Question

Daily demand for a product is 130 units, with a standard deviation of 20 units. The review period is 10 days and the lead time is 10 days. At the time of review there are 40 units in stock.

If 99 percent service probability is desired, how many units should be ordered?

Ordered quantity units.

Answer 1

Review period(T) = 10 days

Lead time (L) = 10 days

Average daily demand(d) = 130 units

Standard deviation of daily demand( d) = 20 units

Service level = 99%

With a service level of 99% the value of Z = 2.33

On hand inventory(I) = 40 units

So,order quantity = [d(T+L)] + [Z X d X √(T+L)] - I

= [130(10+10)] + [2.33 X 20 X √(10+10)] - 40

= (130 X 20) + (2.33 X 20 X √20) - 40

= 2600 + (2.33 x 20 x 4.47) - 40

= 2600 + 208.40 - 40

= 2808.40 - 40

= 2768.40 or rounded off to 2768 units