Chapter 20.9, Problem 9P

Explanation of Solution

Given:

It is given that the Army has 200 tanks. Tanks need maintenance 10 times per year, the maintenance takes an average of 2 days. To have an average of at least 180 tanks working, the number of repairmen needed is to be determined. It is assumed that the inter arrival and service times are exponentially distributed.

Determining the number of repairmen required:

This is similar to a machine repair problem containing “K” machines and “R” repair people is a $M\left|M\right|R\left|GD\right|K|K$queuing system where the inter arrival times, service times are exponential, R servers, general queue discipline, first “K” implies that at any time, no more than “K” machines may be present, the second “K” implies that the arrivals are drawn from a finite source of size “K”.

The steady state probability at state j is as follows:

$\begin{array}{l}{\pi }_j=(C{}_j){\rho }^j{\pi }_0\text{ (j=0,1,}\mathrm{....}\text{,R)}\\ \text{ = }\frac{(C{}_j){\rho }^{j}j!{\pi }_0}{R!R^{(j-R)}}\text{ (j=R+1,R+2,}\mathrm{....}\text{K) }\end{array}$

Here,

• R represents the number of servers,
• K represents the number of machines,
• $\lambda$ represents the arrival time, and
• $\mu$represents the service rate

$\rho =\frac{\lambda }{\mu }$

Let the number of servers or repairmen be R=1, and the number of machines or tanks, K=200.

The arrival rate is, $\lambda =10times/year$

The service rate is, $\mu =\frac{365}{2}times/year$

Therefore,

$\begin{array}{l}\rho =\frac{\lambda }{\mu }\\ \text{ =}\frac{10}{\raisebox{1ex}{$365$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ \text{ =0}\text{.05479}\end{array}$

This machine repair problem is solved using a spreadsheet. In cell B2, enter the value of arrival rate $\lambda$, in cell C2, enter $\mu$, and in cell D2, enter the number of repairers, say 1