Solution to HW problems – Waiting Lines

6. Use Model 1.

100 per hour 120 per hour

a. = 5

= .05 hours or 3 minutesb. Now, 180 per hour

= 1.25

= .0125 hours or .75 minutes or 45 seconds

c. Using model 3, 100 per hour 120 per hour

S =2 , and , from Exhibit 8A.9, = .1756

= 1.01

= .0101 hours or .605 minutes or 36.3 seconds

10.Use model 1. 20 per hour 30 per hour

a. = 2 people in the system

b. = .10 hours or 6 minutesc. Probability of 3 or more is equal to 1 – probability of 0, 1, 2

= .3333

= .2222

= .1481

Total of P0 + P1 + P2 = (.3333 + .2222 + .1481) = .7036

Therefore, the probability of three or more is 1 - .7036 = .2964

d. = .67 or 67%e. Use model 3.

=.6667

From Exhibit 8A.9, = .0093

= .0093 + 20/30 = .676

= .0338 hours or 2.03 minutes

11.Use model 1

5 per hour 6 per hour

a. = 5 people in the system

= 1 hours

b. = 4.17 people on average

c. It would be busy =.833 or 83.3% of the time, for a 12 hour day .833(12) = 10 hours

a. = .167 or 16.7%

b. .75 and = 5 per hour,

, therefore, the service rate must be at least 6.33 customers per hour

13. Use model 1.

2 per hour 3 per hour

a. = 1.333 customers waiting

b. = .667 hours or 40 minutes

c. = 1 hour

d. = .67 or 67% of the time

14.Use model 1.

15.Use model 3.

2 per hour 3 per hour

, from Exhibit 8A.9, = .0837

a. = .0837/2 = .0418 hours or 2.51 minutes

b. = .3751 hours or 22.51 minutes

16.Use model 1.

6 per hour 10 per hour

a. = 1.5 people

= .25 hours or 15 minutes

b. = .60 or 60%c. Probability of more then 3 people is equal to 1 – probability of 0, 1, 2

= .4000

= .2400

= .1440

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

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

c. Use model 3.

= .60, from Exhibit 8A.9, = .0593

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

19. Use model 1.

4 per hour 6 per hour

a. = .667 or 66.7%

b. = 2.00 students in the system

c. = .50 hours or 30 minutesd. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3

= .3333

= .2222

= .1481

= .0988

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

Therefore, the probability of three or more is 1 - .8024 = .1976 or 19.76%

e.

21. Use model 3.

10 per minute 12 per minute

a. = .8333, from Exhibit 8A.9, = .175 cars

b. = .175 + .8333 = 1.008

c. = .298Since there are two lines, the total number in line is 2 times .298, which is .596 cars

d. = .7143

= .143 minutes or 8.58 seconds