1 / 21

a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss =0, compute ROP.

ROP Problems. 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?

gilles
Download Presentation

a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss =0, compute ROP.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ROP Problems 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.

  2. Demand Forecast, and LeadTimes 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 timeplus 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.

  3. ROP; Fixed d, Fixed LT, Zero SS 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

  4. ROP; Fixed d, Fixed LT, Positive SS 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

  5. ROP; total demand during lead time is variable 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

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

  7. Common z Values Risk of a stockout Average demand Safety stock z-scale Service level Probability of no stockout ROP Quantity 0 z Risk Service level z value 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33

  8. Relationship between z and Normal Variable x Risk of a stockout Average demand Safety stock z-scale Service level Probability of no stockout ROP Quantity 0 z z = (x-Average x)/(Standard Deviation of x) x = Average x +z (Standard Deviation of x) μ = Average x σ = Standard Deviation of x Risk Service z value level 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33  x = μ +z σ

  9. ROP and ss Risk of a stockout Average demand Safety stock z-scale Service level Probability of no stockout ROP Quantity Risk Service z value level 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33 0 z 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

  10. ROP; total demand during lead time is variable Service level Risk of a stockout Probability of no stockout ROP Expected demand Safety stock x 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 Risk Service z value level 0.10.9 1.28 0.050.95 1.65 0.010.99 2.33

  11. ROP; total demand during lead time is variable 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

  12. ROP; Variable d, Fixed LT 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.

  13. ROP; Variable d, Fixed LT 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.

  14. ROP; Variable d, Fixed LT 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

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

  16. ROP; Variable d, Fixed LT

  17. Now it is transformed into our previous problem where total demand during lead time is variable 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.

  18. ROP; Variable d, Fixed LT 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

  19. ROP; Variable d, Fixed LT 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

  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

More Related