The wheat harvesting season in the American Midwest is? short, and farmers deliv
ID: 357401 • Letter: T
Question
The wheat harvesting season in the American Midwest is? short, and farmers deliver their truckloads of wheat to a giant central storage bin within a 2?-week span. Because of? this, wheat-filled trucks waiting to unload and return to the fields have been known to back up for a block at the receiving bin. The central bin is owned? cooperatively, and it is to every? farmer's benefit to make the? unloading/storage process as efficient as possible. The cost of grain deterioration caused by unloading delays and the cost of truck rental and idle driver time are significant concerns to the cooperative members. Although farmers have difficulty quantifying crop? damage, it is easy to assign a waiting and unloading cost for truck and driver of ?$19 per hour. During the 2?-week harvest? season, the storage bin is open and operated 16 hours per? day, 7 days per? week, and can unload 35 trucks per hour according to a negative exponential distribution. Full trucks arrive all day long? (during the hours the bin is? open) at a rate of about 31 per? hour, following a Poisson pattern.
(a) average number of trucks in the unloading system.
(b) average time per truck in the system.
(c) utilization rate for the bin area.
(d) probability that there are more than three trucks in the system at any given time.
(e) total daily cost to the farmers of having their trucks tied up in the unloading process.
Explanation / Answer
This is a M/M/1 queue model with following parameters
Arrival rate, ? = 31 per hour
Service rate, ? = 35 per hour
(a) Average number of trucks in the unloading system (L) = ?/(?-?) = 31/(35-31) = 7.75 trucks
(b) Average time per truck in the system, (W) = L/? = 7.75/31 = 0.25 hour
(c) Utilization rate for the bin area, ? = ?/? = 31/35 = 0.886
(d) Pn is the probability of n trucks in the system, Pn =(1-?)*?n
P0 = (1-0.886)*0.8860 = 0.1143
P1 = (1-0.886)*0.8861 = 0.1012
P2 = (1-0.886)*0.8862 = 0.0897
P3 = (1-0.886)*0.8863 = 0.0794
Probability of more than 3 trucks in the system = 1 - (P0+P1+P2+P3)
= 1 - (0.1143+0.1012+0.0897+0.0794)
= 0.6154
(e) Total daily cost to the farmers of having their trucks tied up in the unloading system = L*Cw = 7.75*19*16 = $ 2356
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