Question

Annual demand for a product is 13,000 units; weekly demand is 250 units with a standard deviation of 40 units. The cost of placing an order is $100, and the time from ordering to receipt is four weeks. The annual inventory carrying cost is $0.65 per unit.a. To provide a 98 percent service probability, what must the reorder point be?b. Suppose the production manager is told to reduce the safety stock of this item by 100 units. If this is done, what will the new service probability be?

Answer 1

Answer:

a. Reorder point is 1,164 units to provide a 98 percent service probability.

b. the new service probability will be 79% if production manager reduces the safety stock by 100 units.

Explanation:

a. To provide a 98 percent service probability, what must the reorder point be?

This can be calculated as follows:

Step 1: Calculation of optimal order quantity

The optimal order quantity also known as economic order quantity (EOQ) using the following formula:

$EOQ = \sqrt{\frac{2 *D*O}{C} }$ ........................................... (1)

Where,

EOQ = Optimal order quantity = ?

D = Annual demands = 13,000

O = Ordering cost = $100

C = Carrying cost of annual inventory = $0.65 per unit

Substituting the values into equation (1), we have:

$EOQ = \sqrt{\frac{2*13,000*100}{0.65} }$

$EOQ = \sqrt{\frac{2,600,000}{0.65} }$

$EOQ = \sqrt{4,000,000}$

EOQ = 2,000 units

Step 2: Calculation of standard deviation during the lead time

This can be calculated using the following formula:

$SL = \sqrt{L*(S)^{2} }$ ................................................. (2)

Where;

SL = Standard deviation during the lead time = ?

L = Lead time = 4

S = Standard deviation = 40

Substituting the values into equation (2), we have:

$SL = \sqrt{4 *(40)^{2} }$

$SL = \sqrt{4*1,600}$

$SL =\sqrt{6.400}$

SL = 80

Also, z = 2.05 from the standard normal distribution

Step 3: Calculation of reorder point

Total calculate reorder point, we use the following formula:

R = (d * L) + (z * SL) ............................................ (3)

Where;

R = Reorder point = ?

d = Weekly demand = 250

L = Lead time = 4

z = 2.05

SL = Standard deviation during the lead time = 80

Substituting the values into equation (3), we have:

R = (250 * 4) + (2.05 * 80)

R = 1,000 + 164

R = 1,164 units

Therefore, reorder point is 1,164 units to provide a 98 percent service probability.

b. Suppose the production manager is told to reduce the safety stock of this item by 100 units. If this is done, what will the new service probability be?

ISS = Initial safety stock = z * SL = 2.05 * 80 = 164

If the safety stock is reduced by 100 units, we have:

NSS = New safety stock = ISS - 100 = 164 - 100 = 64

The new z (nz) can be obtained as follows:

NSS = nz * SL ................................................. (4)

Where;

NSS = 64

nz = new z = ?

SL = Standard deviation during the lead time = 80

Substituting the values into equation (4) and solve for nz, we have:

64 = nz * 80

nz = 64 / 80

nz = 0.80

For the new z, nz = 0.80, from Standard Normal distribution, the new service probability is 79%.

Therefore, the new service probability will be 79% if production manager reduces the safety stock by 100 units.