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Question

The manager of the fuel distribution center for the Jacksonville area has to define the inventory

management policy for fuel. The average demand in the area is 220,000 gallons per day, with a standard

deviation of 50,000 gallons. It takes 2 days for processing and delivering a new order from the refinery.

The manager wants to achieve a service level of 99.6%.

* How much is the z value for this example?

A. 2.5

B. 1.75

C. 2.65

D. None of the above

* How many gallons should the safety stock be?

A. Less than 180,000

B. 180,000199,999

C. 200,000219,999

D. 220,000 or more

*The owner of the distribution center decided to invest in a new fuel tank. The total safety stock is

now 230,000 gallons. What will the resulting service level if the standard deviation and lead time

remains the same?

A. 99.6099.69%

B. 99.7099.79%

C. 99.8099.89%

D. 99.9099.99%

* Due to a change in the standard deviation of demand, the manager changed the safety stock to

112,000 gallons. The demand per day remains the same. What will the reorder point?

A. 100,000299,999

B. 300,000499,999

C. 500,000699,999

D. 700,000899,999

Solution

a)

For maintaining the service level of 99.6%, we should have a Z value of
Z = P^-1(0.996)

= 2.65

b)

Thus, safety stock = 2.65 * sqrt(2) * 50,000

Thus, option B is correct

Pls. note there are many other ways to determine the safety stock and hence there can be multiple correct answers.

c)

Assuming the same equation as above:

230,000 = Z-score * sqrt(2) * 50000

Thus, z-score = 3.2526

Thus, this score corresponds to level of 99.90 - 99.99% of service level

d)

Reorder point = lead time demand + safety stock

new lead time demand = (220,000 + (2.65 * new stdev) )

new stdev = 112000 / (sqrt(2) *2.65) = 29885 ~ 30,000

Thus, new lead time demand = (220,000 + (2.65 * 30000) ) = 299500

new safety stock = 112000

Thus, reorder point = 299500 + 112000 = 411500

hence option B is correct