At Main Street Dealer, there is a one pet manager who takes care of whatever par
ID: 454571 • Letter: A
Question
At Main Street Dealer, there is a one pet manager who takes care of whatever parts the five mechanics might need while servicing cars. Whenever a mechanic needs an auto part, they would go to the part counter and wait until the part manager gives him what he needs. On average a mechanic needs some pets every 20 minutes (exponentially distributed) and it takes a average of 3 minutes for the part manager to retrieve any request. Then probability of nobody waiting at the counter is 0.40447.What type of queuing system is this? A. M/M/1 system B. M/M/s C. M/M/1/FP with finite population D. M/M/1/C with finite system capacity
On average, how many mechanics are there waiting at the part counter (including the one waiting for his request)? A. 0.9917 B. 0.7292 C. 1.4004 D. 1.0298
On average how long does each mechanic have to wait for the part manager to take his request? A. 1.459 minutes B. 2.593 C. 2.188 D. 3.089
On average how many times does each mechanic spend at the part counter away from servicing cars? A. 6.089 minutes B. 5.188 C. 4.459 D. 6.593
On average how many mechanics are busy servicing the cars at any time? A. 3.9702 mechanics B. 3.0083 C. 4.5996 D. 3.2708
At Main Street Dealer, there is a one pet manager who takes care of whatever parts the five mechanics might need while servicing cars. Whenever a mechanic needs an auto part, they would go to the part counter and wait until the part manager gives him what he needs. On average a mechanic needs some pets every 20 minutes (exponentially distributed) and it takes a average of 3 minutes for the part manager to retrieve any request. Then probability of nobody waiting at the counter is 0.40447.
What type of queuing system is this? A. M/M/1 system B. M/M/s C. M/M/1/FP with finite population D. M/M/1/C with finite system capacity
On average, how many mechanics are there waiting at the part counter (including the one waiting for his request)? A. 0.9917 B. 0.7292 C. 1.4004 D. 1.0298
On average how long does each mechanic have to wait for the part manager to take his request? A. 1.459 minutes B. 2.593 C. 2.188 D. 3.089
On average how many times does each mechanic spend at the part counter away from servicing cars? A. 6.089 minutes B. 5.188 C. 4.459 D. 6.593
On average how many mechanics are busy servicing the cars at any time? A. 3.9702 mechanics B. 3.0083 C. 4.5996 D. 3.2708
What type of queuing system is this? A. M/M/1 system B. M/M/s C. M/M/1/FP with finite population D. M/M/1/C with finite system capacity
On average, how many mechanics are there waiting at the part counter (including the one waiting for his request)? A. 0.9917 B. 0.7292 C. 1.4004 D. 1.0298
On average how long does each mechanic have to wait for the part manager to take his request? A. 1.459 minutes B. 2.593 C. 2.188 D. 3.089
On average how many times does each mechanic spend at the part counter away from servicing cars? A. 6.089 minutes B. 5.188 C. 4.459 D. 6.593
On average how many mechanics are busy servicing the cars at any time? A. 3.9702 mechanics B. 3.0083 C. 4.5996 D. 3.2708
Explanation / Answer
What type of queuing system is this?
M/M/1/FP with finite population
Let Arrival rate be given by A (3 per hour),Service rate by U (20 per hour)
Number of mechanics as N (5) and P0 (0.40447) as the probability of nobody waiting in the corner
On average, how many mechanics are there waiting at the part counter (including the one waiting for his request)?
The formulas are as below
Lq=N- (A+U)(1-P0)/A
Lq=0.43427
L=Lq+1-P0 =1.0298
On average how long does each mechanic have to wait for the part manager to take his request?
Wq=Lq/(N-L)A= 0.43427/(5-1.0298)3 = 0.36461 hr = 2.188 minutes
On average how much time does each mechanic spend at the part counter away from servicing cars?
W= Wq+ 1/U =2.188 +60/20 =5.188 minutes
On average how many mechanics are busy servicing the cars at any time?
Mechanics busy servicing the cars= N- L =5-1.0298 = 3.9702
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