Let A= event of 0 failure per day
B= event of 1 failure per day
C= event of 2 failure per day
Given P(A)=.64
P(B)=.32
P(C)=.04
If we hire 1 repairman then his cost is $50
Maintenance Cost= (Downtime cost when 0 machine fails)+(Downtime cost when 1 machine fails)+(Downtime cost when 2 machine fails)
= 0 + P(B)*(time for repair* cost) + P(C)*(time for repair when both machines down* cost + time when one machine repaired)
=0 +.32* 4 *80 + .04 *(4*120 + 4*80)
=134.4
Expected Cost = Downtime cost + Repairman Cost
=134.4+ 50
=$184
If we hire 2 repairmen then their cost = 2*50= $100
Maintenance Cost= (Downtime cost when 0 machine fails)+(Downtime cost when 1 machine fails)+(Downtime cost when 2 machine fails)
Maintenance cost = 0 + .32*3*80(both men repair together) + .04*(4*120)(both men repair a machine each)
= $96
Total Expected Cost = $100+$96
=$196
With No repairman available even when one machine breaks down for 8 hours
Downtime Cost becomes = .32*8*80
=$204
which exceeds the total expected cost with 1/2 repairmen on duty
Thus 1 repairmen should be hired and the expected values are as calculated above.
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