Practice Exam 2 Solutions
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McNeese State University *
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Course
382
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
Pages
8
Uploaded by tdavis4751234
Practice Exam 2 Solutions
1
The M/M/
s
Queue
An insurance company has a call center handling customer inquiries. Data shows that on average
50 calls per hour need to be handled (with the exponential distribution as a reasonable picture of
interarrival times). The time needed to help a customer (service or processing time) is exponentially
distributed with a mean of three minutes.
1. How many operators are needed such that the mean waiting time is less than one minute?
(Hint: a waiting time of
E
[
W
q
]
<
1 minute corresponds to a utilization of
ρ
= 0
.
625.)
Answer
: First, we solve for capacity as
µ
= (1
/
E
[
T
])
×
60 = (1
/
3)
×
60 = 20 customers per
hour. Then, we can manipulate the equation for
ρ
to solve for
s
as:
s
=
λ
ρµ
=
50
0
.
625 (20)
= 4 operators
2. Using the above inputs of
λ
and
ρ
along with your derived values for
µ
and
s
, what is the
expected number of customers in the system?
Answer
: We derive the appropriate metric as follows:
E
[
L
] =
E
[
L
q
] +
λ
µ
=
ρ
√
2(
s
+1)
1
−
ρ
+
λ
µ
=
0
.
625
√
2(4+1)
1
−
0
.
625
+
50
20
= 0
.
6 + 2
.
5
= 3
.
1 customers
1
2
Putting it All Together
A bank is determining the staffing level for their branch located within the local university. The
beginning of the fall semester is their busiest time, with incoming freshmen and transfer students
opening accounts. On average, it takes a teller
E
[
T
] = 8 minutes to successfully open an account.
Currently, the average arrival rate has been eight customers per hour. As always, the interarrival
and service times are exponentially distributed.
3. Suppose the bank hired two tellers with dedicated queues (that is, there are two M/M/1
queues). Calculate the average time in line (hours) if each served
λ/
2 of the arrivals.
Answer
: Using
µ
= 60
/
8 = 7
.
5 customers per hour (or its equivalent
µ
= (1
/
E
[
T
])
×
60 =
(1
/
8)
×
60 = 7
.
5), we calculate the three desired metrics for each server in the form:
ρ
=
λ
2
sµ
=
8
2
1 (7
.
5)
= 0
.
53
E
[
L
q
] =
ρ
√
2(
s
+1)
1
−
ρ
=
0
.
53
√
2(1+1)
1
−
0
.
53
= 0
.
6 customers
E
[
W
q
] =
E
[
L
q
]
λ
=
0
.
6
4
= 0
.
15 hours
4. Consider now a pooled queue (that is, an M/M/2 queue) and calculate the average time in
line (hours). Comparing to the above layout, what did you find?
Answer
: Repeating the above procedure, we get:
ρ
=
8
2 (7
.
5)
= 0
.
53
E
[
L
q
] =
0
.
53
√
2(2+1)
1
−
0
.
53
= 0
.
45 customers
E
[
W
q
] =
0
.
45
8
= 0
.
06 hours
2
Remember we are essentially comparing “
s
M/M/1 queues vs. one M/M/
s
queue”. Although
a single line “looks” longer, a pooled queue is better as it reduces variability and idling as
waiting time decreases. Therefore, one M/M/
s
queue is better than
s
M/M/1 queues.
5. The bank is now considering adding a third teller to reduce wait times and customer com-
plaints. Their goal is to solve the following argument:
arg
min
s
∈{
2
,
3
}
$12
s
+ $9
E
[
L
q
]
Here, the term “arg min” or “argument of the minima” gives the solution
s
such that the
function
f
(
s
) = $12
s
+ $9
E
[
L
q
] attains its minimum value. What is the optimal number of
tellers to staff?
Answer
: Initially, we need to calculate
E
[
L
q
] for
s
= 3 tellers:
ρ
=
8
3 (7
.
5)
= 0
.
36
E
[
L
q
] =
0
.
36
√
2(3+1)
1
−
0
.
36
= 0
.
09 customers
Finally, using the cost function provided, we calculate:
f
(2) = $12 (2) + $9 (0
.
45)
= $28
.
05 per hour
f
(3) = $12 (3) + $9 (0
.
09)
= $36
.
81 per hour
Therefore, the optimal service level is
s
∗
= 2.
Notice that there was no need to calculate
any metrics for
s
= 3 since the cost of providing the service for three tellers is $12 (3) = $36,
which is higher than the total cost of
s
= 2. Therefore, the bank should hire two tellers.
3
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3
Tandem Queues
Subway (located near the McNeese campus) is analyzing its simplified, three-step process for a
given time period. A customer arrives and waits in the bread queue (where a single server in an
M/M/1 layout prepares the bread, meat, and cheese), then proceeds to an M/M/1 toppings queue
(toppings and wrapping the completed sandwich), and finally ends in an M/M/1 payment queue
(packaging and payment) before exiting the system. The manager has provided us with an arrival
rate of
λ
= 0
.
4 customers per minute and the following metrics:
Bread queue
Toppings queue
Payment queue
Queue structure
one M/M/1 queue
one M/M/1 queue
one M/M/1 queue
Arrival rate
λ
per minute
λ
per minute
λ
per minute
Service time
E
[
T
] = 2 minutes
E
[
T
] = 1
.
5 minutes
E
[
T
] = 0
.
5 minutes
6. What is the mean service rate
µ
bread
(in customers per minute) for the
bread queue
? How
about the mean number of customers in line for the
bread queue
,
E
[
L
q,
bread
]?
Answer
: Using the appropriate formulas, we calculate:
µ
bread
=
1
E
[
T
bread
]
=
1
2
= 0
.
5 customers
ρ
bread
=
λ
µ
bread
=
0
.
4
0
.
5
= 0
.
8
E
[
L
q,
bread
] =
ρ
2
bread
1
−
ρ
bread
=
0
.
8
2
1
−
0
.
8
= 3
.
2 customers
4
7. Determine the distribution of the variable
L
bread
for the
bread queue
. What are the de-
scriptive statistics (mean
E
[
L
bread
] and standard deviation
σ
L
bread
) for the average number
of customers (waiting and being served) at this
bread queue
?
Answer
: With the utilization factor
ρ
bread
=
λ/µ
bread
= 0
.
4
/
0
.
5 = 0
.
8 (calculated in the
above question) and the identity of the geometric parameter equal to 1
−
ρ
bread
, we have that
L
bread
∼
Geometric (0
.
2). It follows that:
E
[
L
bread
] =
ρ
bread
1
−
ρ
bread
=
0
.
8
1
−
0
.
8
= 4 customers
σ
L
bread
=
q
σ
2
L
bread
=
r
ρ
(1
−
ρ
)
2
=
√
ρ
1
−
ρ
=
√
0
.
8
1
−
0
.
8
= 4
.
5 customers
8. What is the probability of less than-or equal-to three customers in the system,
P
(
L
bread
≤
3),
for the
bread queue
?
Answer
: We can simply subtract the complement (given) on the formula sheet from one and
get the desired solution:
P
(
L
bread
≤
3) = 1
−
P
(
L
bread
>
3)
= 1
−
ρ
3+1
= 1
−
0
.
8
4
= 0
.
59
5
9. What is the average number of customers in the three-step process (
bread queue
,
toppings
queue
, and
payment queue
),
E
[
L
bread
+
L
toppings
+
L
payment
]? Here, I’m using this no-
tation instead of the generalized “
E
[
L
]” from the two-step McDonald’s example (see the extra
problems file), but it means the same thing.
Answer
:
To begin, we know that both the
toppings queue
and
payment queue
are
M/M/1 layouts. Therefore, the total number in the system,
L
toppings
, follows a geometric
distribution with parameter 1
−
ρ
toppings
. Given
µ
toppings
= 1
/
E
[
T
toppings
] = 1
/
1
.
5 = 0
.
67
and
ρ
toppings
=
λ/µ
toppings
= 0
.
4
/
0
.
67 = 0
.
6, we have
L
toppings
∼
Geometric (0
.
4) and:
E
[
L
toppings
] =
ρ
toppings
1
−
ρ
toppings
=
0
.
6
1
−
0
.
6
= 1
.
5 customers
Of course a similar argument follows for the
payment queue
, resulting in
µ
payment
=
1
/
E
[
T
payment
] = 1
/
0
.
5 = 2,
ρ
payment
=
λ/µ
payment
= 0
.
4
/
2 = 0
.
2, and
L
payment
∼
Geometric (0
.
8), accordingly.
From the above information, we can then derive the average
number in the three-step system as follows:
E
[
L
bread
+
L
toppings
+
L
payment
] =
E
[
L
bread
] +
E
[
L
toppings
] +
E
[
L
payment
]
= 4 + 1
.
5 +
ρ
payment
1
−
ρ
payment
= 5
.
5 +
0
.
2
1
−
0
.
2
= 5
.
5 + 0
.
25
= 5
.
75 customers
10. How long does an average customer spend in the three-step process (in minutes) from the
time they arrive,
E
[
W
bread
+
W
toppings
+
W
payment
]?
Answer
: The simplest way to solve this problem is to use Little’s Law in its usual form:
E
[
W
bread
+
W
toppings
+
W
payment
] =
E
[
L
bread
+
L
toppings
+
L
payment
]
λ
=
5
.
75
0
.
4
= 14
.
375 minutes
6
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4
Combining Tasks
Recall the Subway example from the previous section. The metrics are as follows:
Bread queue
Toppings queue
Payment queue
Queue structure
one M/M/1 queue
one M/M/1 queue
one M/M/1 queue
Arrival rate
λ
per minute
λ
per minute
λ
per minute
Service time
E
[
T
] = 2 minutes
E
[
T
] = 1
.
5 minutes
E
[
T
] = 0
.
5 minutes
An employee has called-in sick to work, so the manager now has to combine the toppings and
payment queues. As such, the layout is now a two-step process.
11. Calculate the utilization for the new
toppings/payment queue
.
Answer
: Given
µ
toppings/payment
= 1
/
E
[
T
toppings
+
T
payment
] = 1
/
(1
.
5 + 0
.
5) = 1
/
2 = 0
.
5
customers per minute, we get:
ρ
toppings/payment
=
λ
µ
toppings/payment
=
0
.
4
0
.
5
= 0
.
8
12. What is the average number of customers in the two-step process (
bread queue
and
top-
pings/payment queue
),
E
L
bread
+
L
toppings/payment
? Comparing your answer to ques-
tion nine, what do you notice?
Answer
: Since the utilization is the same for
toppings/payment queue
as it was for the
bread queue
, we calculate:
E
L
bread
+
L
toppings/payment
=
E
[
L
bread
] +
E
L
toppings/payment
= 4 + 4
= 8 customers
Notice this is larger than the three-step process calculated in question nine. Remember that
queue length (in the form
E
[
L
q
] or
E
[
L
]) is strongly nonlinear with respect to utilization—so,
adding workers has a tremendous effect on the length of the line (even if it adds another queue
to the process)!
7
13. How long does an average customer spend in the two-step process (in minutes) from the time
they arrive,
E
W
bread
+
W
toppings/payment
?
Answer
: Applying Little’s Law, we get:
E
W
bread
+
W
toppings/payment
=
E
L
bread
+
L
toppings/payment
λ
=
8
0
.
4
= 20 minutes
8