Question

In a continuous review inventory​ system, the lead time for door knobs is weeks. The standard deviation of demand during the lead time is units. The desired​ cycle-service level is percent. The supplier of door knobs streamlined its operations and now quotes a 1 week lead time. Refer to the standard normal tableLOADING... for​ z-values. How much can the safety stock be reduced without reducing the percent​ cycle-service level? The safety stock can be reduced by nothing door knobs. ​(Enter your response rounded to the nearest whole​ number.)

Answer 1

Answer:

The answer is "116 doorknobs".

Explanation:

The standard deviation of the demand before the (four weeks) protection intervals = $\sigma-d \times (\sqrt{L}) = 100 \ units$

The desired cycle service level is $99\%$.Therefore, $z = 2.33$

The safety stocks for the four-weeks protecting interval are:

Safety stock $= z\times [ \sigma-d \times (\sqrt{L})]$

                     $= 2.33 \times 100 \\\\= 233\ door\ knobs$

The safety stocks require for the one-week protection interval are: $\sigma-dLT = \sigma-dt \times (\sqrt{L}) = \sigma-dt \times (\sqrt{4}) = 100\ door\ knobs\\\\\sigma-d = \frac{100}{(\sqrt{4})} = \frac{100}{2} = 50 \ door\ knobs$

Safety stock $= z\times \sigma-dt = 2.33 \times 50 = 116.5 \ or\ 117 \ door\ knobs$

Safety stock reduction$= 233 -117 = 116 \ door\ knobs$