Question

The Europa Company has a large warehouse in Florida to store its inventory of goods until they are needed by the company’s many furniture stores in that area. A single crew with four members is used to unload and/or load each truck that arrives at the loading dock of the warehouse. Management currently is downsizing to cut costs, so a decision needs to be made about the future size of this crew.

Trucks arrive randomly at the loading dock at a mean rate of 2 per hour. The time required by a crew to unload and/or load a truck has an exponential distribution with a mean given by (60/n) minutes where n is crew size. The cost of providing each member of the crew is $20 per hour. The cost that is attributable to having a truck not in use (i.e., a truck standing at the loading dock) is estimated to be $30 per hour.

1. What is the minimum crew size needed to achieve steady state?
2. What is the crew-size needed to minimize hourly total cost? Try out different values of crew-size. Create a table showing the costs of the different crew-sizes.
3. For the crew-size you determine in (b), calculate the following:

1. Average utilization of crew
2. Average number of trucks waiting to be unloaded
3. Average time in queue
4. Probablity at least 2 trucks will be waiting to be unloaded
5. Probability a truck will have to wait at least 1 hour prior to unloading
6. Probability that a truck will be in and out of the dock in 1 hour

4. Using the same crew-size as in (c), assume that with the use of modern lifting equipment, the standard deviation of unloading time is reduced by 50%. This means service time no longer has an exponential distribution. Calculate:
   1. The average number of trucks waiting to be unloaded
   2. The average time in queue
   3. Compare your answers with those for part (c). What is the explanation for the difference?

Answer 1

Solution :

Arrival rate (λ) = 2 per hour

Service rate (µ) = 60/ (60/n) = n per hour

Number of servers (c) = 1 because we will consider the entire crew as a single server

Cost of waiting (Cw) = 30 per hour

Cost of service (Cs) = 20 per hour

The single server queue system formulas are

P = Prob Po = Probſsystem is 1., a Frob [empty (idle) = 1 - average number la in the queue Mine - 2) · ū - à L = average number in the system 2 W = average time in the queue ulu – 2) W = average time 1 =- in the systemu - À

a.

In order for the crew to achieve steady state, the utilization λ/c µ should not be more than 1. This is only possible if at least the crew size is 2. However this will barely be sufficient hence we should have a crew size of 3 at least.

b.

The total cost of the operation in queuing system is given by Cw*L + Cs*c

At crew size of 3, the service rate is 3. This means L = 2/(3-2) = 2. This means the cost is 30*2 + 20*3 = 120

At crew size of 4, the service rate is 4. This means L = 2/(4-2) = 1. This means the cost is 30*1 + 20*4 = 110

At crew size of 5, the service rate is 5. This means L = 2/(5-2) = 0.66. This means the cost is 30*0.66 + 20*5 = 119.8

Since the price has started rising, we can stop. The dipping point is at crew size 4 with a total cost of 110

c.

Optimum crew size = 4

i. Average utilization of the crew = λ/ µ = 2/4 = 0.5 or 50%

ii. Average # of trucks waiting = Lq = 2^2 / (4*(4-2)) = 0.5

iii. Average time in queue = Wq = 2/(4*(4-2)) = 0.25 hours or 15 minutes

iv. At least 2 trucks waiting means there is at least three trucks in the system. Probability of empty queue P0 = 1- λ/ µ = 0.5
Probability of 1 truck in the system = (λ/ µ)^1 * P0 = 0.5^1 * 0.5 = 0.25

Probability of 2 trucks in the system = (λ/ µ)^2 * P0 = 0.5^2 * 0.5 = 0.125

Probability of 3 trucks or more in the system = 1 – 0.5 – 0.25 – 0.125 = 0.125

Probability of at least 2 trucks waiting is 0.125 or 12.5%

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