Question

Problem 4: A make to order manufacturer has documented the four stages of its order fulfillment process, obtaining the following information: Process Stage Mean Duration (weeks Standard Deviation of Duration Order Acceptance and Price 0.25 Negotiation Design Work Contracting for Manufacturing Manufacturing Shipment 2 2 2 2 a) Assuming that the duration at each stage is normally distributed, determine the mean and variance of the total duration from initial customer contact to shipment b) Once the order has been accepted and the price agreed upon, the manufacturer must quote a delivery time to the customer. How far in the future, after the completion of price negotiations, should the delivery time be quoted to ensure that the probability of the order being delivered no later than that time is at least 0.96? c) If the firm quotes a delivery lead time of 16 weeks, what is the probability a given customer's order will be delivered late?

Answer 1

a)

Order Acceptance and Price Negotiation ~ N(1,0.25²)

Design Work ~ N(4,2²)

Contracting for Manufacturing ~ N(2,1²)

Manufacturing ~ N(6,2²)

Shipment ~ N(2,1²)

total time duration from initial customer contact to shipment ~ N(1+4+2+6+2 , 0.25²+2²+1²+2²+1²)

total time duration from initial customer contact to shipment ~ N(15,10.0625)

Mean of total time duration from initial customer contact to shipment = 15 weeks

Variance of total time duration from initial customer contact to shipment = 10.0625 sq. weeks

b)

Let total time duration from initial customer contact to shipment be denoted by T.

T ~ N(15,10.0625)   (T-15)/ ~ N(0,1)

Let t be the delivery time quoted by the manufacturer to the customer.

According to the problem ;

[We get the Z value from the standard normal table]

The delivery time should be quoted as 20.5544 weeks, i.e approximately 20 weeks 4 days.

c)

Probability that a given customer's order will be delivered later than 16 weeks

[We get the probability value from the standard normal table]