Question

You work in a factory that runs 24 hr a day, 7 days a week. The boss insists that all 10 of the machines in the factory get a good servicing before he goes on vacation. Each of these machines has a part replaced that has an MTTF(µ) of 10,000hrs. The boss will then take the next 2 months (60 days) off for his “well deserved” vacation. The purchasing department will not purchase any expensive items while the boss is on vacation. Each of the parts they just replaced costs so much, they will not be bought without the boss’s permission. And the boss wont take a call during his vacation.

What is the probability that he will replace 2 or less of these parts in a year due to failure? Use the binomial distribution.

If all 10 parts are replaced before his annual vacation, and any part that fails in a year is replaced when it fails, how many parts does he replace in a year on average? (10 + E(x))

E(x) = 10*(1-P(x))(from part ii)

If each device costs $10,000, how much does he spend on average, in a year, replacing parts?

Redo the calculations for part 3 to part 6 using a better part that has an MTTF of 100,000 and costs $80,000.

Use the above information to explain to your boss that he is a fool for using the cheap stuff.

Answer 1

P(x) = 1- exp (-x/mu)

which is the probability of failure of a part in x hours where mu = MTBF

P( 1440) = 1- exp ( -0.144)

=0.1341

Require probability = P(0) + P(1) + P(2)

= 10! / 0! x 10! X P⁰ X(1-P)¹⁰ + 10! / 1! x9! xP x (1-P)⁹ + 10! / 2! x8! xP² x( 1-P)⁸

= 0.2369+0.366+ 0.2557 = 0.8586

Parts replaced during one year

= 10+ 10 x probability of failure

P(x) = 1 - exp ( 8640/10000)

= 1-1/2.372 = 0.5785

Number of parts replaced = 10+ 10x0.5785 = 16

Cost of replacement of parts = 160000

Probability of failure of the new part in 2 months

P(x) = 1- exp ( 1440 /100000)

= 1-1/1.0145 = 0.0142

Probability of failure of the part in a year

= 1- 1/ exp( 8640/100000)

= 1-1/1.014 =0.014

Parts replaced per year = 10x0.014 = 0.14

Cost of replacement = 80000x0.14 = 11200

which is significantly less then the annual cost of replacement of older parts.

Hence new marts are recommneded, as they will pay for their initial cost in few years.