A maintenance plan for a machine will be prepared. This machine can be in n diff

Question

A maintenance plan for a machine will be prepared. This machine can be in n different states where i = 1,2, ..., n. When the machine is in any i state, there is operational cost that is represented by g (i). There is a relation between the costs g (1) g (2) ... g (n) associated with the situation and the best case to find the machine and the worst case.

The machine in any state goes to (i + 1) state with the probability of p (i, i + 1)> 0 in the next period, if it does not receive maintenance, otherwise it remains in the same state with probability of 1-p(i, i+1) in the next period.

If i = n, the machine will be in n state in the next period. In all other cases, the transition probabilities are zero.

Assuming that the machine state is known at the beginning of each period, one of the following two decisions must be given:

1-)Allowing the machine to work one more period without any intervention

2-)Receiving machine to maintenance with R cost in order to transit it into state 1 that is the best state for the machine.

a) Considering the infinite time horizon, write the dynamic programming model in order to minimize the average cost per period.

b) For the best solution to this problem, show that an i * value exists and i> i * is the best solution if machine receives maintenance, otherwise keeping the machine running is best solution.

Explanation / Answer

Solution:

let,

X(t)= number of operational (both used and back up) machines at time t.

Now at any given point of time, a machine could work or break with probability 0.5 each

At max, all 10 machines could be operational and at min, 0 machine could be operational.

Hence X(t) ~ Bin(10, 0.5)

a. One machine can have one of the following 3 states at any given point of time.

1. Broken- A machine can be broken with probability 0.5 at any point of time.

2. Operational- a machine can work good with probability 0.5 at any point of time. Now, within the machines that works good, a machine can either be used or can be put as back up. Hence,

2.1. Operational and used with probability 0.5* (8/10)= 0.4 (M)

2.2. Operational and back up wiith probability 0.5*(2/10)= 0.1 (X(t)-M)

X(t) is the number of operational machines and ranges from 0 to 10. M is the number of used machines and ranges from 0 to 8. X(t)~ Bin(10, 0.5), M~ Bin(8,0.4)

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