A small town with one hospital has 3 ambulances to supply ambulance service. Req

Question

A small town with one hospital has 3 ambulances to supply ambulance service. Requests for ambulances during nonholiday weekends average .80 per hour and tend to be Poisson-distributed. Travel and assistance time averages 2 hours per call and follows an exponential distribution Use Table 1.

a. Find system utilization. (Round your answer to the nearest whole percent. Omit the "%" sign in your response.)

System utilization %

b. Find the average number of customers waiting. (Round your answer to 3 decimal places.)

Average number of customers

c. Find the average time customers wait for an ambulance. (Round your answer to 3 decimal places.)

Average time hour

d. If there are 3 ambulances, find the probability that all ambulances will be busy when a call comes in. (Round intermediate calculations to 3 decimal places and final answer to 2 decimal places.)

Probability

Explanation / Answer

s = 3 = 2 hours µ = 0.8 hours Average Utilization of the system = p = /S µ = 2 / 3*0.8 = 83.33% Probability that there are 0 customers in the system (P0 = 1 / [1 + / µ + {( / µ)^s / s!} * (1 / 1-p)] = 1 / [ 1 + 2/0.8 + ((2/0.8)^3 / 3!) * (1 / 1 - 0.8333) ] = 1 / [ 1 + 2.5 + (2.604167 * 6)] = 1 / (3.5 + 15.625) = 1 / 19.125 = 0.052288 Average number of customers = Lq = (P0*(/µ)^s*p)/(s!* (1 -p)^2) = (0.052288 * 2.5^3 * 0.8333) / (3! * 0.1667^2) = 0.680828 / 6 * 0.027778 = 4.085 Average waiting time of customers in line = Lq / = 4.085 / 2 = 2.042

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