PracticeProblems_QueuingManagement_Solution.pdf

8/18/2019 PracticeProblems_QueuingManagement_Solution.pdf

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Solutions to Practice Questions on Queuing Management

1. Use M/M/1.

=λ 4/hour = µ 6/hour

a.

3333.6

4

111 =−=−=− µ

λ

ρ or 33.33%

b.

)46(6

4

)(

22

−=

−=

λ µ µ

λ q L =1.33,

4

33.1==

λ

q

q

LW = 1/3 hour or 20 minutes

c.

)46(6

4

)(

22

−=

−=

λ µ µ

λ q L = 1.33 students

d. At least one other student waiting in line is the same as at least two in the

system. This probability is 1-(P0+P1).

n

n P

−=

µ λ

µ λ 1 ,

0

064

641

−= P = .3333,

1

164

641

−= P = .2222

Probability of at least one in line is 1-(.3333 + .2222) = .4444

2.

Use M/M/1.

a. =λ 4/hour = µ 10/hour

)410(10

4

)(

22

−=

−=

λ µ µ

λ q L = .267 students

Waiting cost per day = number in line times goodwill loss per hour times

number of hours per day. (This is a formula you need to remember!)

=.267(10)8 = $21.33 per day

Total cost = waiting cost + additional service cost

=$21.33 + $99.50 = $120.83 per day

b.

Use M/M/S.

S=2,6

4==

µ

λ ρ =.667

Interpolating from q L Table, q L = .0837

Therefore, waiting cost per day = number in line times goodwill loss per hour

times of hour per day

= .0837(10)8 = $6.70 per day

Total cost = waiting cost + additional service cost

= $6.70 + $75.00 = $81.70 per day


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3/6

b. )56(6

5

)(

22

−=

−=

λ µ µ

λ q L = 4.17 people on average

c. It would be busy6

5==

µ

λ ρ =.833 or 83.3% of the time, for a 12 hour

day .833(12) = 10 hours

d.

00

065

6511

−=

−=

µ λ

µ λ P = .167 or 16.7%

e.

=W .75 and λ = 5 per hour,

75.35

75. === L L L

W λ

33.65

575.3 =

−=

−= µ

µ λ µ

λ L , therefore, the service rate must be at least

6.33 customers per hour

11. Use M/M/1.

=λ 2 per hour = µ 3 per hour

a. )23(3

2

)(

22

−=

−=

λ µ µ

λ q L = 1.333 customers waiting

b. 2

333.1==

λ

q

q

LW = .667 hours or 40 minutes

c. 2

2,2

23

2===

−=

−=

λ λ µ

λ LW L = 1 hour

d.

3

2== µ

λ

ρ = .67 or 67% of the time

12.

Use M/M/S, S=2


a.

3

2==

µ

λ ρ , From q L Table, q L is around .0837

b. λ qq LW = = .0837/2 = .0418 hours or 2.51 minutes

c. 2/7504./,7504.3/20837. ===+=+= λ µ λ LW L L q = .3751 hours or

22.51 minutes

13. Use M/M/1.


a. 610

6

−=

−=

λ µ

λ L = 1.5 people

6

5.1==

λ



4/6

b. 10

6==

µ

λ ρ = .60 or 60%

c. Probability of more then 3 people is equal to 1 – probability of 0, 1, 2

n

n P

−=

µ

λ

µ

λ 1 ,

0

010

6

10

61

−= P = .4000,

1

110

6

10

61

−= P = .2400,

2

210

6

10

61

−= P = .1440

Total of P0 + P1 + P2 = (.4000 + .2400 + .1440) = .7840

Therefore, the probability of three or more is 1 - .7840 = .2160

d.

Use M/M/S, S=2

10

6== µ

λ

ρ = .60, from Table, q L = .0593

6593.10/60593. =+=+= µ λ q L L

λ LW = = .6593/6 = .1099 hours or 6.6 minutes

15. Use M/M/1.


a.

30

25==

µ

λ ρ = .833 or 83.3%

b.

2530

25

−=−= λ µ

λ L = 5.00 document in the system

c.

25

00.5==

λ


d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3

n

n P

−=

µ

λ

µ

λ 1 ,

0

030

25

30

251

−= P = .1667

1

1

30

25

30

251

−= P = .1389,

2

2

30

25

30

251

−= P = .1157

3

330

25

30

251

−= P = .0965

Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157 + .0965) = .5178

Therefore, the probability of three or more is 1 - .5178 = .4822 or 48.22%


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e. ∞→−

=−

=)3030(30

30

)(

22

λ µ µ

λ q

L

16. Use M/M/1.


a.

6

4== µ

λ

ρ = .667 or 66.7%

b.

24

4

−=

−=

λ µ

λ L = 2.00 students in the system

c.

4

00.2==

λ


d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3

n

n P

−=

µ

λ

µ

λ 1 ,

0

06

4

6

41

−= P = .3333

1

16

4

6

41

−= P = .2222,

2

26

4

6

41

−= P = .1481

3

36

4

6

41

−= P = .0988

Total of P0 + P1 + P2 + P3 = (.3333 + .2222 + .1481 + .0988) = .8024


e. ∞→−=−= )66(6

6

)(

22

λ µ µ

λ q L

17. Use M/M/1.

=λ 10 per minute = µ 12 per minute

a. )1012(12

10

)(

22

−=

−=

λ µ µ

λ q L = 4.17 vehicles

b.

1012

10

−=

−=

λ µ

λ L = 5,

10

5==

λ

LW = .5 minute or 30 seconds

c.

12

10== µ

λ

ρ = .833 or 83.3%

d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2

n

n P

−=

µ

λ

µ

λ 1 ,

0

012

10

12

101

−= P = .1667

1

112

10

12

101

−= P = .1389,

2

212

10

12

101

−= P = .1157


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Total of P0 + P1 + P2 = (.1667 + .1389 + .1157) = .4213


18. Use M/M/2.

=λ 10 per minute = µ 12 per minute

a.

12

10== µ

λ

ρ = .8333, from q L Table, q L is around .175 cars

b. µ λ += q L L = .175 + .8333 = 1.008

10

008.1==

λ

LW = .101 minutes or 6.06 seconds

When there are two dedicated lines for the two servers, the system becomes two

independent M/M/1 with half of the arrivals. When applying the fomular, cut

λ by half.

c.

)512(12

5

)(

22

−=

−= λ µ µ

λ q L = .298.

d.

512

5

−=

−=

λ µ

λ L = .7143,

5

7143.==

λ

LW = .143 minutes or 8.58 seconds

19. M/M/1. Note that since repair persons need to work in a group, there is only one

server even though there can be multiple repair people.

=λ 2 per hour

With one repair person: = µ 2, ∞→−

=−

=22

2

λ µ

λ L

With two repair people: = µ 3, 223

2=

−=

−=

λ µ

λ L cars

With three repair people: = µ 4, 124

2=

−=

−=

λ µ

λ L car

Number of

repair

personnel

Service rate

per hour

(µ)

snCost of

waiting per

hour 1

Cost of

service per

hour 2

Total

cost per

hour

1 2 ∞ $ ∞ $ 20 $ ∞

2 3 2 80 40 120

3 4 1 40 60 100

Note: 1 = cost of waiting is number in system times downtime cost of $40 per hour.

2 = cost of service is number of repair personnel times wage rate ($20 per hour).

From the table, Should use three repair persons.

PracticeProblems_QueuingManagement_Solution.pdf

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