3. Consider a job shop that consists of 3 identical machines and 2 technicians.

Question

3. Consider a job shop that consists of 3 identical machines and 2 technicians. Suppose that, the amount of time each machine operates before breaking down is exponentially distributed with mean 10 and, a technician takes to fix a machine is exponentially distributed with parameter 0.4. Suppose that all the times to breakdown and times to repair are independent random variables and let X(t) be the umber of operating machines at time t. (a)3] Write the Kolmogorov's forward differential equations in terms of p,(t)P,) P(X(t)-jlX (0) 0), for j 0,1,2,3 (b) [2] Obtain the equilibrium probabilities pi = InnxP)(t) (c)1] What is the average number of busy technicians in the long-run?

Explanation / Answer

a)Now first of all at time "t" there can be 3 possibiities.Either there can be 2 machines presnt ,1 machine present or no machine present.So the possible states od the markov chain X(t) are 0,1,2.

Also the point to be noted is that only One machine can be repaired at a single time as there is only 1 repairman.

so PROB(of going from 0 to 2 machines)=0

PROB(2 to 0 machines)=0

PROB(of going from 0 to 1 machine) means that 1 machine is being repaired

and if one 1 machine repairs at rate "l    " so the probability of going from 0 to 1 machine is “(l)/(m+l)”

similarly PROB(1 to 0 machine) means that 1 machine has failed and the rate of failure of machine is “m”   so the PROB(1-0)=m/m+l

The possible transitions from state 0 is to state 1 only with prob=(l)/(m+l)”

c)

Prob (0-2)=0

So prob(0-0)= -(l)/(m+l) (as the sum of rows of a generator matrix =0)

So the continuous time markov chain is ,

And the generator matrix is

states

0

1

2

0

-(l)/(m+l)

(l)/(m+l)

0

1

m/m+l

-(m+l)/(m+l)=-1

(l)/(m+l)

2

0

m/m+l

-m/m+l

where m represent "mu" and

l represents 'lambda"

states

0

1

2

0

-(l)/(m+l)

(l)/(m+l)

0

1

m/m+l

-(m+l)/(m+l)=-1

(l)/(m+l)

2

0

m/m+l

-m/m+l

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