An on-site mechanic is responsible for maintaining 5 machines. On any given day
ID: 3069327 • Letter: A
Question
An on-site mechanic is responsible for maintaining 5 machines. On any given day each will fail independently with probability 20%. Suppose that she can easily work with 1 machine a day, but any additional would require a service of an outside contractor i. What is the probability that an outside contractor will be required on any given day? ii. Suppose that the contractor services cost $1000 per call (regardless of the number of ma chines failed), and by changing some production practices you can reduce the probability of failure on all of the machines to 10%. How much would you be willing to pay to implement these changes? iii. Without changes described in (b) how much would you allocate for contractor services in an annual budget? Justify your answer. Can you estimate the probability that your value will be too much or too little?Explanation / Answer
i.) P(outside contractor required) = P(more than 1 machine fails on a day) = 1 - P(less than or equal to 1 machine fail) = 1 - [P(0 machine fail) + P(1 machine fail)]
=> P(outside contractor req.) = 1 -[0.8^5 + 5*(0.8^4)*0.2^1] = 1 - 0.7373 = 0.2627
ii.) P(outsider contractor req at 10% failure chance) = 1 - [(0.9^5) + 5*(0.9^4)*0.1^1] = 1 - 0.9185 = 0.0815
Saved Money = Expected Cost at 20% failure - Expected Cost at 10% failure
= 1000*0.2627 - 1000*0.0815 = $181.2
Therefore, we'd be willing to spend anything less tha $181.2 to make the failure rate less
iii.) Without the changes, expected cost per day = 1000*0.2627 = $262.7
therefore, annually = 365*262.7 = $95885.5
This will be too much in Best case => Less than machine fails on all days => P(no machine fails any day) = (1 - 0.2627)^365 = 4.9*10^-49
This will be too little in the worst case => More than 1 machine fail every day => 0.2627^365 = 1.2674*10^-212