Truck Queue at Warehouse Submit Assignn Due Dec 1 by 6pm Points 10 Submitting a

Question

Truck Queue at Warehouse Submit Assignn Due Dec 1 by 6pm Points 10 Submitting a file upload At the warehouse, trucks arrive at the rate of 10 per shift (in a Poisson distribution). If the warehouse single dock can handle 12 unload/reload per shift: 1. What is the average length of the waiting line? 2. What is the average percentage of time (shift) a truck spends in the waiting line? 3. What is the utilization of the dock? 4. What is the probability that there are 3 or more trucks ahead of you, if were arriving at the warehouse? Hint- use Example 12.7 in text Previous Next How to use Canvas STOP Meet the p most respor for your sa

Explanation / Answer

We assume that unloading / loading of trucks at the warehouse single dock follows negative exponential distribution.

Accordingly,

Arrival rate of trucks = a = 10 / shift

Handling of truck = s = 12 / shift

Average length of the queue ( waiting line )

= a ^2 / S x ( S – a )

= 10 x 10/ 12 x ( 12 – 10)

= 100/ 24

= 4.17 ( rounded to 2nd decimal place )

AVERAGE LENGTH OF THE WAITING LINE = 4.17

Time the truck spends in the system

= a/ s x ( s -a ) + 1/s shift

= 10/12 x ( 12 – 10)   + 1/12

= 5/12 + 1/12

= 6/12

= 1 / 2 shift

Time the truck spends in the waiting line

= a/ s x ( s -a ) shift

= 10/ 12 x ( 12 – 10)

= 5/12 shift

Therefore , percentage of time the truck spends in the waiting line

= 5/12 / ( 6/12) x 100

= 500/ 6

= 83.33 %

PERCENTAGE OF TIME THE TRUCK SPENDS IN THE WAITING LINE = 83.33 %

Utilization of the dock

= a/s x 100

= 10 /12 x 100

= 83.33 %

UTILIZATION OF THE DOCK = 83.33 %

Probability that there are ZERO trucks waiting = Po = 1 – a/s = 1 – 5/6 = 0.166

Probability that there are 1 truck waiting = P1 = ( a/s) x P0 = ( 10/12) x 0.166 = 0.138

Probability that there are 2 trucks waiting = P2 = ( a/s)^2 x Po = ( 10/12)^2 x 0.166 = 0.115

Hence , probability that there are less than 3 trucks waiting

= P0 + P1 + P2

= 0.166 + 0.138 + 0.115

= 0.419

Therefore, Probability that there are 3 or more number of trucks

= 1 – Probability that there are 3 or more number of trucks

= 1 – 0.419

= 0.581

PROBABILITY THAT THERE ARE 3 OR MORE NUMBER OF TRUCKS AHEAD = 0.581

AVERAGE LENGTH OF THE WAITING LINE = 4.17

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