• Author / Uploaded
• Jonas Ciabis

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Supplement

C

Waiting Lines

PROBLEMS 1. Solomon, Smith and Sanson a. Single-server model, average utilization rate. λ 8 ρ = = = 0.8 or 80% utilization µ 10 b. The probability of four or fewer documents in the system is 0.6723 as shown following. Therefore, the probability of more than four documents in the system is 1 – 0.6723 = 0.3277. Pn = (1 − ρ )( ρ )

n

P4 = (1 − 0.8 )( 0.8 ) = 0.0819 4

P3 = (1 − 0.8 )( 0.8 ) = 0.1024 3

P2 = (1 − 0.8 )( 0.8 ) = 0.1280 2

P1 = (1 − 0.8 )( 0.8 ) = 0.1600 1

P0 = (1 − 0.8 )( 0.8 ) = 0.2000 0

= 0.6723

c. The average number of pages of documents waiting to be typed, λ λ 8 8 Lq = ρ L = = = 3.2 pages µ µ −λ 10 10 − 8 2. Benny’s Arcade Because there are only six machines, we must use the finite source model. a. To calculate the Jimmy’s utilization, we need to compute the probability that he will have no machines to repair. λ = 1/ 50 = 0.02 machines per hour µ = 1/15 = 0.0667 machines per hour 6

6! Po = n =o ( 6 − n ) !

0.02 0.0667

n −1

= [13.92] = 0.0718 −1

ρ = Jimmy’s utilization = 1 − 0.0718 = 0.9282

Waiting Lines

SUPPLEMENT C

b. Average number of machines out of service. 0.0667 L = 6− (1 − 0.0718) = 6 − 3.095 = 2.905 machines 0.02

c. Average time a machine is out of service. W = 2.905 ( 6 − 2.905 )( 0.02 ) = 46.93 hours 3. Moore, Akin, and Payne (dental clinic). Multiple-server model. λ 5 s = 3, λ = 5, µ = 2 , ρ = = = 0.8333 sµ 3 ( 2 ) a. Probability of no patients, P0 =

( λ µ )n + ( λ µ ) s

s −1

P0 =

n!

n =0

s!

(5 2) + (5 2) + (5 2) 0! 1! 2! 0

=

−1

1 1− ρ

1

2

=

( 5 2 )n + ( 5 2 ) s n!

n =0

(5 2) + 3!

{

= [1 + 2.5 + 3.125] + 2.604 ( 6 ) =

2

}

3

3!

1 1 − 56

−1

−1

1 1 − 56

−1

1 = 0.04494 ( 6.625) + (15.625)

b. The probability of 6 or more customers in the clinic is: P0 = 0.04494 (from part a), s = 3 for n < s for n ≥ s

( ) P = λ µ

n

n

n!

P0

Pn =

( ) λ µ

n

s ! s n− s

P0

(5) P1 = 2 ( 0.04494 ) = 0.11235 1!

5 ( ) P3 = 2 0 ( 0.04494 ) = 0.11703

( ) ( 0.04494 ) = 0.14044 P2 = 2!

5 ( ) P4 = 2 1 ( 0.04494 ) = 0.09753

1

5 2 2

3

3!3

4

3!3

5 ( ) P5 = 2 2 ( 0.04494 ) = 0.08127 5

3!3

1 − ( P0 + P1 + P2 + P3 + P4 + P5 )

= 1 − ( 0.04494 + 0.11235 + 0.14044 + 0.11703 + 0.09753 + 0.08127 ) = 0.40644

173

174

PART 2

Managing Processes

c. The average number of patients waiting in the lobby, Lq =

P0

( ) λ µ

s

ρ

s !(1 − ρ )

2

=

0.04494 ( 52 ) 3!(1 −

( ) = 0.58516 = 3.5109

3 5 6 2 5 6

)

1 6

d. The average time spent in the clinic, 1 Lq 1 3.5109 1 W = wq + = + = + = 1.2022 hours µ λ µ 5 2 4. Fantastic Styling Salon a. A common queue, two-server model. λ = 8, µ = 5, s = 2

ρ=

λ 8 = = 0.8 2µ 2 ( 5)

Customer’s average waiting time in queue = Wq = Lq =

Po =

Po

( ) λ µ

s

2

( λ µ )n + ( λ µ ) s

n =0

λ

ρ

s !(1 − ρ ) 1

Lq

n!

s!

( λ µ )0 + ( λ µ )1 + ( λ µ )2 0!

= 1 + (8 5)

1!

2 8 5) ( +

2

−1

1 1− ρ

2!

1 1 − ( 0.80 )

1 1− ρ

−1

−1

= [1 + 1.6 + 6.4]

−1

= 1 9 = 0.1111 2 0.1111)( 8 5 ) ( 0.80 ) 0.2275 ( Lq = = = 2.844 2 0.08 2!(1 − 0.80 )

Wq =

2.844 = 0.36 hours or 21.6 minutes 8

b. Two separate, single-server models. The results for Jenny and Jill are identical.

Waiting Lines

λ = 0.5 ( 8 ) = 4 , µ = 5 , ρ =

175

4 = 0.80 5

Waiting in line is given by Wq = ρW = ρ Wq =

SUPPLEMENT C

1 µ −λ

0.8 = 0.80 hours or 48 minutes 5−4

c. On average, a customer’s waiting will be more than twice that of the multiple server design. The reason is that a server will serve any customer as they enter the system in the multiple-server system, thereby reducing the wait time on average. 5. Local Bank Service rate µ = 60 3 min. per customer = 20 customers/hour. a. Average utilization, ρ = λ sµ = 50 ( 3 ( 20 ) ) = 0.8333 . s −1

b. P = o

( λ µ )n + ( λ µ ) s n!

n =0 2

=

s!

1 1− ρ

( 50 20 )n + ( 50 20 )3 n!

n =0

3!

−1

1 1 − 0.8333

−1

= [1 + 2.5 + 3.125 + 15.622]

−1

= 0.0449 Po ( λ µ ) ρ

3 0.0449 )( 50 20 ) ( 0.8333) ( = = 3.5063 2 3!(1 − 0.8333) s !(1 − ρ ) s

c. Lq =

d. Wq =

Lq

λ

=

3.5063 = 0.0701 hours, or 4.2 minutes 50

(

)

e. L = λW = λ Wq + (1 µ ) = 50 ( 0.0701 + 1 20 ) = 6.005 or 6 customers

6. KRAN radio. Single-server model. 60 min hr µ= = 6 calls hr 10 min call

λ=

60 min hr = 2.4 calls hr 25 min call

ρ=

λ 2.4 = = 0.4 µ 6

176

PART 2

Managing Processes

A caller will not receive a busy signal when there are zero, one, or two callers in the system. Therefore, the probability of receiving a busy signal is one minus the probability of two or fewer callers in the system.

Pn = (1 − ρ )( ρ )

n

P2 = (1 − 0.4 )( 0.4 ) = 0.096 2

P1 = (1 − 0.4 )( 0.4 ) = 0.240 1

P0 = (1 − 0.4 )( 0.4 ) = 0.600 0

= 0.936 1 − 0.936 = 0.064 Jake’s callers will get busy signals 6.4 percent of the time. 7. Precision Machine Shop. Single-server model. With the junior attendant, the average number of idle machinists, L λ 8 L= = =4 µ − λ 10 − 8 Average hourly idle machinist cost = $20(L) = $20(4) = $80 With the senior attendant, average number of idle machinists, L λ 8 L= = =1 µ − λ 16 − 8 Average hourly cost of idle machinists drops to $20(L) = $20(l) = $20 Adding the attendant pay gives a total cost of $85 per hour ($80 + $5) for the junior attendant and $32 per hour ($20 + $12) for the senior attendant. The best choice is the senior attendant. 8. Hasty Burgers. Single-server model, λ = 20 a. Find µ resulting in L = 4. L=

λ

µ −λ

20 µ − 20 4 µ − 80 = 20 4=

4 µ = 100

µ = 25 The required service rate is 25 customers per hour. b. Find the probability that more than four customers are in the system. This is one minus the probability of four or fewer customers in the system. First we calculate average utilization of the drive-in window.

ρ=

Waiting Lines

λ 20 = = 0.8 µ 25

SUPPLEMENT C

177

The probability that more than four customers are in line and being served is: P = 1 − ( P0 + P1 + P2 + P3 + P4 ) where

Pn = (1 − ρ )( ρ )

n

P = 1 − { (1 − ρ )( ρ ) + (1 − ρ )( ρ ) + (1 − ρ )( ρ ) 0

1

+ (1 − ρ )( ρ ) + (1 − ρ )( ρ ) 3

}

4

P = 1 − {(1 − ρ ) ( ρ ) + ( ρ ) + ( ρ ) + ( ρ ) + ( ρ ) 0

1

2

3

4

2

}

when ρ = 0.8

P = 1 − {( 0.2 ) [1 + 0.8 + 0.64 + 0.512 + 0.4096]}

P = 0.3277 Consequently, there is about a 33 percent chance of more than four customers in the system. c. Find the average time in line. 1 Wq = ρW = ρ µ −λ 1 25 − 20 Wq = 0.16 hours or 9.6 minutes = 0.8

Ten minutes borders on being unbearable, particularly in the atmosphere of exhaust fumes. Keep in mind that this is an average, and some people must wait longer. ADVANCED PROBLEMS

9. Pinball Wizard. Multiple-server model. In this analysis we determine the expected total labor and machine failure costs for the existing complement of three employees and then compare it to larger maintenance complements until costs begin to rise. Three maintenance people: s = 3, λ = 0. 333, µ = 0. 125, Average utilization λ 0.333 ρ= = = 0.888 sµ 3 ( 0.125 )

178

PART 2

Managing Processes

Probability of an empty system s −1

P0 =

( ) +( )

(

λ µ

n!

n =0

=

n

λ µ

s!

) +(

−1

s

1 1− ρ

=

) +(

2

(

n =0

)

0.333 0 0.125

0.333 1 0.125

0.333 2 0.125

0!

1!

2!

+

(

{

= [1 + 2.664 + 3.548] + 3.151( 8.929 ) P0 =

) +(

)

0.333 n 0.125

0.333 3 0.125

n!

3!

)

0.333 3 0.125

3!

}

−1

1 1 − 0.888 −1

1 1 − 0.888

−1

1 = 0.0283 ( 7.212 ) + ( 28.135)

Average number of machines in waiting line Lq = Lq =

P0

( ) λ µ

s

ρ

s !(1 − ρ )

2

0.333 0.0283 ( 0.125 ) ( 0.888) 3

=

3!(1 − 0.888 )

2

0.4751 = 6.310 0.0753

The average time waiting in line Lq 6.310 Wq = = = 18.949 hrs. λ 0.333 Average time in system 1 W = Wq +

µ

1 0.125 = 26.949 hrs = 18.949 +

Average number of machines in system L = λW = 0.333 ( 26.949 )

L = 8.974 machines The total expected hourly costs for the crew size of three employees is: Labor: 3 ($8 per hour) $ 24.00 Machine downtime: 8.974 ($10 per hour) 89.74 TOTAL $113.74

Waiting Lines

SUPPLEMENT C

Four maintenance people: Average utilization

ρ=

λ 0.333 = = 0.666 sµ 4 ( 0.125 )

Probability of an empty system s −1

P0 =

( ) +( ) n

λ µ

λ µ

n!

n =0

s!

−1

s

1 1− ρ

3

=

0.333 0.333 ( 0.125 ) + ( 0.125 )

n =0

n

n!

4!

( 2.664 )0 + ( 2.664 )1 + ( 2.664 )2 + ( 2.664 )3

=

0!

1!

2!

3!

{

= [1 + 2.664 + 3.548 + 3.151] + 2.099 ( 2.994 )

P0 =

1 = 0.0601 (10.363) + ( 6.284 )

Average number of machines in waiting line Lq = Lq =

P0

( ) λ µ

s

ρ

s !(1 − ρ )

2

0.0601( 2.664 ) ( 0.666 ) 4

=

4!(1 − 0.666 )

2.016 = 0.753 2.677

The average time waiting in line Lq 0.753 Wq = = = 2.261 hrs. λ 0.333 Average time in system 1 W = Wq +

µ

= 2.261 +

1 0.125

= 10.261 Average number of machines in system L = λW = 0.333 (10.261)

L = 3.417

2

}

+ −1

4

−1

1 1 − 0.666

( 2.664 )4 4!

1 1 − 0.666

−1

179

180

PART 2

Managing Processes

The total expected hourly costs for the crew size of four employees is: Labor: 4 ($8 per hour) $ 32.00 Machine downtime: 3.417 ($10 per hour) 34.17 TOTAL $ 66.17 Five maintenance people: Average utilization λ 0.333 = = 0.5328 ρ= sµ 5 ( 0.125 )

Probability of an empty system s −1

P0 =

( ) +( ) n

λ µ

λ µ

n!

n =0

−1

s

s!

1 1− ρ

4

=

0.333 0.333 ( 0.125 ) + ( 0.125 )

n=0

n

n!

5!

0!

1!

2!

3!

{

= [1 + 2.664 + 3.548 + 3.151 + 2.099] + 1.118 ( 2.140 ) 1 = 0.0673 (12.462 ) + ( 2.393)

Average number of machines in waiting line Lq = Lq =

1 1 − 0.5328

( 2.664 )0 + ( 2.664 )1 + ( 2.664 )2 + ( 2.664 )3 + ( 2.664 )4

=

P0 =

−1

5

P0

( ) λ µ

s

ρ

s !(1 − ρ )

2

0.0673 ( 2.664 ) ( 0.5328 ) 4

=

5!(1 − 0.5328 )

4.811 = 0.184 26.193

The average time waiting in line Lq 0.184 Wq = = = 0.553 λ 0.333 Average time in system 1 W = Wq +

µ

= 0.553 + = 8.553

1 0.125

2

4!

}

−1

+

( 2.664 )5 5!

1 0.4672

−1

Waiting Lines

SUPPLEMENT C

181

Average number of machines in system L = λW = 0.333 ( 8.553)

L = 2.848 The total expected hourly costs for the crew size of five employees is: Labor: 5 ($8 per hour) $ 40.00 Machine downtime: 2.848 ($10 per hour) 28.48 TOTAL $ 68.48 This total is higher than that for employing four maintenance people. Therefore, the manager of the Pinball Wizard should add only one more maintenance person. 10. Benton University, Finite Source Model a. To calculate the utilization in a finite source waiting line situation, we must first compute the probability that the maintenance person will have no customers. N

N! λ P0 = n =0 ( N − n ) ! µ

=

5! 0.4 1+ ( 5 − 1)! 2.5 5! 0.4 + ( 5 − 4 )! 2.5

n −1

1

4

5! 0.4 + ( 5 − 2 )! 2.5 5! 0.4 + ( 5 − 5)! 2.5

2

5

= {1 + 0.8 + 0.512 + 0.246 + 0.079 + 0.013} 1 = 0.3774 2.65 ρ = 1 − P0 = 1 − 0.3774 = 0.6226

P0 =

b. Copy machines in repair system L=N−

µ (1 − P0 ) λ

2.5 (1 − 0.3774 ) 0.4 = 5 − 3.891 = 5−

= 1.109

5! 0.4 + ( 5 − 3)! 2.5

−1

3

−1

182

PART 2

Managing Processes

c. Time spent in repair system W =L

=

( N − L) λ

−1

1.109 ( 5 − 1.109 ) 0.4

= 0.712 days = 5.7 hours, assuming an 8-hour day

11. Quarry a. Current System: Single-server model λ = 9 hour ; µ = 10 hour Average waiting line in the system W =

1 1 = = 1 hour or 60 minutes µ − λ 10 − 9

b. First Alternative: Improved single-server model λ = 6 hour ; µ = 15 hour 1 1 Average waiting line in the system W = = = 0.1111 hour or 6.67 minutes 15 − 6 9 c. Second Alternative: Two-server model λ = 9 hour ; µ = 10 hour; s = 2

ρ = ( λ 2µ ) = ( 9 2 (10 ) ) = 9 20 = 0.45

(

)

W = Wq + (1 µ ) = Lq λ + (1 µ ) Lq = Po ( λ µ ) ρ

s !(1 − ρ )

s

Po =

=

s −1

( λ µ )n + ( λ µ ) s

n =0

n!

s!

2 0.90 ) ( 1 + 0.90 +

2

1 1− ρ

1 1 − 0.45

2!

−1

−1

= [1 + 0.90 + 0.7364]

−1

= 0.3793 2 0.3793)( 9 10 ) ( 0.45 ) ( = 0.2285 Lq = 2 2!(1 − 0.45 ) W = ( 0.2285 9 ) + (1 10 )

= 0.1254 hours or 7.52 minutes

Waiting Lines

SUPPLEMENT C

183

The second alternative results in an 87 percent reduction in waiting time relative to the current system, however, the first alternative dominates the second alternative in time and cost. Nonetheless, to make a final determination between the current system and alternative 1, the cost of the waiting time of the trucks must be considered.