A graphics reproduction firm has four units of equipment that are automatic but occasionally become inoperative because of the need for supplies, maintenance, or repair. Each unit requires service roughly twice each hour, or, more precisely, each unit of equipment runs an average of 30 minutes before needing service. Service times vary widely, ranging from a simple service (such as pressing a restart switch or repositioning paper) to more involved equipment disassembly. The average service time, however, is five minutes. Equipment downtime result in a loss of $20 per hour. The one equipment attendant is paid $6 per hour.

Using finite queuing analysis, answer the following questions:

a. What is the average number of units in line?

b. What is the average number of units still in operation?

c. What is the average number of units being serviced?

(a) Average number of unit in line = 0.256

(b) Average number of unit in operation = 3.209

(c) Average number of unit being service in operation = 0.535

Explanation:

Given dа ta:

Number of machine N = 4

Number of attendant (S) = 1

Service time (T) = 5 mins

Time required by the machine before servicing = 30 mins

Calculating the service factor (X) using the formula;

X = T / (T+U)

= 5 / (5+30)

= 5/35

= 0.1429

(a) Calculating the average number of unit in line (L) using the formula;

L = N * (1-F)

where, N = Number of unit

F = efficiency factor

L = average number of unit in line

Using the finite queuing table at X = 0.1429 and S = 1,

Efficiency factor = 0.936

Substituting, we have;

L = 4 * (1-0.936)

= 4 * 0.064

= 0.256

(c) Calculating the average number of unit being service in operation (H) using the formula;

H = N*F*X

= 4 * 0.936*0.1429

= 0.535

(b) Calculating the average number of unit in operation using the formula;

Average number of unit in operation = Number of unit-down unit

But down unit = L+H

The formula becomes;

Average number of unit in operation = Number of unit - (L+H)

= 4 - (0.256+0.535)

= 4-0.791

= 3.209