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A company currently has 200 units of a product on hand that it orders every 2 weeks when the salesperson visits the premises. Demand for the product averages 20 units per day with a stand deviation of 5 units. Lead time for the product to arrive is 7 days. Management has a goal of a 95% probability of not seeking out for this product. The salesperson is due to come in late this afternoon when 180 units are left in stock (assuming that 20 are sold today). How many units should be ordered

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Answer: 289 units

Step-by-step explanation:

Given the following :

Inventory (I) = 180

Lead time (L) = 7 days

Review time (T) = 2 weeks = 14 days

Demand (D) = 20

Standard deviation (σ) = 5

Zscore for 95% probability = 1.645

Units to be ordered :

D(T + L) + z(σT+L)

(σT+L) = √(T + L)σ²

= √(14 + 7)5²

= √(21)25

= 22.9

D(T + L) + z(σT+L) - I

20(14 + 7) + 1.645(22.9 + 7) - I

= 420 + 49.1855 - 180

= 289.1855

= 289 quantities

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