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Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 6th Edition Test Bank Cliff Ragsdale
Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 6th Edition Test Bank Cliff Ragsdale
Sample
Chapter 13—Queuing Theory
MULTIPLE CHOICE
1.Which of the following best describes queuing theory?
a. 
The study of arrival rates. 
b. 
The study of service times. 
c. 
The study of waiting lines. 
d. 
The evaluation of service time costs. 
ANS: C PTS: 1
2.Which of the following is a reason to employ queuing theory?
a. 
To reduce customer wait time in line. 
b. 
To reduce service times. 
c. 
To generate more arrivals to the system. 
d. 
To reduce worker idle time in line. 
ANS: A PTS: 1
3.If the service rate decreases as the arrival rate remains constant, then, in general
a. 
customer waiting time increases. 
b. 
customer waiting time decreases. 
c. 
service costs increase. 
d. 
customer dissatisfaction decreases. 
ANS: A PTS: 1
4.Which type of queuing system are you likely to encounter at an ATM?
a. 
Single waiting line, single service station. 
b. 
Multiple waiting lines, single service station. 
c. 
Single waiting line, multiple service stations. 
d. 
Multiple waiting lines, multiple service stations. 
ANS: A PTS: 1
5.Which type of queuing system are you likely to encounter at a grocery store?
a. 
Single waiting line, single service station. 
b. 
Multiple waiting lines, single service station. 
c. 
Single waiting line, multiple service stations. 
d. 
Multiple waiting lines, multiple service stations. 
ANS: D PTS: 1
6.Which type of queuing system are you likely to encounter at a Wendy’s restaurant?
a. 
Single waiting line, single service station. 
b. 
Multiple waiting lines, single service station. 
c. 
Single waiting line, multiple service stations. 
d. 
Multiple waiting lines, multiple service stations. 
ANS: C PTS: 1
7.What is the service policy in the queuing systems presented in this chapter that is considered “fair” by the customers?
a. 
FIFO 
b. 
LIFO 
c. 
FILO 
d. 
Priority 
ANS: A PTS: 1
8.If the arrival process is modeled as a Poisson random variable with arrival rate λ, then the average time between arrivals is
a. 
1/μ 
b. 
1/λ 
c. 
1/λ^{2} 
d. 
σ 
ANS: B PTS: 1
9.For a Poisson random variable, λ represents the ____ number of arrivals per time period
a. 
maximum 
b. 
minimum 
c. 
average 
d. 
standard deviation of 
ANS: C PTS: 1
10.What is the formula for the probability of x arrivals, p(x), under a Poisson distribution with arrival rate λ?
a. 

b. 

c. 

d. 
ANS: B PTS: 1
11.A Poisson distribution shape can be described as
a. 
slightly skewed to the left. 
b. 
symmetric around the parameter λ. 
c. 
skewed to the right. 
d. 
discrete so it lacks any definable shape. 
ANS: C PTS: 1
Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 6th Edition Test Bank Cliff Ragsdale
12.If the number of arrivals in a time period follow a Poisson distribution with mean λ then the interarrival times follow a(n) ____ distribution with mean ____.
a. 
normal; μ 
b. 
constant; λ 
c. 
exponential; λ 
d. 
exponential; 1/λ 
ANS: D PTS: 1
13.The number of arrivals to a store follows a Poisson distribution with mean λ = 10/hour. What is the mean interarrival time?
a. 
6 seconds 
b. 
6 minutes 
c. 
10 minutes 
d. 
10 hours 
ANS: B PTS: 1
14.An arrival process is memoryless if
a. 
the time until the next arrival depends on the time elapsed since the last arrival. 
b. 
the time until the next arrival is based on the time elapsed since the last arrival. 
c. 
the time until the next arrival does not depend on the time elapsed since the last arrival. 
d. 
the time until the next arrival is based on the arrival rate. 
ANS: C PTS: 1
15.The memoryless property is also referred to as the ____ property.
a. 
Markov 
b. 
Erlang 
c. 
Poisson 
d. 
Normal 
ANS: A PTS: 1
16.What is the mean arrival rate based on the following 8 arrival rate observations?
Number of arrivals per hour: 6, 5, 3, 4, 7, 6, 4, 5
a. 
3 
b. 
4 
c. 
5 
d. 
6 
ANS: C PTS: 1
17.What is the formula for P(t ≤ T) under the exponential distribution with rate μ?
a. 
1 − e^{μ}^{T} 
b. 
e^{μ}^{T} 
c. 
1 − e−^{μ}^{T} 
d. 
1 − e^{T} 
ANS: C PTS: 1
18.If cell B2 contains the value for μ and cell A5 contains the value for T, what formula should go in cell B5 to compute the P(Service time) ≤ T for this exponential distribution?
a. 
=1EXP($B$2*A5) 
b. 
=EXP($B$2*A5) 
c. 
=1EXP($B$2) 
d. 
=1EXP($B$2*A5) 
ANS: D PTS: 1
19.Which of the following is the typical operating characteristic for average number of units in a queue?
a. 
W 
b. 
W_{q} 
c. 
L 
d. 
L_{q} 
ANS: D PTS: 1
20.Which of the following is the typical operating characteristic for average time a unit spends waiting for service?
a. 
W 
b. 
W_{q} 
c. 
L 
d. 
L_{q} 
ANS: B PTS: 1
21.Which of the following is the typical operating characteristic for the probability an arriving unit has to wait for service?
a. 
W_{p} 
b. 
P_{0} 
c. 
P_{w} 
d. 
P_{n} 
ANS: C PTS: 1
22.The amount of time a customer spends with the server is referred to as
a. 
system time. 
b. 
queue time. 
c. 
service time. 
d. 
served time. 
ANS: C PTS: 1
23.What is the probability that it will take less than or equal to 0.25 hours to service any call based on the following exponential probability distribution with rate μ = 5?
Average Service Rate 

μ = 
5 
per hour 
T 
P(service time ≤ T) 
0.00 
0.00 
0.05 
0.22 
0.10 
0.39 
0.15 
0.53 
0.20 
0.63 
0.25 
0.71 
0.30 
0.78 
0.35 
0.83 
0.40 
0.86 
0.45 
0.89 
0.50 
0.92 
0.55 
0.94 
0.60 
0.95 
0.65 
0.96 
a. 
0.00 
b. 
0.71 
c. 
0.92 
d. 
1.00 
ANS: B PTS: 1
24.A company has recorded the following list of service rates (customers/hour) for one of its servers. What is the mean service time for this server?
Customers / hour: 4, 4, 5, 6, 5, 4, 3, 4, 3, 5, 5, 6
a. 
0.22 min 
b. 
1.11 min 
c. 
4.5 min 
d. 
13.3 min 
ANS: D PTS: 1
25.The standardized queuing system notation such as M/M/1 or M/G/2 is referred to as
a. 
Kendall notation. 
b. 
Erlang notation. 
c. 
Poisson notation. 
d. 
Queuing notation. 
ANS: A PTS: 1
26.The M in M/G/1 stands for
a. 
Markovian interarrival times. 
b. 
Mendelian interarrival times. 
c. 
Mean interarrival times. 
d. 
Mathematical interarrival times. 
ANS: A PTS: 1
27.A barber shop has one barber, a Poisson arrival rate and exponentially distributed service times. What is the Kendall notation for this system?
a. 
M/M/E 
b. 
M/M/1 
c. 
M/E/1 
d. 
P/M/1 
ANS: B PTS: 1
28.If a service system has a constant service time, Poisson arrival rates and 2 servers its Kendall notation is
a. 
P/D/2 
b. 
M/D/2 
c. 
M/D/1 
d. 
G/D/2 
ANS: B PTS: 1
29.To find steadystate values for the M/M/S queuing system, which of the following statements must be true about the arrival rate?
a. 
λ < s μ 
b. 
λ − s = μ 
c. 
λ > s μ 
d. 
λ = s μ 
ANS: A PTS: 1
Exhibit 13.1
The following questions are based on the output below.
A store currently operates its service system with 1 operator. Arrivals follow a Poisson distribution and service times are exponentially distributed. The following spreadsheet has been developed for the system.
M/M/s queuing computations 

Arrival rate 
6 

Service rate 
8 

Number of servers 
1 
(max of 40) 

Utilization 
75.00% 

P(0), probability that the system is empty 
0.2500 

Lq, expected queue length 
2.2500 

L, expected number in system 
3.0000 

Wq, expected time in queue 
0.3750 

W, expected total time in system 
0.5000 

Probability that a customer waits 
0.7500 
30.Refer to Exhibit 13.1. What is the probability that a customer must wait in queue before being served?
a. 
0.00 
b. 
0.25 
c. 
0.75 
d. 
1.00 
ANS: C PTS: 1
31.Refer to Exhibit 13.1. What is average amount of time spent waiting in line?
a. 
0.375 
b. 
0.50 
c. 
2.25 
d. 
3.00 
ANS: A PTS: 1
32.Refer to Exhibit 13.1. What is the probability that a customer can go directly into service without waiting in line?
a. 
0.00 
b. 
0.25 
c. 
0.75 
d. 
1.00 
ANS: B PTS: 1
33.Refer to Exhibit 13.1. How many customers will be in the store on average at any one time?
a. 
0.375 
b. 
0.50 
c. 
2.25 
d. 
3.00 
ANS: D PTS: 1
34.If a company adds an additional identical server to its M/M/1 system, making an M/M/2 system, what happens to a customer’s average service time?
a. 
increases 
b. 
decreases 
c. 
it is unchanged 
d. 
depends on the arrival rate 
ANS: C PTS: 1
35.A balk refers to
a. 
a customer who refuses to join the queue. 
b. 
a customer who refuses service by a specific server. 
c. 
a customer who joins the queue but leaves before service is complete. 
d. 
a customer who requires extra service time. 
ANS: A PTS: 1
36.A doctor’s office only has 8 chairs. The doctor’s service times and customer interarrival times are exponentially distributed. What type of system is it?
a. 
M/M/1 
b. 
M/M/8 
c. 
M/M/1 with Finite Queue 
d. 
M/M/1 with Finite Population 
ANS: C PTS: 1
37.Joe’s Copy Center has 10 copiers. They break down and require service quite often. Time between breakdowns follows an exponential distribution for each copier. The repair person services machines as quickly as possible, but the service time follows an exponential distribution. What type of system is it?
a. 
M/M/1 with Finite Population 
b. 
M/M/1 with Finite Queue 
c. 
M/M/1 
d. 
M/M/10 
ANS: A PTS: 1
38.Joe’s Copy Center has 10 copiers. They break down at a rate of 0.02 copiers per hour and are sent to the service facility. What is the average arrival rate of broken copiers to the service facility?
a. 
0.02 
b. 
0.2 
c. 
10 
d. 
It cannot be determined from the information provided. 
ANS: B PTS: 1
39.The service times for a grocery store with one checkout line have a mean of 3 minutes and a standard deviation of 20 seconds. Customer arrivals at the checkout stand follow a Poisson distribution. What type of system is it?
a. 
M/G/1 
b. 
M/D/1 
c. 
G/M/1 
d. 
M/M/1 
ANS: A PTS: 1
40.A store is considering adding a second clerk. The customer arrival rate at this new server will be
a. 
twice the old rate. 
b. 
half the old rate. 
c. 
the same as the old rate. 
d. 
unpredictable. 
ANS: B PTS: 1
41.The M/D/1 model results can be derived from which of the following systems?
a. 
M/M/1 with λ = 0 
b. 
M/G/1 with μ = 0 
c. 
M/G/1 with σ = 0 
d. 
M/M/2 with finite queue length. 
ANS: C PTS: 1
42.A renege refers to
a. 
a customer who refuses to join the queue. 
b. 
a customer who refuses service by a specific server. 
c. 
a customer who joins the queue but leaves before service is complete. 
d. 
a customer who requires extra service time. 
ANS: C PTS: 1
43.A jockey refers to
a. 
a customer who refuses to join the queue. 
b. 
a customer who refuses service by a specific server. 
c. 
a customer who joins the queue but leaves before service is complete. 
d. 
a customer who switches between queues in the system. 
ANS: D PTS: 1
PROBLEM
44.What is the mean arrival rate based on the following 10 arrival rate observations?
Number of arrivals per hour 
16 
15 
14 
15 
17 
16 
14 
15 
13 
16 
ANS:
15.1 arrivals per hour
PTS: 1
45.A company has recorded the following list of service rates (customers/hour) for one of its servers. What is the mean service time for this server?
Customers 
/ hour 
2 
3 
3 
4 
2 
3 
3 
3 
4 
2 
4 
3 
2 
4 
ANS:
3 customers per hour gives a mean service time of 20 minutes per customer
PTS: 1
46.Customers arrive at a store randomly, following a Poisson distribution at an average rate of 90 per hour. How many customers arrive per minute, on average?
ANS:
1.5 customers per minute
PTS: 1
47.Customers arrive at a store randomly, following a Poisson distribution at an average rate of 90 per hour. How many customers would you expect to arrive in a 20 minute period?
ANS:
30 customers per 20 minute period
PTS: 1
48.Customers arrive at a store randomly, following a Poisson distribution at an average rate of 20 per hour. What is the probability of exactly 0, 1 2, and 3 arrivals in a 15 minute period?
ANS:
λ = 5 customers/15 minute period
Arrivals 
Probability 
0 
0.0067 
1 
0.0336 
2 
0.0842 
3 
0.1404 
PTS: 1
49.A grocery clerk can serve 20 customers per hour on average and the service time follows an exponential distribution. What is the expected service time per customer?
ANS:
3 minutes per customer
PTS: 1
50.A grocery clerk can serve 20 customers per hour on average and the service time follows an exponential distribution. What is the probability that a customer’s service time is less than 2 minutes?
ANS:
T = 2/60 = 0.033 hours
P(t < T) = 1 − e^{−μ}^{T} = 1 − e^{−}^{20*0.033} = 1 − 0.5134 = 0.4866
PTS: 1
51.A grocery clerk can serve 20 customers per hour on average and the service time follows an exponential distribution. What is the probability that a customer’s service time is more than 4 minutes?
ANS:
T = 4/60 = 0.0667 hours
P(t > T) = e^{−μ}^{T} = e^{−}^{20*0.0667} = 0.2636
PTS: 1
52.The customer service desk at Joe’s Discount Electronics store receives 5 customers per hour on average. On average, each customer requires 10 minutes for service. The customer service desk is staffed by a single person. What is the average time a customer spends in the customer service area if modeled as an M/M/1 queuing system?
ANS:
W = 1 / (μ − λ) = 1 / (6 − 5) = 1
PTS: 1
53.The customer service desk at Joe’s Discount Electronics store receives 5 customers per hour on average. On average, each customer requires 10 minutes for service. The customer service desk is staffed by a single person. What is the average number of customers in the customer service area, if modeled as an M/M/1 queuing system?
ANS:
W = 1 / (μ − λ) = 1 / (6 − 5) = 1, so L = λ W = (5)(1) = 5
PTS: 1
Exhibit 13.2
The following questions refer to the information and output below.
A barber shop has one barber who can give 12 haircuts per hour. Customers arrive at a rate of 8 customers per hour. Customer interarrival times and service times are exponentially distributed. The following queuing analysis spreadsheet was developed from this information.
Arrival rate 
8 

Service rate 
12 

Number of servers 
1 
(max of 40) 

Utilization 
66.67% 

P(0), probability that the system is empty 
0.3333 

Lq, expected queue length 
1.3333 

L, expected number in system 
2.0000 

Wq, expected time in queue 
0.1667 

W, expected total time in system 
0.2500 

Probability that a customer waits 
0.6667 
54.Refer to Exhibit 13.2. What is the Kendall notation for this system?
ANS:
M/M/1
PTS: 1
55.Refer to Exhibit 13.2. Based on this report what percent of the time is the barber busy cutting hair?
ANS:
66.67%
PTS: 1
56.Refer to Exhibit 13.2. Based on this report what is the average number of customers waiting for a haircut?
ANS:
1.3333
PTS: 1
57.Refer to Exhibit 13.2. Based on this report what is the average waiting time before the barber begins a customer’s haircut?
ANS:
0.1667
PTS: 1
58.Refer to Exhibit 13.2. Based on this report what is the average total time spent waiting for a haircut and getting a haircut?
ANS:
0.2500
PTS: 1
Exhibit 13.3
The following questions refer to the information below.
A company has recorded the following customer interarrival times and service times for 10 customers at one of its single teller service lines. Assume the data are exponentially distributed and the 10 data points represent a reasonable sample.
All time in minutes 

Customer 
Interarrival 
Service 
1 
11.08 
2.20 
2 
2.50 
2.50 
3 
6.00 
1.10 
4 
5.75 
14.50 
5 
8.50 
2.00 
6 
4.15 
2.70 
7 
15.50 
5.00 
8 
13.00 
8.50 
9 
10.50 
5.00 
10 
6.00 
1.50 
59.Refer to Exhibit 13.3. What is the mean arrival rate per hour?
ANS:
Mean interarrival rate is 8.30 so that 60 / 8.30 = 7.23 is mean arrival rate per hour (λ)
PTS: 1
60.Refer to Exhibit 13.3. What is the mean service rate per hour?
ANS:
The mean service time is 4.50 minutes so the 60 / 4.5 = 13.33 is the mean service rate (μ).
PTS: 1
61.Refer to Exhibit 13.3. What is the average time a customer spends in the service line?
ANS:
λ = 7.23, μ = 13.33, W = 1 / (μ − λ) = 1 / (13.33 − 7.23) = 0.164
PTS: 1
62.Refer to Exhibit 13.3. What is the average number of customers in the service line?
ANS:
λ = 7.23, μ = 13.33, W = 1 / (μ − λ) = 1 / (13.33 − 7.23) = 0.164, L = λW = 7.23*0.164 = 1.186
PTS: 1
Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 6th Edition Test Bank Cliff Ragsdale
Exhibit 13.4
The following questions refer to the information and output below.
A grocery store can serve an average of 360 customers per hour. The service times are exponentially distributed. The store has 4 checkout lines each of which serves 90 customers per hour. Customers arrive at the store at a Poisson rate of 240 customers per hour. The following queuing analysis spreadsheet was developed from this information.
Arrival rate 
240 

Service rate 
90 

Number of servers 
4 
(max of 40) 

Utilization 
66.67% 

P(0), probability that the system is empty 
0.0599 

Lq, expected queue length 
0.7568 

L, expected number in system 
3.4235 

Wq, expected time in queue 
0.0032 

W, expected total time in system 
0.0143 

Probability that a customer waits 
0.3784 
63.Refer to Exhibit 13.4. What is the Kendall notation for this system?
ANS:
M/M/4
PTS: 1
64.Refer to Exhibit 13.4. Based on this report what percent of the time is a grocery clerk busy serving a customer?
ANS:
66.67%
PTS: 1
65.Refer to Exhibit 13.4. Based on this report what is the average number of customers waiting for a checker?
ANS:
0.7568
PTS: 1
66.Refer to Exhibit 13.4. Based on this report how long does a customer wait before the checker begins serving them?
ANS:
0.0032
PTS: 1
67.Refer to Exhibit 13.4. Based on this report what is the average total time spent in line and being checked out?
ANS:
0.0143
PTS: 1
Exhibit 13.5
The following questions refer to the information and output below.
A computer printer in a large administrative office has a printer buffer (memory to store printing jobs) capacity of 3 jobs. If the buffer is full when a user wants to print a file the user is told that the job cannot be printed and to try again later. There are so many users in this office that we can assume that there is an infinite calling population. Jobs arrive at the printer at a Poisson rate of 55 jobs per hour and take an average of 1 minute to print. Printing times are exponentially distributed. The following queuing analysis spreadsheet was developed from this information.
Arrival rate 
55 

Service rate 
60 

Number of servers 
1 
(max of 40) 

Maximum queue length 
3 
(max of 40 combined) 

Utilization 
76.38% 

P(0), probability that the system is empty 
0.2362 

Lq, expected queue length 
1.0628 

L, expected number in system 
1.8265 

Wq, expected time in queue 
0.0193 

W, expected total time in system 
0.0360 

Probability that a customer waits 
0.7638 

Probability that a customer balks 
0.1668 
68.Refer to Exhibit 13.5. What is the Kendall notation for this system?
ANS:
M/M/1/ with finite queue length
PTS: 1
69.Refer to Exhibit 13.5. Based on this report what is the probability that a computer user will be told to resubmit a print job at a later time?
ANS:
0.1688
PTS: 1
70.Refer to Exhibit 13.5. Based on this report what is the average number of jobs waiting to be printed?
ANS:
1.0628
PTS: 1
71.Refer to Exhibit 13.5. Based on this report how long does a computer user have to wait for his/her job to be completed?
ANS:
0.0360
PTS: 1
Exhibit 13.6
The following questions refer to the information and output below.
The university computer lab has 10 computers which are constantly being used by students. Users need help from the one lab assistant fairly often. Students ask for help at a Poisson rate of with an average of 4 requests per hour for any one computer. The assistant answers questions as quickly as possible and the service time follows an exponential distribution with mean of 1 minute per help session. The following queuing analysis spreadsheet was developed from this information.
Arrival rate 
4 

Service rate 
60 

Number of servers 
1 
(max of 40) 

Population Size 
10 
(max of 100) 

Utilization 
58.97% 

P(0), probability that the system is empty 
0.41034 

Lq, expected queue length 
0.56545 

L, expected number in system 
1.15511 

Wq, expected time in queue 
0.01598 

W, expected total time in system 
0.03265 

Probability that a customer waits 
0.58966 
72.Refer to Exhibit 13.6. What is the Kendall notation for this system?
ANS:
M/M/1 with finite population
PTS: 1
73.Refer to Exhibit 13.6. Based on this report what is the probability that a student will not get instantaneous help?
ANS:
0.58966
PTS: 1
74.Refer to Exhibit 13.6. Based on this report what is the average number of students waiting to be helped?
ANS:
0.56545
PTS: 1
75.Refer to Exhibit 13.6. Based on this report how much time do students spend getting help before they can resume work on their computers?
ANS:
0.03265
PTS: 1
Exhibit 13.7
The following questions refer to the information and output below.
A tax accountant has found that the time to serve a customer has a mean of 30 minutes (or 0.5 hours) and a standard deviation of 6 minutes (or 0.1 hours). Customer arrivals follow a Poisson distribution with an average of 60 minutes between arrivals. The following queuing analysis spreadsheet was developed from this information.
average 

Arrival rate 
1 
service rate 

Average service time 
0.5 
2 

Standard dev. of service time 
0.1 

Utilization 
50.00% 

P(0), probability that the system is empty 
0.5000 

Lq, expected queue length 
0.2600 

L, expected number in system 
0.7600 

Wq, expected time in queue 
0.2600 

W, expected total time in system 
0.7600 
76.Refer to Exhibit 13.7. What is the Kendall notation for this system?
ANS:
M/G/1
PTS: 1
77.Refer to Exhibit 13.7. Based on this report what is the probability that a customer does not have to wait for assistance with his or her taxes?
ANS:
0.5000
PTS: 1
78.Refer to Exhibit 13.7. Based on this report what is the average number of customers waiting to be helped?
ANS:
0.2600
PTS: 1
79.Refer to Exhibit 13.7. Based on this report how long does a customer spend at the tax accountant’s office?
ANS:
0.7600
PTS: 1
PROJECT
80.Project 13.1 − Internet Sales, Inc.
Internet Sales, Inc. is establishing a new ordering system to handle its online sales. Customers arrive randomly at an average of one every minute. On average it takes 45 seconds to process a customer’s order. The current system employs one network server to process orders. This server processes orders requests one at a time in the order received. The Internet Sales, Inc., Analysis Team has determined that customer orders arrive according to a Poisson process and the average time spent processing an order is exponentially distributed.
Internet Sales’ management knows that offering prompt, reliable online service is critical to their continued success. Upgrading the current online ordering capacity will establish their image as a “cutting edge” online sales company. On the other hand, too much unused capacity is inefficient and will ultimately cause company profits to decline.
Recently, the management group established a goal of serving an online customer immediately, 90% of the time. This means that 90% of the time a customer does not have to wait to place an order online after using the Internet Sales, Inc. web page. A heated discussion ensued over the current system’s capability of meeting this goal.
Under the current system, a customer waits if the system is busy processing another order. Network server upgrades are available in increments of $5000. Each upgrade adds the capability to process an additional order in parallel. For example if two upgrades were purchased at a cost of $10,000, orders will be processed by three independent network servers each serving at an exponential rate of one order per 45 seconds. The Analysis Team was asked to consider the alternatives and recommend a course of action to the management group. What upgrade (if any) will meet the service goal set by the management group? What are the cost and performance tradeoffs?
ANS:
Answer not provided.
PTS: 1
Spreadsheet Modeling & Decision Analysis: A Practical Introduction to Management Science 6th Edition Test Bank Cliff Ragsdale
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