Question

In a large law firm, the demand for copier paper has normally distributed with an average of 10 packages of a day with the standard deviation of 2. Each package contains 500 sheets. The firm operates 255 days a year. Holding cost for the paper is $1 per year per package, and ordering cost is $10 per order.

a) What order quantity would minimize total annual ordering and holding cost?

b) What is the total annual inventory cost associated with the EOQ?

c) How many orders per year should be placed if the order size set to EOQ?

d) What will be the average inventory level is the order size is set to EOQ?

e) Before adopting EOQ order quantity, the law firm used to have the order size of 200 packages for each individual order. How much do you think the firm would save by switching to the EOQ-based order size?

f) If it takes exactly 3 days to receive a newly placed order in full, and the managers would like to avoid stock out with %95 chance, what would be the reorder point? (Note: Z- value for %95 is 1.65)

g) Managers of the law firm are interested to study the
situation in which the lead time **averages** on 3
days with the standard deviation of 1. In this situation, if the
managers would like to avoid stock out with %95 chance what would
be the right reorder point? (Note: Z-value for %95 is 1.65)

Answer #1

Annual demand, D = 255*10 = 2550 packages

Ordering cost, S = $ 10

Holding cost, H = $ 1

a) Economic order quantity, Q = SQRT(2DS/H)

= SQRT(2*2550*10/1)

= 226 packages

b) Total annual inventory cost = SD/Q + HQ/2

= 10*2550/226 + 1*226/2

= $ **225.83**

c) Number of orders per year = D/Q = 2550/226 = 11.3

d) Average inventory level = Q/2 = 226/2 = **113**
packages

e) Annual inventory cost of previous policy (Q=200) = 10*2550/200 + 1*200/2 = $ 227.5

Annual savings by switching to the EOQ based order size = 227.5
- 225.83 = $ **1.67**

f) z value for 95% service level = 1.65

Reorder point = average daily demand * lead time in days + z * Std deviation of daily demand * SQRT(lead time)

= 10*3 + 1.65*2*SQRT(3)

= **36** packages

g)

d = 10 packages per day

L = 3 days

_{L}
= 1

_{d}
= 2

Reorder point = d*L + z*SQRT(_{d}^{2}L
+
_{L}^{2}d^{2})

= 10*3 + 1.65*SQRT(2^{2}*3 +
1^{2*}10^{2})

= **47.5**

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