1. Outline Multi period stochastic Inventory control
Continuous review
(Q,R) model
Service level; Cycle service level and Fill rate
Periodic Review
Order-up-to policy
(s,S) policy
Single period stochastic inventory control – Newsboy model
ABC analysis for multi item inventory control

2. Sources of uncertainity
Demand
Could be correlated or independent across time, across several items.
Could be stationary or nonstationary (trend, seasonality)
We will assume single item, independenly identically distributed across time, and stationary demand
Lead time
Product quality

3. Multi period stochastic inventory problem – Continuous review (Q,R)
Policy; Order Q whenever inventory posistion reaches R
Decision variables; Q,R
Order
received
Order
received
Order
received
Inventory
Order
received Q Q Q On
Hand R
Order
placed
Order
placed
Order
placed
Time τ τ τ
TBO1
TBO2
TBO3

4. Terminology Inventory level: Stock that is physically on the shelf
Inventory Position = On-hand + On-order – Backorders
Reorder point is based inventory position.
Safety Stock: Avg level of the inventory just before a replenishment arrives
In case of shortage;
Complete backordering: backordered demand is filled as soon as an adequate-size replenishment arrives
Complete lost sales: when out of stock, demand is lost, customers go somewhere else

5. Multi period stochastic inventory problem – Continuous review (Q,R)
Order
received
Order
received
Order
received
Inventory
Order
received Q Q Q On
Hand R
Order
placed
Order
placed
Order
placed
Time τ τ τ
TBO1
TBO2
TBO3

6. Average (expected) Inventory Profile with stochastic demand
Multi period stochastic inventory problem – Continuous review (Q,R)
Average (expected) Inventory Profile with stochastic demand
Inv
level Q Q
Avg. Cycle Inv=Q/2
Safety Stock (ss)
Time

7. (Q,R) model cost function
G(Q,R): Expected annual cost of [Fixed order cost + Inventory holding cost + Shortage cost]
Trade-offs;
As Q increases?
Avr. inventory increases, number of orders in a year decreases
As R increases?
Avr inventory increases since safety stock increases, shortage costs decreases since the probability of running out of stock decreases.

8. (Q,R) model cost function
G(Q,R) = Exp ( annual fixed ordering cost + annual inventory holding cost + annual shortage cost)
Exp. annual fixed ordering cost;
Exp. annual inventory holding cost;

9. (Q,R) model cost function
Expected shortage cost;
Expected total annual cost;

10. Reorder point with random demand
Pdf of demand during lead time
Inventory Level
P(Stockout) = 1- F(R)
Freq ss μ
R = μ+ss
Reorder Point , R
Safety Stock (SS)
Place order
Receive order
Time
Lead Time

11. (Q,R) model cost function
Safety stock = Reorder level – exp. demand during lead time
Expected total annual cost;

12. Exp. Number # of shortages in a cycle; n(R)
n(R) = Exp. # of shortages in a cycle n(R) = E[ max (x_R,0)] x: Random demand during lead time f(x) = pdf for x F(x) = Pr. dist. function for X 𝑛 𝑅 = 0 ∞ max 𝑥−𝑅,0 𝑓 𝑥 𝑑𝑥
𝑛 𝑅 = 0 ∞ 𝑥−𝑅 𝑓 𝑥 𝑑𝑥

13. (Q,R) model cost function: minimization
Expected Cost Function:
Partial Derivatives:
(1)
(2)

14. (Q,R) model cost function: minimization
Partial Derivatives:
(2)

15. Exp. shortage in a cycle when demand during lead time follows normal distribution
𝑛 𝑅 = 𝑅 ∞ 𝑥−𝑅 𝑓 𝑥 𝑑𝑥
L(z): standardized loss function
= 𝜎 𝑧= 𝑅−𝜇 𝜎 ∞ 𝑡−𝑧 𝜙 𝑡 𝑑𝑡
𝜙 𝑡 :𝑝𝑑𝑓 𝑓𝑜𝑟 𝑠𝑡𝑑. 𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
Tabulated values we have
𝑛 𝑅 = 𝜎𝐿 𝑧 = 𝜎𝐿( 𝑅−𝜇 𝜎 )
𝜇=𝑒𝑥𝑝. 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑑𝑒𝑚𝑎𝑛𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑙𝑒𝑎𝑑 𝑡𝑖𝑚𝑒
𝜎=𝑠𝑡𝑑. 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑒𝑚𝑎𝑛𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑙𝑒𝑎𝑑 𝑡𝑖𝑚𝑒

16. Service Levels in (Q,R) Systems
In many circumstances, the penalty cost, p, is difficult to estimate.
For this reason, it is common business practice to set inventory levels to meet a specified service objective instead.
Cycle (Type 1) service level (alpha): Choose R so that the probability of not stocking out during the lead time is equal to a specified value.
Appropriate when a shortage occurrence has the same consequence independent of its time and amount.
Fill rate (Type 2) service level (beta): Choose both Q and R so that the proportion of demands satisfied directly from stock equals a specified value.
Appropriate when a shortage amount is important.

17. Comparison of Type 1 and Type 2 Services
Order Cycle Demand Shortage
For a type 1 service (cycle service level) objective there are two cycles out of ten in which a stock-out occurs, so the type 1 service level is 80%.
For type 2 service (fill rate), there are a total of 1,450 units demand and 55 total shortage (which means that 1,395 demand are satisfied). This translates to a 1395/1450= 0,96 => 96% fill rate.
Fill rate is always greater than cycle service level

18. Solution to (Q,R) Systems with cycle service level (Type 1 Service) Constraint
For type 1 service, if the desired service level is α then one finds R from F(R)= α and Q=EOQ
Specify a, which is the proportion of cycles in which no stockouts occur. This is equal to the probability that the entire demand is satisfied in a cycle.

19. Solution to (Q,R) Systems with Fill rate (Type 2 Service) Constraint
1. Set Q=EOQ and
2. Find R to satisfy n(R) = (1-β)Q since;
Calculate n(R) using Q and beta
Calculate L(z)
Find the corresponding z from the table
Using this z, calculate R

20. Example Example 5.4. Havey shop buys and sells mustard sauce
c = 10$ /jar, T = 6 months lead time
I = 0,2/ year, q= 25$/ jar (shortage cost per unit)
K = 50 $
Demand during lead time follows a Normal Dist. With
μ = σ = 25 jars 𝑄 ∗ , 𝑍 ∗ = ?
λ= 100.(2) = 200 units / year (expected demand per year)

21. Example Example 5.5. for the problem in ex. 5.4
If we want cycle service level α=0,98, what would be the solution
1) Q = EOQ = 2𝐾𝜆 ℎ = =100
2) R = μ + 𝑍 α . σ
R = 𝑍 0, =151
𝑄,𝑅 = 100,151
𝑆𝑆= 𝑍 0, =51

22. Example b) If we want fill rate β = 0,98 what would be the solution?
𝑄=𝐸𝑂𝑄=100
𝐿 𝑧 = 𝑄 (1−𝛽) 𝜎 = 100 (1−0,98) 25 =0,08
𝑧=1,02 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒
𝑧= 𝑅 − 𝜇 𝜎 𝑅= 𝜇+𝑧 .𝜎
𝑅=100+ 1, =126
𝑠=1,02 25 = 𝑠𝑎𝑓𝑒𝑡𝑦 𝑠𝑡𝑜𝑐𝑘

23. Example c) What is the cycle service level for the solution (100,126)
𝛼=𝐹 𝑅 =𝐹 126 = 𝐹 𝑠 − = 𝐹 𝑠 1,02 =0,846
What is the fill rate for the solution (100,151)
𝑛 𝑅 =𝜎𝐿 𝑧 =𝑄 1−𝛽
𝛽=1− 𝑛 𝑅 𝑄 =1 − 𝑛
𝑛 151 =25 . 𝐿 151− =25 . 𝐿 2,05 =0,185
𝛽=1− 0, ≥0,99

24. Distribution of demand during lead time
Constant lead time of T periods
γ=𝑆𝑡𝑑. 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑒𝑚𝑎𝑛𝑑 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑
λ=𝐸𝑥𝑝. 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑑𝑒𝑚𝑎𝑛𝑑 𝑝𝑒𝑟 𝑝𝑒𝑟𝑖𝑜𝑑
Expected value of demand during lead time
𝜇= 𝜆 . 𝑇
Std. Deviation of demand during lead time
σ=γ . 𝑇
𝜎 2 = 𝑇=1 𝑇 γ 𝑖 2
(𝐷𝑒𝑚𝑎𝑛𝑑 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑝𝑒𝑟𝑖𝑜𝑑 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑟.𝑣)

25. Distribution of demand during lead time
2. Lead time is a random variable
𝜇 𝜏 =𝑒𝑥𝑝. 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑒𝑎𝑑 𝑡𝑖𝑚𝑒
𝜎 𝜏 2 =𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑒𝑎𝑑 𝑡𝑖𝑚𝑒
Expected value of demand during lead time
𝜇 =𝜆 ∗ 𝜇 𝜏
Variance of demand during lead time
𝜎 2 = 𝜇 𝜏 ∗ γ 2 + 𝜆 2 ∗ 𝜎 𝜏 2
σ= 𝜇 𝜏 ∗ γ 2 + 𝜆 2 ∗ 𝜎 𝜏 2

26. Distribution of demand during lead time
Example 5.6 Constant Lead Time
A mechanic shop orders buji
Demand per week ~ 𝑁 ( 𝜇 𝐻 =34, 𝜎 𝐻 =12)
𝑇= 6 weeks (lead time)
Distribution of demand during lead time?
𝜇=𝑇 . 𝜆=(6)(34)=204
𝜎=γ 𝑇 =12 6 =29,39
Dist. of demand during lead time ~ 𝑁 (𝜇=204 ,𝜎=29,39)

27. Distribution of demand during lead time
Example 5.7 (Lead time is R.V)
Harvey shop orders a special olive
Lead time is a R.V. and 𝜇 𝜏 =4 𝑚𝑜𝑛𝑡ℎ𝑠, 𝜎 𝜏 =6 𝑤𝑒𝑒𝑘𝑠 (1,5 𝑚𝑜𝑛.)
Demand per month ~ 𝑁 ( 𝜇 𝐴 =15 𝑗𝑎𝑟𝑠, 𝜎 𝐴 =6 𝑗𝑎𝑟𝑠)
Distribution of demand during lead time?
𝜇 =𝜆 ∗ 𝜇 𝜏 =4∗15=60 𝑗𝑎𝑟𝑠
σ= 𝜇 𝜏 ∗ γ 2 + 𝜆 2 ∗ 𝜎 𝜏 2
4 ∗ ∗ (1,5) 2
σ 2 =620,25
Distribution of demand during lead time?
~ 𝑁 (𝜇=60 , 𝜎 2 =650,25)

28. Imputed shortage cost
From Ex. 5.5; We had 𝛽=0,98 and corresponding solution 𝑄,𝑅 = 100,126 What would be the value of penalty per shortage (p) that would give us the same solution?
𝑝= 100(2) 200 (1−𝐹 126 )
1−𝐹 𝑅 = 𝑄 ∗ ℎ 𝑝 ∗ 𝜆
𝐹 126 = 𝐹 𝑠 126− =0,86
𝑝= 𝑄 ∗ ℎ 𝜆 (1−𝐹 𝑅 )
𝑝= 100(2) 200(1−0,86) =7,14