5. (25 points) A factory manager must decide whether to stock a particular spare

Question

5. (25 points) A factory manager must decide whether to stock a particular spare part. The part is absolutely essential to the operation of certain machines in the plant. Stocking the part costs S10 per day in storage and cost of capital. If the part is in stock, a broken machine can be repaired immediately, but if the part is not in stock, it takes one day to get the part from the distributor, during which time the broken machine sits idle. The cost of idling one machine for a day is S65. There are 50 machines in the plant that require this particular part. The probability that any one of them will break and require the part to be replaced on any one day is only 0.004 (regardless of how long since the part was previously replaced). The machines break down independently of one another. (5 points) If you wanted to use a probability distribution for the number of machines that break down on a given day, would you use the binomial or Poisson distribution? Why? a. b. (5 points) Whichever theoretical distribution you chose in part a, what are the appropriate parameters? That is, if you chose the binomial, what are the values of p and n? If you chose the Poisson, what is the value of ? (15 points) If the plant manager wants to minimize his expected cost, should he keep zero, one, or two parts in stock? (Do not forget that more than one machine can fail in one day!) c.

Explanation / Answer

a) We will use binomial distribution, because the problem involves determining the probability of a number of machines down for part replacement out of a given number of machines. Further the machine breakdown is independent. Therefore, binomial distribution is the most appropriate.

b) Parameters of binomial distribution are:

p = 0.004

n = 50

Parameters of poisson distribution = 0.004*50 = 0.20

c) With 0 parts in stock, expected cost = 65*Pk*k, where Pk is the probability of k machine breakdowns.

Probability of breakdown is calculated by formula =BINOMDIST(k, 50, 0.004, 0), where k is # of machine breakdowns. Cost = 65*Pk*k, for example for 1 machine breakdown, cost = 65*0.16434*1 = 10.68

Total cost per day with 0 parts in stock = 0+10.68+2.1+0.2+0.01 = $ 13

With 1 part in stock, Total cost = Probability of 0 breakdown * cost of keeping 1 part in inventory + 65*Pk*k for k greater than 1 = 0.8184*10 + 65*(0.01617*(2-1)+0.00104*(3-1)+0.00005*(4-1)) = $ 9.38

With 2 parts in stock, Total cost = Probability of 0 breakdown * cost of keeping 2 parts in inventory + Probability of 1 breakdown* cost of keeping 1 part inventory + 65*Pk*k for k greater than 2 = 0.8184*2*10 + 0.16434*1*10 + 65*(0.00104*(3-2)+0.00005*(4-2)) = $ 18.1

Total cost is lowest with keeping 1 part in stock, therefore, he should keep 1 part in stock

# of Machines breakdown (k) Probability of breakdown (Pk) Cost = 65*Pk*k 0 0.81840 0.00 1 0.16434 10.68 2 0.01617 2.10 3 0.00104 0.20 4 0.00005 0.01

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.