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 demand is 5 per day, 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 Assignment Ch12(d)

2. Re-Order Point (ROP) in periodic inventory control is the beginning of each period. (ROP) in perpetual inventory control is the inventory level equal to the 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, and demand was really constant; the ss is zero and ROP is when inventory on hand is equal to the average demand during 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 What is 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. Problem 4(b): If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock. ss = 70 What is 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. RO P 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)

6. Common z Values RiskService levelz value.1.9 1.28.05.95 1.65.01.99 2.33 RO P Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand

7. Relationship between z and Normal Variable x RO P Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand 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.1.91.28.05.951.65.01.99 2.33  x = μ +z σ

8. Relationship between z and Normal Variable ROP RO P Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale Average demand LTD = Lead Time Demand ROP = Average LTD +z (Standard Deviation of LTD) ROP = Average LTD +ss ss = z (Standard Deviation of LTD)

9. Safety Stock and ROP RO P Risk of a stockout Service level Probability of no stockout Safety stock 0z Quantity z-scale RiskService levelz value.1.9 1.28.05.95 1.65.01.99 2.33 Average demand ss = z × (standard deviation of demand during lead time)

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 Risk of a stockout Service level Probability of no stockout Expected demand Safety stock x ROP; total demand during lead time is variable

11. Risk of a stockout Service level Probability of no stockout z RiskService levelz value.1.9 1.28.05.95 1.65.01.99 2.33 ROP; total demand during lead time is variable

12. 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

13. (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

14. 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

15. 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

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

17. ROP; Variable d, Fixed LT

18. 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 ROP Risk of a stockout Service level Probability of no stockout Expected demand Safety stock x Now it is transformed into our previous problem where total demand during lead time is variable

19. 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

20. (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; Variable d, Fixed LT

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

22. 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