Question

A bank of 10 machines requires regular periodic service. Machine running time and service time are both exponential. Machines run for an average of 44 minutes between service requirements, and service time averages six minutes per machine. What is the probability that a machine will have to wait for service with two operators

Answer 1

Answer:

0.346

Step-by-step explanation:

Given:

Number of machines, n = 10

Average run time before servicing, R= 44 minutes

Service time, T= 6 minutes per machine

Required:

Find the probability that a machine will have to wait for service with two operators.

First find the service factor using the formula below:

$X = \frac{T}{R + T}$

$= \frac{6}{44 + 6}$

$= \frac{6}{50}$

$= 0.12$

X = 0.12

Given number of service operators,S = 2.

Using the finite queuing table, efficiency factor at S=2 & X=0.12 is 0.346.

Therefore, the probability that a machine will have to wait for service with two operators is 0.346

Answer 2

The probability that a machine will have to wait for service with two operators is 0.3455

The given parameters are:

$\mathbf{n = 10}$ --- number of machines

$\mathbf{T = 6}$ --- service time

$\mathbf{R = 44}$ --- average run time

$\mathbf{S = 2}$ --- number of service operator

Start by calculating the service factor (X)

$\mathbf{X = \frac{T}{R + T}}$

So, we have:

$\mathbf{X = \frac{6}{44 + 6}}$

$\mathbf{X = \frac{6}{50}}$

$\mathbf{X = 0.12}$

Next, we determine the efficiency factor from the finite queuing table, a

The efficiency factor at $\mathbf{X = 0.12}$ and $\mathbf{S = 2}$, is 0.3455

Hence, the probability is 0.3455

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