10 per hour the second mechanic b takes about 4

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5 minutes on an average to repair a machine and demands wages Rs. 10 per hour. The second mechanic B takes about 4 minutes on an average to repair a machine and demands wages Rs. 15 per hour. Assuming that the rate of machine breakdown is Poisson-distributed and the repair rate is exponentially distributed, which of the two mechanics should be engaged? Solution: Given problem is M/M/1 Model. The ideal time cost of machine = Rs.20 /hr Number of working hours in factory = 8 hrs Total cost = (Total Wages + Cost of non-productive time) /day = (Hourly rate x No. of hours) +{(Average No. of machines in the system)x(Cost of idle machine hour)x(No. of hours)} Mechanic A: Average arrival rate of breakdown machines ( λ ) = 10 machines /hr Average service rate of breakdown machines ( μ ) = 12 machines /hr Service cost = Rs. 10/hr
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Total cost = Total Wages + Cost of non-productive time = (Hourly rate x No. of hours) +{(Avg. no. of machines in the system)x(Cost of idle machine hour)x(No. of hrs)} = (10x8) + {( λ /( μ - λ ))x20x8} = 80 + {(10/(12-10))x160} = 80 + 800 = Rs.880 /day Mechanic B: Average arrival rate of breakdown machines ( λ ) = 10 machines /hr Average service rate of breakdown machines ( μ ) = 15 machines /hr Service Cost = Rs. 15/hr Total cost = Total Wages + Cost of non-productive time = (Hourly rate x No. of hours) +{(Avg. no. of machines in the system)x(Cost of idle machine hour)x(No. of hrs)} = (15x8) + {( λ /( μ - λ ))x20x8} = 120 + {(10/(15-10))x160} = 120 + 320 = Rs.440 /day Interpretation: The incurring cost with Mechanic A is Rs.880 and with Mechanic B is Rs. 440. Mechanic B is employed because with him incurring cost is less compare to Mechanic A.
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