n a continuous review system, the average daily demand for a part is Normally distributed with mean 20 and standard deviation of 4. The lead time to receive the part from the time it is ordered is 9 days. The appropriate re-order point for this part if we want a 95% service level is a. 200 b. 20 c. 180 d. 184

200

Explanation:

Based on the information given in the question, the appropriate re-order point for this part if we want a 95% service level will be:

Mean demand = 20

Standard deviation of demand = 4

Lead time = 9 days

Service level = 95% = 95/100 = 0.95

Re-order Point will be:

= (demand × lead time) + (z* × std dev × ✓leadtime)

= (20 × 9) + (1.645 × 4 × ✓9)

= (180) + (1.645 × 4 × 3)

= 199.74

= 200 approximately

The re-order point is 200

y=mx+b is the right slope of velocity carry the 4 over