The following waiting-line problems are all based on the as- sumptions of Poisso
ID: 365755 • Letter: T
Question
The following waiting-line problems are all based on the as- sumptions of Poisson arrivals and exponential service times. 1.* Trucks using a single-server loading dock have a mean arrival rate of 14 per day. The loading/ unloading rate is 19 per day. a. What is the probability that the truck dock will be idle? b. What is the average number of trucks waiting for c. What is the average time a truck waits for the d. What is the probability that a new arrival will e. What is the probability that more than three service? loading or unloading service? have to wait? trucks are waiting for service?Explanation / Answer
Mean arrival rate = 14 trucks per day
Mean service rate = 19 trucks per day
The queue type is M/M/1
a) The probability that no trucks are in the system is:
Calculate P0 and Pn for n = 1
P0 = 1 - Mean arrival rate/ mean service rate = 1 - (14/19)
P0 = 0.2632
Whereas, Pn = (1 - (mean arrival rate/ mean service rate)) (mean arrival rate/ mean service rate)n
Pn = (1 - 14/19) (14/19)
Pn = 0.1939
b) The average number of trucks waiting for service are
L and Lq
L = mean arrival rate/ (mean services rate - mean arrival rate)
L = 14/(19-14)
L = 2.8
Lq = (mean arrival rate)2 / (mean service rate (mean services rate - mean arrival rate)
Lq = (14)2 / (19 (19 - 14)
Lq = 196/ 95 = 2.063
c) The average time the truck waits for loading and unloading is given by Wq and W
Wq = mean arrival rate/ (mean service rate (mean service rate - mean arrival rate))
Wq = 14 / 19 (19-14)
Wq = 14 / 95 = 0.1473
W = 1/ mean service rate - mean arrival rate = 1/ 19 - 14 = 1/ 5 = 0.2
d) The probability that the new arrival will have to wait
Pw = mean arrival rate / mean service rate = 14/ 19 = 0.7368
e) Probability that more than three trucks are waiting for service
Pk = mean arrival rate / 3 ( mean service rate)
Pk = 14 / 3(19)
Pk = 0.2456
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