Hello, I am currently working on the problem below, however I am completely lost

Question

Hello, I am currently working on the problem below, however I am completely lost on how to approach it, please help me.

A food company is attempting to set the customer service level (in-stock probability in its warehouse) for a particular product line item. Annual sales for the items are 100,000 boxes, or 3,846 boxes biweekly. The product cost in inventory is $10 per box, to which $1 is added as profit margin. Stock replenishment is every two weeks and the demand during this time is assumed normally distributed with a standard deviation of 400 boxes. Inventory carrying costs are 30 percent per year of item value. Management estimates that a 0.156 percent change in total revenue would occur for each 1 percent change in the in-stock probability. Based on the information, find the optimal in-stock probability for the item.

SL Levels in % for Various z Values

SL (%) Zu - ZL = z

U L   

87-86 1.125 -1.08 = 0.045

88-87 1.17 -1.125 = 0.045

89-88 1.23 -1.17 = 0.05

90-89 1.28 -1.23 = 0.05

91-90 1.34 -1.28 = 0.06

92-91 1.41 -1.34 = 0.07

93-92 1.48 -1.41 = 0.07

94-93 1.55 -1.48 = 0.07

95-94 1.65 -1.55 = 0.10

96-95 1.75 -1.65 = 0.10

97-96 1.88 -1.75 = 0.13

98-97 2.05 -1.88 = 0.17

99-98 2.33 -2.05 = 0.28

Developed from entries in a normal distribution table

Explanation / Answer

The optimal service level will be achieved at a point where the change in cost is equal to change in profit i.e. C= p

p = trading margin * rate of sales response * total annual sales

p = $1*0.00156*100,000 = $156 per year ( per 1% change in the service level)

C= annual carrying cost* product cost* z*demand std. deviation

C=0.30*$10* z*400

C=$1200 z

As we know that C= p

Thus 156 =1200 z

z =0.13

From the z table given, for the value of z=0.13, the probability is= 97-96

So the answer is 97-96

PLEASE RATE THE ANSWER AS THUMBS UP. THANKS A LOT

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