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Chapter 5 Inventory Control Subject to Uncertain Demand. Timing Decisions. Quantity decisions made together with decision When to order? One of the major decisions in management of the inventory systems. Impacts : inventory levels, inventory costs, level of service provided Models:
Quantity decisions made together with decision
When to order?
One of the major decisions in management of the inventory systems.
Impacts: inventory levels, inventory costs, level of service provided
(S, T) System
(Q, R) System
(s, S) System
Structure of timing decisions
“Christmas tree” modelOne-Time Decision
Situation is common to retail and manufacturing environment
Consider seasonal goods, which are in demand during short period only.
Product losses its value at the end of the season. The lead time can be longer than the selling season if demand is higher than the original order, can not rush order for additional products.
Christmas ornament retailer
finished good inventory
Trivial problem if demand is known (deterministic case), in practical situations demand is described as random variable (stochastic case).
Mrs. Kandell has been in the Christmas tree business for years. She keeps track of sales volume each year and has made a table of the demand for the Christmas trees and its probability (frequency histogram).
Q – order quantity; Q* - optimal
D – demand: random variable with probability density function f(D)
F(D) – cumulative probability function:
F(D) = Pr (demand ≤D)
co – cost per unit of positive inventory
cu – cost per unit of unsatisfied demand
Economics marginal analysis:
overage and underage costs are balanced
finding the expected profit of ordering one more unit.
Probability of not selling
Your Last item in stock and having extra inventory on hand at the end on the period
P(X < Q)
Selling everything, and facing shortage
P(X ≥ Q)
Single Period Inventory Model Marginal Analysis:
E (revenue on last sale) = E (loss on last sale)
Shortages = lost profit + lost of goodwill
Overage = unit cost + cost of disposal of the overage
Either ignore the purchase cost, because it does not impact the optimal solution or implicitly consider it in the overage and underage costs.
Expected overage cost of the order Q* is
P(Demand < Q*) co = F(Q*)co
Expected shortage cost is
P(Demand > Q*) cu = (1-F(Q*)) cu
For order Q* those two costs are equal: F(Q*)co= (1-F(Q*))cu
- probability of satisfying demand during
the period, also is known as critical ratio
To calculate Q* we must use cumulative probability distribution.
Mrs. Kandell estimates that if she buys more trees than she can sell, it costs about $40 for the tree and its disposal. If demand is higher than the number of trees she orders, she looses a profit of $40 per tree.
Cu = 7.50 = lost profit
Co = 14.50 - 8.50 = cost – salvage value = 6
CR = Cu/(Cu + Co) = 7.5 / (7.5 + 6) = .56
Suppose that we represent demand as
D = Ddeterministic + Drandom
If the random component is small compared to the deterministic component, the models used in chapter 4 will be accurate. If not, randomness must be explicitly accounted for in the model.
In chapter 5, assume that demand is a random variable with cumulative probability distribution F(D) and probability density function f(D).
D - continuous random variable, N(μ, σ)
The critical ration can also be derived mathematically.
At the start of each day, a newsvendor must decide on the number of papers to purchase. Daily sales cannot be predicted exactly, and are represented by the random variable D with normal distribution N(μ, σ), where μ = 11.73 and σ = 4.74
It can be shown that the optimal number of papers to purchase is given
by F(Q*) = cu / (cu + co),
where cu = 75 – 25 = 50, and c0 = 25 – 10 =15
unrealized profit per unit = (selling price – purchase price);
loss per excess = (purchase price – disposal price);
F(Q*) = cu / (cu + co) = 0.77 Pr ( D < Q* ) = 0.77
How to find Q* ?
Using table A-4 find z = 0.74, with μ = 11.73 and σ = 4.74
Newsvendor has to order 15 copies every week.
Avg. Cycle Inv=300
On-hand + On-order - Backorders - Committed
The ABC classification, devised at General Electric during the 1950s, helps a company identify a small percentage of its items that account for a large percentage of the dollar value of annual sales. These items are called Type A items.
Percentage of dollar value
Percentage of items
The ABC Inventory Classification System
Percentage of demand that is satisfied from inventory
Order Cycle Demand Stock-Outs
1 180 0
2 75 0
3 235 45
4 140 0
5 180 0
6 200 10
7 150 0
8 90 0
9 160 0
10 40 0
For a type 1 service 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, there are a total of 1,450 units demand and 55 stockouts (which means that 1,395 demand are satisfied). This translates to a 96% fill rate.
R + Q
Safety Stock (SS)
Reorder Point , R
Safety Stock (SS)
1. Compute Q = EOQ.
2. Substitute Q in to Equation (2) and compute R.
3. Use R to compute n(R) in Equation (1).
4. Solve for Q in Equation (1).
5. Go back to Step 2, continue until convergence.
With lead time equal to 2 weeks:
SS = R – lt =190-800(2/52)=159
= 2 E(demand in a single week) = 2 μ = 80
Standard deviation over 2 weeks is σ = (2*8)0.5 = 4
This is the unit normal loss expression. Table A - 4 gives values.
and substitute into the equation:
(s, S) system
(Q,,R) systemOther Continuous Review System: Order-Up-To-Level (s, S) vs (Q, R) System
R + Q
Order up to S
every T periods of time.
Order arrives. Cycle continues.
Lead Time passes…
At level R*,on average, order Q =S-R* units.
τ periods later, units arrive.
S-lt units present when
Q arrive (expected) as lt units consumed over
S - lt
lT units removed
Average Inventory Level =
holding, setup, penalty and ordering (per unit) costs
h = Ic = .30 (122.50) = 36.75 / 52 = $.7067 per week
Define two levels, s < S, and let u be the starting inventory at the beginning of a period.
Then, if u is less or equal to s, order (S – u),
if u is more than s, do not place an order.
In general, computing the optimal values of s and S is extremely difficult, so few systems operate using optimal (s,S) values.
Compute optimal values for (Q,R) model and set s = R and S = R + Q