Suppose a product must go through an assembly line made up of five sequential op

Question

Suppose a product must go through an assembly line made up of five sequential operations. The time it takes to complete each operation is normally distributed with a mean of 180 seconds and standard deviation of 5 seconds. Let X denote the cycle time for the line, so that after X seconds, each operation is supposed to be finished and ready to pass the product to the next operation in the assembly line. If the cycle time X = 180 seconds, what is the probability that all five operations will be completed? What cycle time will ensure that all operations are finished 98% of the time? Suppose that the company wants all operations to be completed within 190 seconds 98% of the time. Further suppose that the standard deviation of an operation can be reduced at a cost of $5,000 per second of reduction (from 5), and any or all operations may be reduced as desired by up to 2.5 seconds. By how much should the standard deviations be reduced to achieve the desired performance level, and how much would that cost? Suppose a product must go through an assembly line made up of five sequential operations. The time it takes to complete each operation is normally distributed with a mean of 180 seconds and standard deviation of 5 seconds. Define the flow time to be

Explanation / Answer

a). X = 180. Probability that any one operation will be completed = NORMDIST(180,180,5,1) = 0.5 . Probability that all five operations will be completed = 0.55 = 0.03125 or 3.125%

b). For all operations to finish 98% of the time, each operation need to be finished with a probability equal to 0.981/5 = 0.99597 . X corresponding to this probability (0.99597) =NORMINV(0.99597,180,5) = 193.248 seconds

c) given that, X = 190 seconds, in part (b) we determined the probability 0.99597 as that of individual operations for all 5 operations to finish 98% of the time. Using excel goal seek function, we can determine the standard deviation so that individual operation's cycle time is 190 seconds with a confidence level of 0.99597 , we can also find Z-value from the standard normal table and calculate the desired standard deviation using this formula, 190 = z* + µ, where Z = 2.64955, corresponding to 0.99597. µ = 180 , Therefore = (190-180)/2.64955 = 3.77423 sec.

Standard deviation required to achieve the desired performance level = 3.77423 seconds

Cost = (5 - 3.77423)*5000 = $ 6129

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