Any one have an idea on how to start this problemSolutionQue

Any one have an idea on how to start this problem?

Solution

Queuing Theory :

a )

Parts arrive for processing at a bottleneck tool at an avg rate = a = 15 per hour = 60 / 15 = 4 /min

Tool process part =p = 3.75 min , standard deviation = 1.3 min

What is avg no of parts waiting at bottleneck station =

u= p / a

= 3.75 / 4

= 0.9375

average time spent in the queue waiting =T = p * (u/1-u)(1+1/2)

=3.75 * (0.9375/1-0.9375)*1

= 56.25

No of customers waiting = (1/a) *(T+P)

= (1/4)*(56.25+3.75)

= 15

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b )

Too, Crib

Exponential Interval & service times

Mean time between arrivals =a = 3.5 min

Single tool crib attedent = p = 3 min avg for service

Tool rib attendant paid = $ 25 /hour

1 ) Utilization of the tool crib attendant = u

u = p / a

=3 / 3.5

= 0.8571

2 ) Hourly cost

1st case Now in this case

Each mechanics is served in 3 min.

That means in hour 20 mechanic will served.

Now the cost of single mechanic is $ 25 / hour =$ 0.41 /min

For single mechanic = 3 * 0.41 = $1.25 per service time

For 20 mechanic =20 * 1.25

= $ 25

2nd case

Tool crib attendant cost = $ 15 / hour

Add 1 & 2 case

= 25 + 15

$ 40

$ 40 will be hourly cost at tool crib.

Because of time shortage i could not solve simulation problem