Kelly’s Tavern serves Shamrock draft beer to its customers. The daily demand for
ID: 424680 • Letter: K
Question
Kelly’s Tavern serves Shamrock draft beer to its customers. The daily demand for beer is normally distributed, with an average of 30 gallons and a standard deviation of 6 gallons. The lead time required to receive an order of beer from the local distributor is 13 days.
Determine the safety stock and reorder point if the restaurant wants to maintain a 90% service level. What would be the increase in the safety stock if a 95% service level were desired? (Round z values and final answer to 2 decimal places, e.g. 25.75.)
Explanation / Answer
Z value for 90% service level = NORMSINV ( 0.90 ) = 1.28
Standard deviation of demand during lead time
= Standard deviation of daily demand x Square root ( Lead time )
= 6 x Square root ( 13 )
= 6 x 3.605
= 21.63
Safety stock for 90% service level
= z value x Standard deviation of demand during lead time
= 1.28 x 21.63
= 27.68
Reorder point
= Daily demand x Lead time ( Gallons ) + safety stock
= 30 x 13 + 27.68
= 390 + 27.68
= 417.68
Z value for 95% service level = NORMSINV ( 0.95 ) = 1.64
Safety stock at 95% service level
= Z value x Standard deviation of demand during lead time
= 1.64 x 21.63
= 35.47
Therefore , increase in safety stock at 95% service level = 35.47 – 27.68 = 7.79
SAFETY STOCK AT 90% SERVICE LEVEL = 27.68
REORDER POINT AT 90% SERVICE LEVEL = 417.68
INCREASE IN SAFETY STOCK AT 95% SERVICE LEVEL = 7.79
SAFETY STOCK AT 90% SERVICE LEVEL = 27.68
REORDER POINT AT 90% SERVICE LEVEL = 417.68
INCREASE IN SAFETY STOCK AT 95% SERVICE LEVEL = 7.79
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