1. Time Inventory Demand During Lead Time

2. LT ROP when demand during lead time is fixed

3. LT Demand During Lead Time is Variable

4. Inventory Time Demand During Lead Time is Variable

5. Average demand during lead time A large demand during lead time ROP Time Quantity Safety stock reduces risk of stockout during lead time Safety Stock Safety stock LT

6. ROP Time Quantity Safety Stock LT

7. When to re-order (ROP) Demand during lead time has Normal distribution. We can accept some risk of being out of stock, but we usually like a risk of less than 50%. If we order when the inventory on hand is equal to the average demand during the lead time; then there is 50% chance that the demand during lead time is less than our inventory. However, there is also 50% chance that the demand during lead time is greater than our inventory, and we will be out of stock for a while. We usually do not like 50% probability of stock out

8. 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)

9. 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

10. μ and σ of demand during lead time Demand of sand during lead time has an average of 50 tons Standard deviation of demand during lead time is 5 tons Assuming that the management is willing to accept a risk no more that 5%. What is the service level. Service level = 1-risk of stockout = 1-.05 =.95 Find the z value such that the probability of a standard normal variable being less than or equal to z is.95 Go to normal table, look inside the table. Find a probability close to.95. Read its z from the corresponding row and column. z=1.64

11. Demand of sand during lead time has an average of 50 tons Standard deviation of demand during lead time is 5 tons Assuming that the management is willing to accept a risk no more that 5% z = 1.64 What ROP? ROP = Average demand during lead time + ss ROP = 50 + 1.64(5) = 58.2 How much safety stock do we have. ROP is 58.2, Average demand during lead time = 50 ss = 58.2- 50 = 8.2 Alternatively ss = 1.64(5) = 8.2 μ and σ of demand during lead time

12. 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

13. 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)

14. 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)

15. Demand During Lead Time is Variable N(μ,σ) Demand of sand during lead time has an average of 50 tons Standard deviation of demand during lead time is 5 tons Assuming that the management is willing to accept a risk no more that 5%. What is the service level. Service level = 1-risk of stockout = 1-.05 =.95 What is the z value corresponding to this service level? z=1.64 What is safety stock (ss) ss = z (standard deviation of demand during lead time) ss = 1.64 (5) = 8.2

16. What is ROP? ROP = Average demand during lead time + ss μ and σ of demand during lead time ROP = 50 + 1.64(5) = 58.2

17. Demand of sand during lead time has an average of 75 tons Standard deviation of demand during lead time is 10 tons Assuming that the management is willing to accept a risk no more that 10%. What is the service level. Service level = 1-risk of stockout = 1-.1 =.9 RiskService levelz value.1.9 1.28.05.95 1.65.01.99 2.33 Example 2; total demand during lead time is variable

18. z = 1.28 What is safety stock? ss = 1.28(10) = 12.8 ROP = Average demand during lead time + ss ROP = 75 + 12.8 = 87.8 The general relationship between service level, risk, and safety stock Service level increases Risk decreases ss increases Example 2; total demand during lead time is variable

19. If demand is variable and Lead time is fixed μ and σ of demand per period and fixed LT

20. Demand of sand has an average of 50 tons per week. Standard deviation of the weekly demand is 3 tons. Lead time is 2 weeks. Assuming that the management is willing to accept a risk no more that 10% z = 1.28 μ and σ of demand per period and fixed LT

21. If lead time is variable and demand is fixed μ and σ of Lead Time and fixed demand

22. Lead Time Variable, Demand fixed Demand of sand is fixed and is 50 tons per week. The average lead time is 2 weeks. Standard deviation of lead time is 0.5 week. Assuming that the management is willing to accept a risk no more that 10%. Compute ROP and SS. Acceptable risk; 10%  z = 1.28