solution to hw problems-waiting lines
TRANSCRIPT
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