Question

Daily Demand for a product is 60 units with a standard deviation of 10 units. The review period is 10 days, and lead time is 2 days. At the time of review there are 100 units inventory stock. If 98 percent serviceprobability is desired, how many units should be ordered?

Answer 1

690 units

Step-by-step explanation:

daily demand = 60 units

standard deviation = 10 units

review period = 10 days

lead time = 2 days

stock = 100 units

service probability = 98%

order quantity using fix time period formula:

`q = \bar{d}(L+R) + z \sigma_(L+R) - I`

• standard deviation =
  `\sigma_(L+R) \\` = √(10 + 2) x 10 = 34.64
• z for 98% = 2.05
• I = stock = 100
• 
  `\bar{d}(L+R)` = 60 x (10 + 2)

order quantity = (60 x 12) + (2.05 x 34.64) - 100 = 690 units