Question

the pre 5.22, The Seaboard Shipping Company has a warehouse terminal in ta Spartanburg, South Carolina. The capacity of each terminal dock is three trucks. As trucks enter the terminal, the drivers receive num- bers, and when one of the three dock spaces becomes available, the truck with the lowest number enters the vacant dock. Truck arrivals are Poisson distributed, and the unloading and loading times (service th of d times) are exponentially distributed. The average arrival rate at the terminal is five trucks per hour, and the average service rate per dock is two trucks per hour (30 minutes per truck). a. Compute L, Lg W, and W b. The management of the shipping company is considering adding extra employees and equipment to improve the average service time per terminal dock to 25 minutes per truck. It would cost the company $18,000 per year to achieve this improved service. Management estimates that it will increase its profit by $750 per year for each minute it is able to reduce a truck's waiting time. Determine whether management should make the investment. c.Now suppose that the managers of the shipping company have decided that truck waiting time is excessive and they want to reduce it. They have determined that there are two alternatives available for reducing the waiting time. They can add a fourth dock, or they can add extra employees and equipment at the existing docks, which will reduce the average service time per location from the original 30 minutes per truck to 23 minutes per truck. The costs of these alternatives are approximately equal. Management desires to implement the alternative that reduces waiting time by the greatest amount. Which alternative should be selected?

Answer 1

From the problem we can identify this as M/M/s queueing system. The parameters are

Arrival rate (lambda) = 5 per hour

Service rate (mu) = 60/30 = 2 per hour

Number of servers (c) = 3

Utilization rate (rho) = lambda/(mu*c) = 5/6 = 0.83

The formulas we use for this problem is shown below

1 prob.no customers in system Po= 71=c-1 12 1 си + (си-2 Po for n>c Pn= clon-c 17 1 Pn Pofor n c prob. of n customers in system L=uL Po+ average customers in the system (с-1)4еи— 2)2 -ll= L W average time customer spends in the system Lq L-average number of customers in the queue Wq-IV-L average time customer is in the queue 1(A Pw= Po-probability customer must wait for service 77CLI-A

In order to solve the problem quickly, we have applied these formulas on excel and will be changing the parameters and obtaining the required values.

a.

L = 6.01

Lq = 3.51

W = 1.2 hours

Wq = 0.70 hours or 42 minutes

A1 F A B C D M N O P 1 2 3 Inputs: 4 Arrival rate (2) 5 per hour 5 Service rate per server (u) 2 per hour 6 Number of servers (c) 3 7 Data 8 10 Probability that the system is empty 11 Average number of customers in queue Po 0.044944 3.511236 12 Average time spent in queue 13 Average time spent in the system 0.702247 1.202247 W 6.011236 14 Average number of customers in system 15 Probability that the time in the queue is 0 16 Probability that the time in the queue is Enter t 0: L We(0) 0.297753 17 no more than t time units. 18 Utilization 0.353901 0.083 0.833333 19 20 21 М-М-s Sheet1 1490 READY

b.

If the service time is reduced to 25 minutes then the value of mu =60/25 = 2.4

This means the waiting time in the queue (Wq) will be 0.22 or 13.2 minutes

f A1 Q P O M N L F D E C B A 1 2 3 Inputs 4 Arrival rate (2) 5 Service rate per server (u) 5 per hour 2,4 per hour 3 Number of servers (c) 6 7 Data 8 9 0.098178 Po 10 Probability that the system is empty 11 Average number of customers in queue 1.100513 Lg 0.220103 Wa 12 Average time spent in queue 0.636769 W 13 Average time spent in the system 14 Average number of customers in system 15 Probability that the time in the queue is 0 3.183847 L 0.515774 Wa(0) 16 Probability that the time in the queue is no more than t time units. 18 Utilization Enter t 0: 0.596887 0.083 17 0.694444 19 20 21 + M-M-s Sheet1 145% READY

The time saved will be 42-13.2 = 28.8 minutes per truck per year. This means the increase in profit is 28.8*750 = 21600 per year. However the cost is only 18000 per year. This means that the ROI is positive and the management should make the investment.

c.

Here we need to consider two options. Adding a fourth dock means that service time will remain 30 minutes but there will be 4 servers. This means the waiting time (Wq) will be 0.10 hours

A1 A B C D F M N O P 1 2 3 Inputs: 4 Arrival rate (2) 5 per hour Service rate per server (u) 5 2 per hour Number of servers (c) 6 4 7 Data 8 10 Probability that the system is empty 0.073695 Po 11 Average number of customers in queue 0.533095 12 Average time spent in queue 13 Average time spent in the system W4 W 0.106619 0.606619 3.033095 14 Average number of customers in system 15 Probability that the time in the queue is 0 16 Probability that the time in the queue is L Wa(0) 0.680143 17 Enter t 0: 0.083 0.750895 no more than t time units. 18 Utilization 0.625000 19 20 21 M-M-s Sheet1 459% READY

The next option is to keep 3 servers but reduce the service time to 23 minutes. This means the service rate will be 60/23 = 2.6. This means the waiting time (Wq) will be 0.14 hours

f A1 C D E F M A B L N O P Q 1 2 3 Inputs 4 Arrival rate (2) 5 Service rate per server (u) 5 per hour 2.6 per hour Number of servers (c) 6 3 7 Data 8 9 10 Probability that the system is empty 0.123852 Po 11 Average number of customers in queue L 0.730279 12 Average time spent in queue 13 Average time spent in the system 14 Average number of customers in system Wa 0.146056 W 0.530671 L 2.653356 Wo(0) 15 Probability that the time in the queue is 0 16 Probability that the time in the queue is no more than t time units. 18 Utilization 0.591044 Enter t 0: 17 0.083 0.676152 0.641026 19 20 21 M-M-s Sheet1 + READY 1459

Considering that the waiting time is lesser with the 4 server option, we should opt for the 4 server with 30 minutes service time.