Question

A manufacturer has seven identical machines for the production. A typical machine breaks down once every 46 days and there are two repair workers, each of whom takes an average of 4 days to repair a machine. Breakdowns of machines follow a Poisson distribution and the repair time follows an exponential distribution.

( a ) What is the average number of machines waiting for repairs?

( b ) What is the average number of machines being served by a repair worker?

( c ) What is the average downtime for a machine that needs repairs?

( d ) Suppose each machine downtime costs $600/day and the wage of each repair worker

is $125/day. What is the expected total costs incurred per day?

Finite Queuing Table

XMDF

0.012 1 .048 .999 0.025 1 .100 .997 0.050 1 .198 .989 0.060 2 .020 .999

1 .237 .983 0.070 2 .027 .999 1 .275 .977 0.080 2 .035 .998 1 .313 .969 0.090 2 .044 .998 1 .350 .960 0.100 2 .054 .997 1 .386 .95

Answer 1

Given

N = machines = 7
U = typical machine breaks down time= 46 days
M = repair workers =2
T = average service time = 4 days

Service factor, X = T / (T+U) = 4 / (4+46) = 0.08

From the table, for X=0.08 and M=2, the value D is 0.044, and F is 0.998

(a)

Average number waiting: L = N*(1 - F) = 7*(1 - 0.998) = 0.014

(b)

Average number being serviced: H = F*N*X = 0.998*7*0.08 = 0.5589

(c)

Average number running: J = N*F*(1 - X) = 7*0.998*(1 - 0.08) = 6.427

So, average number of machines down = N - J

Average downtime for a machine = (N - J) * U = (7 - 6.427)*46 = 26.36 days

(d)

Average cost of downtime per day = (N - J)*600 = (7 - 6.427)*600 = 343.8
Average cost of the servers per day = M*125 = 2*125 = 250

So, average total cost per day = 343.8 + 250 = $593.8

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