ION FOURi(5 marko At Alpha Beta Manufacturing, which average, 40 machines break

Question

ION FOURi(5 marko At Alpha Beta Manufacturing, which average, 40 machines break dowrn day on the average. Now revenue, and the cost per mechanic is $200 per day. Poisson arrival and exponential service are uses a large number of production machines, on the every day. One service mechanic can repair 15 machines per assume that an idle machine costs $1000 per day in lost production assumed. a) Find the optimal repair crew size so that the total cost is minimum. Find the average waiting time in hours of a machine in the system under the economie crew size (assume 8-hour working day). b)

Explanation / Answer

Arrival rate, = 40 per day

Service rate,   = 15 per day

Arrival rate is more than service rate, therefore multiple servers (mechanics) are required.

Number of servers required = / = 40/15 = 2.7 ~ 3

Therefore, a minimum of 3 mechanics servers are required.

This is M/M/s queue model, with s number of servers.

We will analyse the daily operating cost with 3 mechanics and keep on increasing the number of mechanics by 1 as long as the operating cost decreases. We will stop increasing the number of mechanics, when the operating cost starts increasing.

With 3 mechanics

For s=0, (/)^s/s! = (40/15)^0/0! = 1

For s=1, (/)^s/s! = (40/15)^1/1! = 2.6667

For s=2, (/)^s/s! = (40/15)^2/2! = 3.5556

For s=3, (/)^s/s! = (40/15)^3/3! = 3.1605

(/)^s/s! (for s=0 to 2) = 1+2.6667+3.5556 = 7.2222

term 2 = ((/)^3/3!)/(1-/3) = 3.1605/(1-40/(3*15)) = 28.4444

P0 = 1/((/)^s/s!+term 2) = 1/(7.2222+28.4444) = 0.0280

Average number of machines in system (waiting or being repaired), L = P0*((/)^3/3!)*(/s)/(1-(/s))^2 + / = 0.0280*3.1605*(40/(3*15))/(1-(40/(3*15)))^2+40/15 = 9.04

Operating cost per day with 3 mechanics = s*Cs+L*Cw = 3*200+9.04*1000 = $ 9640

With 4 mechanics

For s=0, (/)^s/s! = (40/15)^0/0! = 1

For s=1, (/)^s/s! = (40/15)^1/1! = 2.6667

For s=2, (/)^s/s! = (40/15)^2/2! = 3.5556

For s=3, (/)^s/s! = (40/15)^3/3! = 3.1605

For s=4, (/)^s/s! = (40/15)^4/4! = 2.1070

(/)^s/s! (for s=0 to 3) = 1+2.6667+3.5556+3.1605 = 10.3827

term 2 = ((/)^4/4!)/(1-/4) = 2.1070/(1-40/(4*15)) = 6.3210

P0 = 1/((/)^s/s!+term 2) = 1/(10.3827+6.3210) = 0.0599

Average number of machines in system (waiting or being repaired), L = P0*((/)^3/3!)*(/s)/(1-(/s))^2 + / = 0.0599*2.1070*(40/(4*15))/(1-(40/(4*15)))^2+40/15 = 3.4235

Operating cost per day with 3 mechanics = s*Cs+L*Cw = 4*200+3.4235*1000 = $ 4223.5

With 5 mechanics

For s=0, (/)^s/s! = (40/15)^0/0! = 1

For s=1, (/)^s/s! = (40/15)^1/1! = 2.6667

For s=2, (/)^s/s! = (40/15)^2/2! = 3.5556

For s=3, (/)^s/s! = (40/15)^3/3! = 3.1605

For s=4, (/)^s/s! = (40/15)^4/4! = 2.1070

For s=5, (/)^s/s! = (40/15)^5/5! = 1.1237

(/)^s/s! (for s=0 to 3) = 1+2.6667+3.5556+3.1605+2.1070 = 12.4897

term 2 = ((/)^5/5!)/(1-/5) = 1.1227/(1-40/(5*15)) = 2.4080

P0 = 1/((/)^s/s!+term 2) = 1/(12.4897+2.4080) = 0.0671

Average number of machines in system (waiting or being repaired), L = P0*((/)^3/3!)*(/s)/(1-(/s))^2 + / = 0.0671*1.1237*(40/(5*15))/(1-(40/(5*15)))^2+40/15 = 2.8514

Operating cost per day with 3 mechanics = s*Cs+L*Cw = 5*200+2.8514*1000 = $ 3851.4

Repeating this process for 6 mechanics, we see that operating cost increases.

(a) In the above analysis, we see that Daily operating cost is minimum with 5 service mechanics (servers).

Therefore, optimal repair crew size is 5

(b) Average waiting time of machine in the system (W) = L/ = 2.8514/40= 0.0713 work day = 0.0713*8 = 0.57 hours or 34.22 minutes

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