1. a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss=0, compute ROP. b) If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock. ss=70, compute ROP? c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss. d) If average demand per day is 20 units and standard deviation of daily demand is 5, and lead time is 16 days. Compute ROP at 90% service level. Compute ss. e) If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days. Compute ROP at 90% service level. Compute ss. ROP Problems

2. ROP (Re-Order Point) in periodic inventory control is the beginning of each period. ROP in perpetual inventory control is the inventory level equal to the average demand during the lead time plus a safety stock to cover demand variability. Lead Time is the time interval from placing an order until receiving the order. If there were no variations in demand (it was constant) then safety stock (ss) is zero. In the case of ss=0, ROP is when inventory on hand is equal to the average demand during the lead time. Demand Forecast, and Lead Times

3. a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss = 0 Compute ROP. ROP = demand during lead time + ss ROP = demand during lead time Demand during lead time = (lead time) × (demand per unit of time) Demand during lead time = 20 × 10 = 200 ROP; Fixed d, Fixed LT, Zero SS

4. b) If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock. ss = 70 Compute ROP. ROP = demand during lead time + ss Demand during lead time = 200 ss = 70 ROP = 200+70 = 270 ROP; Fixed d, Fixed LT, Positive SS

5. c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss ROP; total demand during lead time is variable

6. Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Safety Stock and ROP Each Normal variable x is associated with a standard Normal Variable z Average demand There is a table for z which tells us a)Given any probability of not exceeding z. What is the value of z b)Given any value for z. What is the probability of not exceeding z x is Normal (Average x, Standard Deviation x)  z is Normal (0,1) ROP

7. Common z Values RiskService levelz value 0.1 0.9 1.28 0.05 0.95 1.65 0.01 0.99 2.33 Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand ROP

8. Relationship between z and Normal Variable x Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand ROP z = (x-Average x)/(Standard Deviation of x) x = Average x +z (Standard Deviation of x) μ = Average x σ = Standard Deviation of x RiskService z value level 0.1 0.91.28 0.05 0.951.65 0.01 0.99 2.33  x = μ +z σ

9. ROP and ss Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand ROP LTD = Lead Time Demand (Demand during the lead time) ROP = Average LTD +z (Standard Deviation of LTD) ss = z (Standard Deviation of LTD) ROP = Average LTD +ss RiskService z value level 0.1 0.91.28 0.05 0.951.65 0.01 0.99 2.33

10. c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss ROP; total demand during lead time is variable ROP Risk of a stockout Service level Probability of no stockout Expected demand Safety stock x RiskService z value level 0.1 0.91.28 0.05 0.951.65 0.01 0.99 2.33

11. z = 1.28 z = (X- μ )/σ X= μ +z σ μ = 200 σ = 25 X= 200+ 1.28 × 25 X= 200 + 32 ROP = 232 ss = 32 ROP; total demand during lead time is variable

12. d) If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss. ROP; Variable d, Fixed LT

13. Previous Problem: If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss. This Problem: If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss. If we can transform this problem into the previous problem, then we are done, because we already know how to solve the previous problem. ROP; Variable d, Fixed LT

14. d) If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss. What is the average demand during the lead time What is standard deviation of demand during lead time ROP; Variable d, Fixed LT

15. Demand During Lead Time If demand is variable and Lead time is fixed

16. ROP; Variable d, Fixed LT

17. The Problem originally was: If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss. We transformed it to: The average demand during the lead time is 320 and the standard deviation of demand during the lead time is 20. Compute ROP at 90% service level. Compute ss. Which is the same as the previous problem: If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss. Now it is transformed into our previous problem where total demand during lead time is variable

18. z = 1.28 z = (X- μ )/σ X= μ +z σ μ = 320 σ = 20 X= 320+ 1.28 × 20 X= 320 + 25.6 X= 320 + 26 ROP = 346 SS = 26 ROP; Variable d, Fixed LT

19. e) If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days. Compute ROP at 90% service level. Compute ss. What is the average demand during the lead time What is standard deviation of demand during lead time ROP; Variable d, Fixed LT

20. Demand Fixed, Lead Time variable If lead time is variable and demand is fixed

21. Lead Time Variable, Demand fixed e) Demand of sand is fixed and is 20 units per day. The average lead time is 16 days. Standard deviation of lead time is 4 days. Service level is 90%. Service level ; 90%  z = 1.28