Question

Problem 11-20 Daily demand for a product is 100 units, with a standard deviation or 15 unts The review peniod is 20 days and the lead time is 10 days. At the time of review there are 50 units in stock If 98 percent service probability is desired, how many units should be ordered? (Use Excet's NORMSINVO function to find the correct critical value for the given a-level. Do not round intermediate calculations. Round "r value to 2 decimal places and final answer to the nearest whole number.) Ordered quantity References Book & Resources Worksheet Difticulty Intermediate Learning Objective 20-02 Analyze how difterent inventory contol systems won Problem 11-20

Answer 1

Order interval(T) = 20 days

Lead time (L) = 10 days

Average daily demand(d-bar) = 100 units

Standard deviation of daily demand( $\sigma$ d) = 15 units

Service level = 98%

With aservice level of 98% the value of Z = 2.05

On hand inventory(I) = 50 units

So,order quantity = [d-bar(T+L)] + [Z X $\sigma$ d X Sqrt of (T+L)] - I

= [100(20+10)] + [2.05 X 15 X Sqrt of (20+10) - 50

= (100X30) + (2.05 X 15 X sqrt of 30) - 50

= 3000 + (2.05 x 15 x 5.48) - 50

= 3000 + 168.51 - 50

= 3168.51-50

= 3118.51 or rounded to 3119 units

So the ordered quantity = 3119 units