in particular, probability of having a sandwich ready in more than 30 minutes is
P(W>30minutes)
And it is equal to 0.606531
So, around 61% of the customers are getting $5 sandwiches
thus 5*0.39*10+5*0.62*0 is the hourly revenues
34.83673 is hourly revenues
you still need to prepare the sandwiches so you pay
5*3=15 dolars and 4 to the employe
your hourly revenue with the promotion is
15.83673 dollars per hour
not good compared to part d
so do not do the promotion
Dr. Konur parks the cars waiting to be repaired inside the shop; but, if there is no space left inside the shop, he
parks them in the outside parking area. The shop can have 2 cars parked inside the shop plus the car that is
currently being repaired.
b) Since parking in the outside parking area is problematic, Dr. Konur wishes to have the cars parked inside the
shop at least 75% of the time on average. Does Dino-Mec satisfy this criterion? Show your calculations on
how you reached your answer; otherwise you will not get any credit. You can use excel template if needed.
From each repair, Dr. Konur earns $1,000 if the repair is completed within a month. The repair time is the time
between the car’s arrival for repair and the car is completely repaired. On the other hand, if the repair time is
more than a month, Dr. Konur charges the half price of the regular repair; hence, he earns $500.
c) If Dr. Konur earns less than $10,000 per month on average, he will consider closing Dino-Mec. Will Dr. Konur
consider closing Dino-Mec? Show your calculations on how you reached your answer; otherwise you will not
get any credit. That is, calculate the expected monthly earnings and compare to $10,000. You can use excel
template if needed.
Customers: Cars coming to be repaired
Arrrival Rate 10 per month
Mean Interaarival Time 3 days (assuming 30 days in a month)
Servers Mechanic (Dr. Konur)
Service Rate 12 per month
Mean Service Time 2.5 days (assuming 30 days in a month)
M/M/1 (Kendall's Notation)
Template for the M/M/s Queueing Model
Data Results
1 = 10 (mean arrival rate) L = 5
0 = 12 (mean service rate) Lq = 4.166666667
0 s = 1 (# servers)
0 W = 0.5
0 Pr(W > t) = 0.135335283 Wq = 0.416666667
0 when t = 1
0 = 0.833333333
0 Prob(Wq > t) = 0.112779403
0 when t = 1 n Pn So, if there are less than or equal to 3 cars in the
0 0 0.166666667 queueing system, then there are no cars parked
0 1 0.138888889 outside. Thus, the percentage of time cars being
0 2 0.115740741 parked out side is:
0 Expected earning from 1 customer Note that probability of earning 3 0.096450617 P0+P1+P2+P3 = 0.51774691
0 regular price = $1000 is equal to Pr(W<1 month) 4 0.080375514
0 1000 Probability of earning $1000 x $1000 and probability of earning $500 is 5 0.066979595 So, 51.78% of the time, the cars
0 discounted price + equal to Pr(W>1 month) 6 0.055816329 are parked inside on average.
0 500 Probability of earning $500 x $500 We know Pr(W<1 month) from 7 0.046513608 This does not satisfy the criteria
0 = the template and 8 0.03876134
0 932.33$ Pr(W>1 month)=1-Pr(W<1 month) 9 0.032301117
0 10 0.026917597
0 11 0.022431331
0 Expected earning in a month 12 0.018692776
0 = 13 0.015577313
0 Expected earning from a customer 14 0.012981094
0 x 15 0.010817579
0 Expected number of customers arriving in a month 16 0.009014649
= 17 0.007512207
9,323.32$ < $10,000 18 0.006260173
Therefore, Dr. Konur will consider 19 0.005216811
closing Dino-Mec 20 0.004347342
Problem 1-c
Problem 1-b
Apparently, one of his students in his EMGT 365 class analyzed Dr. Konur’s Dino-Mec shop and he found out that
Dr. Konur will consider closing the shop. Dr. Konur does not know if this student is right or wrong, but he heard
the rumors so decided to take action. He is considering the following two actions:
Action 1: Increase repair charge to $1,500 and continue on the promotion that if the repair is
not completed within a month, charge the half price, i.e., $750. With this new price, the number
of customers coming to the shop will reduce and the estimated arrival rate will be 7 per month
(interarrival time will still be exponential).
Action 2: Remove the promotion, that is charge $1,000 for any repair. Without the discount
promotion, the number of customers coming to the shop will reduce and the estimated arrival
rate will be 9 per month (interarrival time will still be exponential).
Dr. Konur will take one of these actions.
d) If he wants to have the shop open, he should earn more than $10,000 per month on average. Would Action
1 guarantee that the shop will remain open? Show your calculations on how you reached your answer;
otherwise you will not get any credit. That is, calculate the expected monthly earnings with Action 1 and
compare to $10,000. You can use excel template if needed.
e) If he wants to have the shop open, he should earn more than $10,000 per month on average. Would Action
2 guarantee that the shop will remain open? Show your calculations on how you reached your answer;
otherwise you will not get any credit. That is, calculate the expected monthly earnings with Action 2 and
compare to $10,000. You can use excel template if needed.
Template for the M/M/s Queueing Model
Data Results
1 = 7 (mean arrival rate) L = 1.4
0 = 12 (mean service rate) Lq = 0.816666667
0 s = 1 (# servers)
0 W = 0.2
0 Pr(W > t) = 0.006737947 Wq = 0.116666667
0 when t = 1
0 = 0.583333333
0 Prob(Wq > t) = 0.003930469
0 when t = 1 n Pn
0 0 0.416666667
0 1 0.243055556
0 2 0.141782407
0 Expected earning from 1 customer Note that probability of earning 3 0.082706404 P0+P1+P2+P3 = 0.88421103
0 regular price = $1500 is equal to Pr(W<1 month) 4 0.048245403
0 1500 Probability of earning $1500 x $1500 and probability of earning $750 is 5 0.028143151 So, 88.42% of the time, the cars
0 discounted price + equal to Pr(W>1 month) 6 0.016416838 are parked inside on average.
0 750 Probability of earning $750 x $750 We know Pr(W>1 month) from 7 0.009576489 Action 1 satisfies the criteria
0 = the template and 8 0.005586285
0 1,494.95$ Pr(W<1 month)=1-Pr(W>1 month) 9 0.003258666
0 10 0.001900889
0 11 0.001108852
0 Expected earning in a month 12 0.00064683
0 = 13 0.000377318
0 Expected earning from a customer 14 0.000220102
0 x 15 0.000128393
0 Expected number of customers arriving in a month 16 7.48958E-05
= 17 4.36892E-05
10,464.63$ > $10,000 18 2.54854E-05
Therefore, Dr. Konur will not 19 1.48665E-05
close Dino-Mec 20 8.67211E-06
Problem 1-c
Will be needed in part f)
f) (5 points) Which action should Dr. Konur take if he wants to keep Dino-Mec open? Will this action satisfy the
criterion that cars are being parked inside the shop at least 75% of the time? You can use excel template if
needed.
12. Dino-Bank (FINAL QUESTION OF Spring 2013)
Dr. Konur is an assistant professor in the Engineering Management and Systems Engineering department at the
Missouri University of Science and Technology. Aside from his work related to the university, he manages a local
bank in Rolla, MO called Dino-Bank of Rolla. He must now decide on how many tellers are needed in the bank.
Specifically, Dr. Konur hires his students to work as tellers. The interarrival time between customers is
exponentially distributed with a mean interarrival time of 30 seconds. Customers arriving at the bank wait in a
single line and go to the tellers in a first-come-first served manner. Dr. Konur estimates that his students can
complete a customer’s transaction in 2 minutes on average and the service time is exponentially distributed.
Template for the M/M/s Queueing Model
Data Results
1 = 9 (mean arrival rate) L = 3
0 = 12 (mean service rate) Lq = 2.25
0 s = 1 (# servers)
0 W = 0.333333333
0 Pr(W > t) = 0.049787068 Wq = 0.25
0 when t = 1
0 = 0.75
0 Prob(Wq > t) = 0.037340301
0 when t = 1 n Pn
0 0 0.25
0 1 0.1875
0 2 0.140625
0 Earning from 1 customer 3 0.10546875 P0+P1+P2+P3 = 0.68359375
0 regular price = 4 0.079101563
0 1000 $1,000 5 0.059326172 So, 68.36% of the time, the cars
0 discounted price 6 0.044494629 are parked inside on average.
0 no discount Expected earning in a month 7 0.033370972 Action 2 does not satisfy the criteria
0 = 8 0.025028229
0 Earning from a customer 9 0.018771172
0 x 10 0.014078379
0 Expected number of customers arriving in a month 11 0.010558784
0 = 12 0.007919088
0 9,000.00$ < $10,000 13 0.005939316
0 Therefore, Dr. Konur will 14 0.004454487
0 close Dino-Mec 15 0.003340865
Problem 1-c
Will be needed in part f)
As is clear from parts d and e, Dr. Konur should
take Action 1 to keep Dino-Mec open.
Furthermore, this action satisfies the criteria.
Cost Inside Parking Cost Inside Parking
10,464.63$ 88.42% 9,000.00$ 68.36%
So, pick Action 1
Action 1 Action 2
a) What is the minimum number of students (tellers) Dr. Konur has to hire so that Dino-Bank will reach a
steady-state, i.e., the Dino-Bank will not ‘blow up’? Briefly explain why.
Now, Dr. Konur decided to hire the minimum number of tellers needed. Each teller is paid at a rate of $12 per
hour. And, Dr. Konur believes that each customer standing in the line for a minute has a delay cost of 10¢.
b) Calculate the expected hourly cost of Dino-Bank with the minimum number of tellers needed. Expected
hourly cost is equal to teller costs plus expected delay costs. You can use excel template if needed.
The interarrival time between customers is exponentially distributed with a mean interarrival time of 30 seconds.
This means that the arrival rate is 2 per minute
Dr. Konur estimates that his students can complete a customer’s transaction in 2 minutes on average and the service time is exponentially distributed.
This means that the service rate is 0.5 per minute
Recall that to reach a steady-state the utilization factor should be less than 1. In multiple server queues, utilization factor is
where s is the number of servers. Thus, to have
s <1 we should have s>4
Therefore, the minimum number of tellers needed is 5
=
c) Would adding one teller decrease or increase the expected hourly cost of Dino-Bank? You can use excel
template if needed.
Template for the M/M/s Queueing Model
Data Results
= 2 (mean arrival rate) L = 6.216450216
= 0.5 (mean service rate) Lq = 2.216450216
s = 5 (# servers)
W = 3.108225108
Pr(W > t) = 0.774573786 Wq = 1.108225108
when t = 1
= 0.8
Prob(Wq > t) = 0.336086253
when t = 1 n Pn
0 0.012987013
1 0.051948052
2 0.103896104
Expected delay cost for 1 customer Note that probability of earning 3 0.138528139
= expected delay for 1 customer is 4 0.138528139
Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.110822511
Per Minute x in the line. This is given by Wq 6 0.088658009
0.10$ Delay cost per minute 7 0.070926407
= 8 0.056741126
0.1108225$ per customer 9 0.0453929
10 0.03631432
11 0.029051456
Expected delay cost per minute 12 0.023241165
= 13 0.018592932
Expected delay cost for 1 customer 14 0.014874346
x 15 0.011899476
Expected number of customers arriving in minute 16 0.009519581
= 17 0.007615665
0.2216450$ per minute 18 0.006092532
19 0.004874026
Expected delay cost per hour 20 0.00389922
= 21 0.003119376
13.2987$ per hour 22 0.002495501
Teller Cost 23 0.001996401
Per hour Teller Cost per hour 24 0.001597121
$12 = 25 0.001277697
Per teller cost per hour
x
Number of tellers
=
$60
Expected Hourly Cost
=
Expected Hourly Delay Cost
+
Hourly Tellers Cost
=
73.30$
Dr. Konur thinks that adding another teller will not help. After thinking about Dino-Bank for days and days, he
came up with the following idea: instead of having a single line with minimum number of tellers needed, he will
have 5 different lines such that each line is for each individual teller. In this case, he will manage the customer
arrivals such that customers arrive at each teller’s line with a mean interarrival time of 2.5 minutes, which is
Template for the M/M/s Queueing Model
Data Results
= 2 (mean arrival rate) L = 4.569521691
= 0.5 (mean service rate) Lq = 0.569521691
s = 6 (# servers)
W = 2.284760845
Pr(W > t) = 0.674489182 Wq = 0.284760845
when t = 1
= 0.666666667
Prob(Wq > t) = 0.104757661
when t = 1 n Pn
0 0.016685206
Problem 1-c 1 0.066740823
2 0.133481646
Expected delay cost for 1 customer Note that probability of earning 3 0.177975528
= expected delay for 1 customer is 4 0.177975528
Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.142380423
Per Minute x in the line. This is given by Wq 6 0.094920282
0.10$ Delay cost per minute 7 0.063280188
= 8 0.042186792
0.0284761$ per customer 9 0.028124528
10 0.018749685
11 0.01249979
Expected delay cost per minute 12 0.008333193
= 13 0.005555462
Expected delay cost for 1 customer 14 0.003703642
x 15 0.002469094
Expected number of customers arriving in minute 16 0.001646063
= 17 0.001097375
0.0569522$ per minute 18 0.000731584
19 0.000487722
Expected delay cost per hour 20 0.000325148
= 21 0.000216765
3.4171$ per hour 22 0.00014451
Teller Cost 23 9.63402E-05
Per hour Teller Cost per hour 24 6.42268E-05
$12 = 25 4.28179E-05
Per teller cost per hour
x
Number of tellers
=
$72
Expected Hourly Cost
= Note that expected delay costs decreased
Expected Hourly Delay Cost with the additional teller, however, the cost increase
+ due to adding one more teller is greater than the
Hourly Tellers Cost cost decrease in delays due to adding one more teller.
= Hence, the costs will increase with the additional
75.42$ teller.
exponentially distributed. Each teller will still have exponential service time distribution with mean service time
of 2 minutes. Still, each customer standing in the line for a minute has a delay cost of 10¢.
d) Will the idea of having separate lines reduce expected costs compared to the expected costs with the single
line and the minimum number of tellers needed (i.e., the expected cost calculated in part b)? You can use
excel template if needed. (hint: each teller is a separate but identical queueing system under the new idea).
Suppose that Dr. Konur decided that the new idea is not good and he will continue with a single line and the
minimum number of servers. Now, answer the following questions about Dino-Bank independent of each other
and briefly explain your answers:
Template for the M/M/s Queueing Model
Data Results
1 = 0.4 (mean arrival rate) L = 4
0 = 0.5 (mean service rate) Lq = 3.2
0 s = 1 (# servers)
0 W = 10
0 Pr(W > t) = 0.904837418 Wq = 8
0 when t = 1
0 = 0.8
0 Prob(Wq > t) = 0.723869934
0 when t = 1 n Pn
0 0 0.2
0 Problem 1-c 1 0.16
0 2 0.128
0 Expected delay cost for 1 customer Note that probability of earning 3 0.1024
0 = expected delay for 1 customer is 4 0.08192
0 Delay Cost Expected delay time for 1 customer (minutes) the expected time a customer waits 5 0.065536
0 Per Minute x in the line. This is given by Wq 6 0.0524288
0 0.10$ Delay cost per minute 7 0.04194304
0 = 8 0.033554432
0 0.8000000$ per customer 9 0.026843546
0 10 0.021474836
0 11 0.017179869
0 Expected delay cost per minute 12 0.013743895
0 = 13 0.010995116
0 Expected delay cost for 1 customer 14 0.008796093
0 x 15 0.007036874
0 Expected number of customers arriving in minute 16 0.0056295
= 17 0.0045036
0.3200000$ per minute 18 0.00360288
19 0.002882304
Expected delay cost per hour 20 0.002305843
= 21 0.001844674
19.2000$ per hour 22 0.00147574
Teller Cost 23 0.001180592
Per hour Teller Cost per hour 24 0.000944473
$12 = 25 0.000755579
Per teller cost per hour
x
Number of tellers
=
$12
Expected Hourly Cost Now, this cost is the expected hourly
= cost of a single line-single teller
Expected Hourly Delay Cost We will have 5 of these, which are all
+ identical.
Hourly Tellers Cost
= Then the total cost will be
31.20$ 156.00$
Compared to single-line-multiple tellers of part b
separating lines increases costs
e) If service time of each teller is reduced by the same amount and still exponentially distributed, will the
average number of people waiting in the line increase or decrease? Explain why?
Since service time is reduced, the service rate will increase, hence, the utilization factor will decrease. This means that the average number of people in the line will decrease
f) If one more additional server, with the same characteristics with the other tellers, is hired, will Dino-Bank be
subject to higher or lower delay costs? Explain why?
g) Suppose that Fridays are busier than the other days; hence, the arrival rate of the customers increases. Will
this increase or decrease the expected delay costs? Explain why?
This will increase the utilization factor This will increase average waiting time in line; hence, this will increase
expected delay cost for 1 customer. Thus, Dino-Bank will have higher delay costs.
h) Instead of working with multiple student tellers, Dr. Konur decided to be the single teller of Dino-Bank as he
can complete a customer transaction way faster than his students and he works for lower salaries. He can
complete 3 customer transactions per minute. The customer interarrival time is still exponentially
distributed with mean of 30 seconds. Since Dr. Konur does not like staying idle, whenever there is no
customer in Dino-Bank, he plays Angry-birds. What percentage of time Dr. Konur is playing Angry birds?
Explain why?
i) Still suppose that Dr. Konur is the single teller and Fridays are busier than the other days. Will this increase
the difference between the expected number of people in Dino-Bank and the expected number of people in
the line? Explain why?
We already know the answer to this by comparing parts b and c. Additional server will decrease the
utilization factor. This will reduce average waiting time in line; hence, this will reduce
expected delay cost for 1 customer. Thus, Dino-Bank will have lower delay costs.
Now, we have a single server system and the idle time percentage of the server is 1-utilization factor.
arrival rate = 2 per minute
service rate = 3 per minute
utilization factor = 2/3 = 0.667
Thus, Dr. Konur will be idle for 0.333, or 33.33% of the time playing Angry-birds.
We know that
L = L q + /
That is, the utilization factor is the difference between the expected number of people in Dino-Bank and the
expected number of the people in line.
Since Fridays are busier, it means that the arrival rate increases, hence, the utilization factor increases. Therefore,
This will increase the difference between the expected number of people in Dino-Bank and the expected number of
people in the line.
