Operations Management Help

Chapter 11 - Practice Problem Solutions

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11.6 a) A bank is a queueing system with people as the customers, and tellers as the servers.

b) Wq = 1 minute

W = Wq + (1/) = 1 + 2 = 3 minutes

Lq = Wq = (40/60 per minute) (1 minute) = 0.667 customers

L = W = (40/60 customers per minute) (3 minutes) = 2 customers

11.7 a) A parking lot is a queueing system for providing parking with cars as the customers, and

parking spaces as the servers. The service time is the amount of time a car spends in a space.

The queue capacity is 0.

b) L = 0P0 + 1P1 + 2P2 + 3P3 = 0(0.2) + 1(0.3) + 2(0.3) + 3(0.2) = 1.5 cars

Lq = 0 cars

W = L /  = 1.5 / 2 = 0.75 hours

Wq = Lq /  = 0 / 2 = 0 hours

c) A car spends an average of 45 minutes in a parking space.

11.8 a) L = 0(1/16) + 1(4/16) + 2(6/16) + 3(4/16) + 4(1/16) = 2, which represents the average number

of customers in the shop, including those getting their hair cut.

b)

n # in queue probability product

0 0

1 0

2 0

3 1 0.25 0.25

4 2 0.0625 0.125

Lq = 0.375 which represents the average number of customers in the shop waiting to get a

haircut.

c) W = L/ = 2/4 = 0.5 hours = 30 minutes

Wq = Lq/l = 0.375/4 = 0.094 hours = 5.625 minutes

These quantities mean that customers will be in the shop an average of half an hour, including

the time to get a haircut, and will have to wait an average of 5.625 minutes before their

haircut will begin.

d) W - Wq = 0.5 – 0.094 = 0.406 hours = 24.375 minutes

11.15 a) The customers are trucks to be loaded or unloaded and the servers are the crews. The system

currently has 1 server.

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b) 4 member crew (M/M/1 model):

c) 3 member crew (M/M/1 model):

d) 2 member crew (M/M/1 model):

e) A one person team should not be considered since that would lead to a utilization factor of

=1 which does not enable the qeueuing system to reach a steady-state condition with a

manageable load for the team.

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f &g) Total cost = ($20)(# on crew)+($30)(Lq)

TC(4 members) = ($20)(4)+($30)(0.0833) = $82.50/hour

TC(3 members) = ($20)(3)+($30)(0.167) = $65/hour

TC(2 members) = ($20)(2)+($30)(0.5) = $55/hour

A crew of 2 people will minimize the expected total cost per hour.

11.16 a) 1/ = 1 minute (M/M/1 model):

 

For 1/ = 1.5 minutes (M/M/1 model):

(1/) = 2 minutes is not a feasible alternative since =1.

b) TC(1/ = 1 minute) = $1.60 + ($0.80)(1) = $2.40

TC (1/ = 1.5 minute) = $0.90 + ($0.80)(3) = $3.30

The grinder should be set so that the mean service time is 1 minute.

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11.23 a) M/G/1 Model:

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b) M/G/1 Model.

Variance = 60%(152) = 135.

Standard deviation = 11.62.

c) The new proposal shows that they will be slightly worse off if they switch to the new queueing

system.

d) Total Cost (status quo) = $40 + (Lq)($20) = $85/hour

Total Cost (proposal) = $40 + (Lq)($20) = $88/hour

11.27 a) Option 1 (M/M/3 Model):

Option 2 (M/M/1 Model):

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b) W and L are smaller for Option 2 because it is a more efficient system. This is true because

when there are only 1 or 2 customers in the system Option 2 is operating at full efficiency,

while Option 1 will have idle servers. Wq and Lq are larger for Option 2 because there are

fewer people in service (only 1 register) and therefore more people in line.

c) W should be the most important measure to customers since they should be most concerned

with the total time spent in the system. Given this, Option 2 is better.

11.28 a) Deposits:

Withdrawals:

Combined expected wait time = (16/30)(0.25) + (14/30)(0.1667) = 0.211 hours = 12.67

minutes.

b) Expected waiting time = 0.114 hours = 6.86 minutes.

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c) An expected processing time of 3.42 minutes (17.55 customers per hour) would cause the

expected waiting times to be the same for the two procedures.

11.29 This year:

Next year:

This year’s system yields smaller values for all measure except L.

11.30 a) L = 1.5

W = L/ = 1.5/0.2 = 7.5 hours

Wq = W – 1/ = 7.5 – 1/0.167 = 1.5 hours

Lq = Wq = (0.2)(1.5) = 0.3

b) M/D/1 Model:

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c) Total Cost(Alternative 1) = $70 + ($100)(L) = $220

Total Cost(Alternative 2) = $100 + ($100)(L) = $205

Alternative 2 should be adopted.

11.31 a) This system is an example of a nonpreemptive priority queueing system.

b) M/M/s Nonpreemptive Priority Model:

c) Wq(first-class) / Wq(coach-class) = 0.0333 / 0.833 = 0.4

d) (12 hours) = 0.6(12 hours) = 7.2 hours

11.32 a) A preemptive priorities queueing model fits this system.

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b) M/M/s Preemptive Priority Model:

Guidelines will be satisfied next year with a single doctor. The average wait time for critical

cases is 0.024 hours = 1.43 minutes; the average wait time for serious cases is 0.154 hours =

9.22 minutes; the average wait time for stable cases is 1.03 hours.

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c) M/M/1 Model:

The guideline for stable cases would be satisfied but the other two would not be.

d)

The guidelines are still met. The average wait time for critical cases is 0.027 hours = 1.62

minutes; the average wait time for serious cases is 0.181 hours = 10.89 minutes; the average

wait time for stable cases is 1.57 hours.

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11.33 a) M/M/s Nonpreemptive Priority Model (s=4):

M/M/s Nonpreemptive Priority Model (s=5):

The guidelines are met with 4 lathes.

b) Total Cost(4 lathes) = $250(4) + $750(L1) + $450(L2) + $150(L3) = $3,227.50

Total Cost(5 lathes) = $250(5) + $750(L1) + $450(L2) + $150(L3) = $3,051.50

5 lathes should be obtained to minimize the expected total cost.

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11.34 a) 1 server is optimal.

b) 2 servers are optimal.

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c) 3 servers are optimal.

11.35 Jim should operate 4 cash registers during the lunch hour.

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11.36 Garrett-Tompkins should have 6 copiers.

Cases

11.1 The operations of the records and benefits call center can be modeled as an M/M/s queueing

system. We, therefore, use the template for the M/M/s queueing model throughout this case.

The mean arrival rate equals 70 per hour, and the mean service rate of every representative

equals 6 per hour. Mark needs at least s=12 representatives answering phone calls to ensure

that the queue does not grow indefinitely.

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a) In order to solve this problem we have to determine the number of servers by "trial and error"

until we find a number s such that the probability of waiting more than 4 minutes in the

queue is above 35%.

For 13 servers, the probability that a customer has to wait more than 4 minutes equals 36.3%.

It appears that Mark currently employs 13 servers.

b) Using the same procedure as in part a we find that for s = 18 servers the probability of waiting

more than 1 minute drops below 5%:

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c) Using the same "trial and error" method as before, we find the minimal number of servers

necessary to ensure that 80% of customers wait one minute or less to be s = 15.

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The minimal number of servers to ensure that 95% of customers wait 90 seconds or less is s =

17.

When an employee of Cutting Edge calls the benefits center from work and has to wait on the

phone, the company loses valuable work time for this customer. Mark should try to estimate

the amount of work time employees lose when they have to wait on the phone. Then he could

determine the cost of this waiting time and try to choose the number of representatives in

such a fashion that he reaches a reasonable trade-off between the cost of employees waiting

on the phone and the cost of adding new representatives.

Clearly, Mark's criteria would be different if he were dealing with external customers. While

the internal customers might become disgruntled when they have to wait on the phone, they

cannot call somewhere else. Effectively, the benefits center holds monopolistic power. On the

contrary, if Mark were running a call center dealing with external customers, these customers

could decide to do business with a competitor if they become angry from waiting on the

phone.

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d) If the representatives can only handle 6 calls per hour, then Mark needs to employ 18

representatives (see part b). If a representative can handle 8 calls per hour, then the minimal

number of representatives equals 14.

The cost of training 14 employees equals (14)($2,500) = $35,000 and saves Mark (4)($30,000)

= $120,000 in annual salary. In the first year alone Mark would save $85,000 if he chose to

train all his employees so that they can handle 8 instead of 6 phone calls per hour.

e) Mark needs to carefully check the number of calls arriving at the call center per hour. In this

case we have made the simplifying assumption that the arrival rate is constant. That

assumption is unrealistic; clearly we would expect more calls during certain times of the day,

during certain days of the week, and during certain weeks of the year. We might want to

collect data on the number of calls received depending on the time. This data could then be

used to forecast the number of calls the center will receive in the near future, which in turn

would help to forecast the number of representatives needed.

Also, Mark should carefully check the number of phone calls a representative can answer per

hour. Clearly, the length of a call will depend on the issue the caller wants to discuss. We

might want to consider training representatives for special issues. These representatives could

then always answer those particular calls. Using specialized representatives might increase the

number of phone calls the entire center can handle.

Finally, using an M/M/s model is clearly a great simplification. We need to evaluate whether

the assumptions for an M/M/s model are at least approximately satisfied. If this is not the

case, we should consider more general models such as M/G/s or G/G/s.

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11.2 a) Status quo at the presses – 7.52 sheets of in-process inventory.

Status quo at the inspection station – 3.94 wing sections of in-process inventory.

Inventory cost = (7.52 + 3.94)($8/hour) = $91.68 / hour

Machine cost = (10)($7/hour) = $70 / hour

Inspector cost = $17 / hour

Total cost = $178.68 / hour

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b) Proposal 1 will increase the in-process inventory at the presses to 11.05 sheets since the mean

service rate has decreased.

The in-process inventory at the inspection station will not change.

Inventory cost = (11.05 + 3.94)($8/hour) = $119.92 / hour

Machine cost = (10)($6.50) = $65 / hour

Inspector cost = $17 / hour

Total cost = $201.92 / hour

This total cost is higher than for the status quo so should not be adopted. The main reason for

the higher cost is that slowing down the machines won’t change in-process inventory for the

inspection station.

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c) Proposal 2 will increase the in-process inventory at the inspection station to 4.108 wing

sections since the variability of the service rate has increased.

The in-process inventory at the presses will not change.

Inventory cost = (7.52 + 4.108)($8/hour) = $93.02 / hour

Machine cost = (10)($7/hour) = $70 / hour

Inspector cost = $17 / hour

Total cost = $180.02 / hour

This total cost is higher than for the status quo so should not be adopted. The main reason for

the higher cost is the increase in the service rate variability and the resulting increase in the

in-process inventory.

d) They should consider increasing power to the presses (increasing there cost to $7.50 per hour

but reducing their average time to form a wing section to 0.8 hours). This would decrease the

in-process inventory at the presses to 5.69.

Inventory cost = (5.69 + 3.94)($8/hour) = $77.04 / hour

Machine cost = (10)($7.50/hour) = $75 / hour

Inspector cost = $17 / hour

Total cost = $169.04 / hour

This total cost is lower than the status quo and both proposals.