Question
Inputs Average demand per unit time R = 280 Lead time,L= 2 SD of demand per unit time sR = 180 Order quantity Q= 1100
Inputs | |
Average demand per unit time R = | 280 |
Lead time,L= | 2 |
SD of demand per unit time sR = | 180 |
Order quantity Q= | 1100 |
Reorder Point ROP | 700 |
Distribution of demand during replenishment lead time | |
Mean demand during lead time RL = | 560 |
SD of demand during lead time sL = | 254.56 |
Outputs | |
Safety Inventory ss = | 140 |
Cycle Service Level CSL = | 70.88% |
Expected shortage per replenishment cycle ESC = | 46.537 |
Fill Rate fr = | 95.77% |
After further analysis, WALMART determines that the supply lead time from EPSON is actually normally distributed, with a mean of 2 weeks and a standard deviation of 1.5 weeks.How much safety inventory should WALMART carry if it wants to provide a CSL of 95 percent? What is the expected shortage per replenishment cycle (ESC)?What is the fill rate (fr)? What impact does the variability in the supply lead time have on the safety stock? What if the variability in the supply lead time were reduced to zero? What would the safety stock be for a desired CSL of 95 percent?
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