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?

Respuesta :

Answer:

690 units

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:

[tex]q = \bar{d}(L+R) + z \sigma_{L+R} - I[/tex]

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

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