Chapter 4 Inventory Control Subject to Known Demand McGraw-Hill/Irwin Copyright © 2005 by The McGraw-Hill Companies, Inc. All rights reserved.

Breakdown of the Total Investment in Inventories in the U. S Breakdown of the Total Investment in Inventories in the U.S. Economy (1999)

Reasons for Holding Inventories Economies of Scale Uncertainty in delivery leadtimes Speculation. Changing Costs Over Time Smoothing. Demand Uncertainty Costs of Maintaining Control System

Characteristics of Inventory Systems Demand May Be Known or Uncertain May be Changing or Unchanging in Time Lead Times - time that elapses from placement of order until it’s arrival. Can assume known or unknown. Review Time - Is system reviewed periodically or is system state known at all times? Treatment of Excess Demand. Backorder all Excess Demand Lose all excess demand Backorder some and lose some Inventory that changes over time perishability obsolescence

Relevant Costs a) Physical Cost of Space (3%) Holding Costs - Costs proportional to the quantity of inventory held. Includes: a) Physical Cost of Space (3%) b) Taxes and Insurance (2 %) c) Breakage Spoilage and Deterioration (1%) *d) Opportunity Cost of alternative investment. (18%) (Total: 24%) Note: Since inventory may be changing on a continuous basis, holding cost is proportional to the area under the inventory curve. (See examples.)

Inventory as a Function of Time

Relevant Costs (continued) Ordering Cost (or Production Cost). Includes both fixed and variable components. slope = c K C(x) = K + cx for x > 0 and =0 for x = 0.

Relevant Costs (continued) Penalty or Shortage Costs. All costs that accrue when insufficient stock is available to meet demand. These include: Loss of revenue for lost demand Costs of bookeeping for backordered demands Loss of goodwill for being unable to satisfy demands when they occur. Generally assume cost is proportional to number of units of excess demand.

Simple EOQ Model Assumptions: 1. Demand is fixed at l units per unit time. 2. Shortages are not allowed. 3. Orders are received instantaneously. (this will be relaxed later). 4. Order quantity is fixed at Q per cycle. (can be proven optimal.) 5. Cost structure: a) Fixed and marginal order costs (K + cx) b) Holding cost at h per unit held per unit time.

Inventory Levels for the EOQ Model Saw structure is typical. First order when inventory is 0. Reordering Q everytime when inventory is 0 must be optimal

The EOQ Model: Notation D = λ is the demand rate (in units per year) c = unit production cost, not counting setup or inventory costs (in dollars per unit) K = setup costs (per placed order) in dollars h = holding cost (in dollars per unit per year), if the holding cost consists entirely of interest on money tied up in inventory, h = ic, where i – is an annual interest rate Q = lot size (order size) in units T = time between orders (cycle length) G(Q) = average annual cost

Relationships C(Q) = K + cQ Holding Cost: Time (t) Inventory (I(t)) Assume Constant Demand T Q Time between orders slope = -l Instantaneous Replenishment Ordering Costs: (Order amount Q) C(Q) = K + cQ Holding Cost: h = Ic =(Interest Rate)(Cost of Inv.) Average Inventory Size? Under constant demand: Q/2 Time Between Orders: l = Q/T T = Q/l Rate of consumption l Q T

Total Costs G(Q) = average order cost + average holding cost What is the average annual cost? G(Q) = average order cost + average holding cost Average ordering cost per time T Average inventory level at any time

Total Costs What is the average annual cost?

The Average Annual Cost Function G(Q)

The Average Annual Cost Function G(Q)

Minimize Annual Costs YES! Take the derivative of G(Q) Is this a minimum? EOQ: YES!

Properties of the EOQ Solution Q is increasing with both K and  and decreasing with h Q changes as the square root of these quantities Q is independent of the proportional order cost, c. (except as it relates to the value of h = Ic)

Properties of the EOQ Solution This formula is well-known economic order quantity, is also known as economic lot size This is a tradeoff between lot size and inventory “Garbage in, garbage out” - usefulness of the EOQ formula for computational purposes depends on the realism of input data Estimating setup cost is not easily reduced to a single invariant cost K

Example Q 240 T = = = . 07 years * (250 working days/year) l 3600 = Uvic requires 3600 gallons of paint annually for scheduled maintenance of buildings. Cost of placing an order is $16 and the interest rate (annual) is 25%. Price of paint is $8 per gallon. How much paint should be ordered, and how often? Q 240 T = = = . 07 years * (250 working days/year) l 3600 = 17.5 working days = 18 days

Order Point for the EOQ Model Does it matter if τ < T or τ > T ? Keep track of time left to zero inventory or set automatic reorder at a particular inventory level, R. R = λ*τ, if τ < T R = λ*MOD(τ/T), if τ > T τ τ Assumption: Delivery is immediate There is no time lag between production and availability to satisfy demand Relax this assumption! Let the order lead time to be equal to τ

Sensitivity Analysis independent of Q Let G(Q) be the average annual holding and set-up cost function given by Holding & Setup costs and let G* be the optimal average annual holding and setup cost. Then it can be shown that: Cost penalties are quite small

Finite Replenishment Rate: Economic Production Quantity (EPQ) Assumptions for EOQ: Production is instantaneous There is no capacity constraint, and entire lot is produced simultaneously Delivery is immediate There is no time lag between production and availability to satisfy demand Example: Parts produced at the same factory – production rate is P (P > λ), arriving continuously.

Inventory Levels for Finite Production Rate Model

The EPQ Model: Notation D = λ is the demand rate (in units per year) c = unit production cost, not counting setup or inventory costs (in dollars per unit) K = setup costs (per placed order) in dollars h = holding cost (in dollars per unit per year), if the holding cost consists entirely of interest on money tied up in inventory, h=ic, where i is an annual interest rate Q = size of each production run (order) in units T = time between initiation of orders arrival (cycle length) T1 = production (replenishment) time T2 = downtime H = maximum on-hand inventory G(Q) = average annual setup & holding cost

The EPQ Model: Formula For EOQ:

Quantity Discount Models One of the most severe assumptions: the unit variable cost c did not depend on the replenishment quantity In practice: quantity discounts exist based on the purchase price or transportation costs – take advantage of these can result in substantial savings All Units Discounts: the discount is applied to ALL of the units in the order. Gives rise to an order cost function such as that pictured in Figure 4-9 in Ch. 4.7 Incremental Discounts: the discount is applied only to the number of units above the breakpoint. Gives rise to an order cost function such as that pictured in Figure 4-10

All-Units Discount Order Cost Function

All-Units Discount Average Annual Cost Function G(Q) Q 500 1,000 G0(Q) G1(Q) G2(Q) Gmin(Q)

Incremental Discount Order Cost Function

Average Annual Cost Function for Incremental Discount Schedule

Properties of the Optimal Solutions For all units discounts, the optimal will occur at the minimum point of one of the cost curves or at a discontinuity point One compares the cost at the largest realizable EOQ and all of the breakpoints succeeding it For incremental discounts, the optimal will always occur at a realizable EOQ value. Compare costs at all realizable EOQ’s.

Example Supplier of paint to the maintenance department has announced new pricing: $8 per gallon if order is < 300 gallons $6 per gallon if order is ≥ 300 gallons Data remains as before: K = 16, I = 25%, l = 3600 Is this a case of all units or incremental discount?

Solution Step 1: For Price 1: Step 2: As Q(1) < 300, EOQ is realizable. Step 3: Price 2: Step 4: As Q(2) < 300, EOQ is not realizable.

Cost Function Realizable G(Q|p1) G(Q|p2) Not Realizable C(Q) 240 277 300 Q

Cost Function Only possible solutions G(Q|p1) G(Q|p2) C(Q) 240 277 300

Solution Step 5: Compare costs of possible solutions. For $8 price, Q=240: For $6 price, Q=300: Q=300 is the optimal quantity.

Resource Constrained Multi-Product Systems Classic EOQ model is for a single item. Setup plan for n items. Option A: Treat one system with multiple items as multiple systems with one item Works if: There are no interactions among items, such as sharing common resources – budget, storage capacity, or both Option B: Modify classic EOQ to insure no violation of the resource constraints Works if: Have not made any mistakes and know how to use Lagrange multipliers 

Resource Constrained Multi-Product Systems Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following budget constraint on the system , where Qi is the order size for product i , where wi is the volume occupied by product i Minimize s.t. and For EOQ:

Resource Constrained Multi-Product Systems Minimize s.t. Budget constraint Space constraint Lagrange multipliers method: relax one or more constraints Minimize by solving necessary conditions:

Resource Constrained Multi-Product Systems: Steps to Find Optimal Solution Single constraint: Solve the unconstrained problem. If constraint is satisfied, this solution is the optimal one. If the constraint is violated, rewrite objective function using Lagrange multipliers Obtain optimal Qi* by solving (n+1) equations

Resource Constrained Multi-Product Systems: Steps to Find Optimal Solution Double constraints: Solve the unconstrained problem. If both constraints are satisfied, this solution is the optimal one. Otherwise rewrite objective function using Lagrange multipliers by including one of the constraints, say budget, and solve one-constraint problem to find optimal solution. If the space constraint is satisfied, this solution is the optimal one. Otherwise repeat the process for the only space constraint. If both single-constraint solutions do not yield the optimal solution, then both constraints are active, and the Lagrange equation with both constraints must be solved. Obtain optimal Qi* by solving (n+2) equations

EOQ Models for Production Planning Problem: determine optimal procedure for producing n products on a single machine Consider n items with known demand rates , production rates , holding costs , and set-up costs . The objective is to minimize the cost of holding and setups, and to have no stock-outs. For the problem to be feasible we must have that Assumption: rotation cycle policy – exactly one setup for each product in each cycle; production sequence stays the same in each next cycle

The method of solution is to express the average annual cost function in terms of the cycle time, T to assure no stock-outs. The optimal cycle time has the following mathematical form, where sj is a setup time And the optimal production quantities are given by: where T = max {T*, Tmin}, see pp.216-217

Homework: Read Ch. 4 Problems 4.5, 4.12, 4.15, 4.16 4.17, 4.18, 4.22, 4.24, 4.25 4.26, 4.27, 4.28, 4.30 Work on appendix 4-A,

References Presentations by McGraw-Hill/Irwin and by Wilson,G.R. “Production & Operations Analysis” by S.Nahmias “Factory Physics” by W.J.Hopp, M.L.Spearman “Inventory Management and Production Planning and Scheduling” by E.A. Silver, D.F. Pyke, R. Peterson “Production Planning, Control, and Integration” by D. Sipper and R.L. Bulfin Jr.

Reorder Point Calculation for Example 4.1

Reorder Point Calculation for Lead Times Exceeding One Cycle

Sensitivity Analysis Let G(Q) be the average annual holding and set-up cost function given by and let G* be the optimal average annual cost. Then it can be shown that:

EOQ With Finite Production Rate Suppose that items are produced internally at a rate P > λ. Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely: Except that h’ is defined as h’= h(1- λ/P)

Inventory Levels for Finite Production Rate Model

Quantity Discount Models All Units Discounts: the discount is applied to ALL of the units in the order. Gives rise to an order cost function such as that pictured in Figure 4-9 Incremental Discounts: the discount is applied only to the number of units above the breakpoint. Gives rise to an order cost function such as that pictured in Figure 4-10.

All-Units Discount Order Cost Function

Incremental Discount Order Cost Function

Properties of the Optimal Solutions For all units discounts, the optimal will occur at the bottom of one of the cost curves or at a breakpoint. (It is generally at a breakpoint.). One compares the cost at the largest realizable EOQ and all of the breakpoints succeeding it. (See Figure 4-11). For incremental discounts, the optimal will always occur at a realizable EOQ value. Compare costs at all realizable EOQ’s. (See Figure 4-12).

All-Units Discount Average Annual Cost Function

Average Annual Cost Function for Incremental Discount Schedule

Resource Constrained Multi-Product Systems Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following constraint on the system: When the condition that is met, the solution procedure is straightforward. If the condition is not met, one must use an iterative procedure involving Lagrange Multipliers.

EOQ Models for Production Planning Consider n items with known demand rates, production rates, holding costs, and set-up costs. The objective is to produce each item once in a production cycle. For the problem to be feasible we must have that

The method of solution is to express the average annual cost function in terms of the cycle time, T. The optimal cycle time has the following mathematical form. And the optimal production quantities are given by: