Question

here is the machine and serviceman problem.

Question 9 (10 points): Consider a job shop that consists of M machines and one serviceman. Suppose that the amount of time each machine nuns before breaking down is exponentially distributed with mean 1/1, and suppose that the amount of time that it takes for the serviceman to fix a machine is exponentially distributed with mean 1/u. Based on given information above (a) 6 points) What is the average number of machines not in use? (b) (5 points) What proportion of time is each machine in use?

Answer 1

Solution:-

Given that

Consider a job shop that consists of M machines and one serviceman. Suppose that the amount of time each machine runs before breaking down is exponentially distributed with mean ,

1/3

Suppose that the amount of time that it takes for the serviceman to fix a machine is exponentially distributed with mean . Based on given information above

1/

In a birth and death process with birth rate $\lambda$ and death rate $\mu$ , the probability of system to be in state 'n' is given as

(1)

Pn λολι....λη+1 λολι...λε-1) μιμ2...μη(1 + ΣΕΙ n=1 μιμ2...μη -1 η Σ1

Where $\lambda_n,\mu_n$ are birth and death rates for state n

(a) What is the average number of machines not in use?

For the given problem, if the state n is defined as that "n machines are not in use", then it can be modeled as a birth and death process having parameters

$\mu_n=\mu\ \ n\geq 1$

$\lambda_n=(M-n)\lambda \ \ n\leq M$

Here failing machine is regarded as a birth process and a fixed machine as a death process.

If any machine fails, serviceman rate for each state $\mu_n=\mu$ .

For state n, if n machines are not in use, then
(M-n
machines are in use and each fail at rate $\lambda$ .

so for state n, $\lambda_n=(M-n)\lambda$

Putting values of $\lambda_n$ and $\mu_n$ in eqn (1)

we get

$P_n=\frac{(M-0)\lambda:(M-1)\lambda...(M-n+1)\lambda}{\mu\mu...ntimes (1+\sum_{n=1}^{M}\frac{(M-0)\lambda(M-1)\lambda...(M-n+1)\lambda}{\mu\mu...n times})}$$=\frac{M(M-1)...(M-n+1)\lambda^n}{\mu^n (1+\sum_{n=1}^{M}\frac{M(M-1)...(M-n+1)\lambda^n}{\mu^n})}$

$P_n=\frac{(\frac{\lambda}{\mu})^n(\frac{M!}{(M-n)!})}{(1+\sum_{n=1}^{M}(\frac{\lambda}{\mu})^n(\frac{M!}{(M-n)!}))}$............(2)

Average number of machines not in use

$\sum_{n=0}^{M}nP_n=\sum_{n=0}^{M}\frac{(\frac{\lambda}{\mu})^n(\frac{M!}{(M-n)!})}{(1+\sum_{n=1}^{M}(\frac{\lambda}{\mu})^n(\frac{M!}{(M-n)!}))}$...........(3)

(b) What proportion of time is each machine in use?

proportion of time each machine in use.

$=\sum_{n=0}^{M}p(machine\ is\ working\ /n machine\ not\ working)P_n$$=\sum_{n=0}^{M}\frac{M-n}{M}P_n$

$=\sum_{n=0}^{M}(1-\frac{n}{M})P_n$

$=1-\frac{1}{M}\sum_{n=0}^{M}nP_n$

Where

$\sum_{n=0}^{M}nP_n$ is given by equation (3)

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