Inventory Control Subject to Deterministic Demand Operations Analysis and Improvement 2017 Spring Dr. Tai-Yue Wang Industrial and Information Management Department National Cheng Kung University

Contents Introduction Types of Inventories Why Inventory? Characteristics of Inventory Systems Relevant Costs The EOQ Model Extension to a Finite Production Rate

Contents Quantity Discount Models Resource-constrained Multiple Product Systems EOQ Models for Production Planning Power of two Policy

Overview of Operations Planning Activities

Introduction -- 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?

Breakdown of the total investment in inventories

Introduction -- Characteristics of Inventory Systems Treatment of Excess Demand. Backorder all Excess Demand Lose all excess demand Backorder some and lose some Inventory that changes over time Perishability – 農產品 Obsolescence – 過期之設備備品

Introduction -- Purposes Demand is known Methods to control individual item inventory

Types of Inventories Raw material Components Work-in-Process (WIP) Finished goods

Reasons for Holding Inventories Economies of Scale Uncertainty in delivery leadtimes, supply Speculation-- Changing Costs Over Time Transportation Smoothing Demand Uncertainty Logistics Costs of Maintaining Control System

Relevant Costs 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%) Note: Since inventory may be changing on a continuous basis, holding cost is proportional to the area under the inventory curve.

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 bookkeeping 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.

The EOQ Model —The Basic 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).

The EOQ Model —The Basic Model Assumptions: 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.

The EOQ Model —The Basic Model

The EOQ Model —The Basic Model Q is the size of the order At t=0, Q is increased instantaneously from 0 to Q The objective is to choose Q to minimize the average cost per unit time. In each cycle, the total fixed plus proportional order cost is C(Q)=K+cQ Since the inventory is consumed by the rate of λ, the cycle length T is computed by Q/ λ

The EOQ Model —The Basic Model In addition, the average inventory level during one order cycle is Q/2. Thus, the annual cost, G(Q)

The EOQ Model —The Basic Model Thus Since G”(q) > 0, G(Q) is a convex function of Q and G’(0)=- ∞ and G’(∞)=h/2, the curve of G(Q) is in next slide.

The EOQ Model —The Basic Model

The EOQ Model —The Basic Model --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)

The EOQ Model —The Basic Model --Properties of the EOQ Solution The optimal value of Q occurs where G’(Q)=0 Q* is known as the economic order quantity(EOQ).

The EOQ Model —The Basic Model --Example Number 2B pencils at campus bookstore are sold at a rate of 60 per week. The pencil cost is two cents each and sell for 15 cents each. It cost the bookstore $12 to initiate an order and the holding cost are based on annual interest rate of 25 percent. Please determine the optimal number of pencils for the bookstore to purchase and the time between placement of orders.

The EOQ Model —The Basic Model --Example The annual demand rate λ=(60)(52)=3,120 The holding cost h=(0.25)(0.02)=0.005 The cycle time is T=Q/λ = 3,870/3,120 =1.24 years

The EOQ Model — Order Lead time In previous example, if the pencils must be ordered four months in advance, we would try to find out when to place order depends on how much inventory on hand. So we want to reorder at inventory on hand, R, the reorder point. where  is the lead time

The EOQ Model — Order Lead time

The EOQ Model — Order Lead time If the lead time exceeds one cycle, it is more difficult to determine the reorder point Let EOQ=25, demand rate = 500/year, lead time = 6 weeks, Cycle time T = 25/500=0.05 year = 2.6 weeks or Lead time = /T = 2.31 cycles  two cycles + 0.31 cycle  0.0155 year R=0.0155*500=7.75  8

The EOQ Model — Order Lead time

The EOQ Model — Order Lead time Procedure: Form the ratio of /T Get the fractional remainder of the ratio Multiply this fractional remainder by cycle length to convert to year Multiply the result of previous step by the demand rate

The EOQ Model — 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:

The EOQ Model — Sensitivity Analysis In general, G(Q) is relatively insensitive to errors in Q If would results lower average annual cost than a value of

The EOQ Model — Example A company produces desks at a rate of 200 per month. Each desk requires 40 screws purchased from a supplier. The screw costs 3 cents each. Fixed delivery charges and cost of receiving and storing equipment of screws amount to $100 per shipment, independently of the size of the shipment. The firm uses 25 percent interest rate to determine the holding cost What standing order size should they use?

The EOQ Model — Example Solution Annual demand=(200)(12)(40)=96,000 Annual holding cost per screw = 0.25*0.03=0.0075 EOQ

EOQ With Finite Production Rate Suppose that items are produced internally at a rate P > λ. The total cost is Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely:

EOQ With Finite Production Rate --Inventory Levels for Finite Production Rate Model

EOQ With Finite Production Rate — Example A company produces EPROM for its customers. The demand rate is 2,500 units per year. The EPROM is manufactured internally at rate of 10,000 units per year. The cost for initiating the production is $50 and each unit costs the company $2 to manufacture. The cost of holding is based on a 30 percent annual interest rate. Please determine the optimal size of a production run, the length of each production run, and the average annual cost of holding and setup. What is the maximum level of the on-hand inventory of the EPROM?

EOQ With Finite Production Rate — Example Solution: h=0.3*20.6 per unit per year The modified holding cost h’=0.6*(1-2,500/10,000)=0.45 the length of each production run T=Q/=745/2500=0.298 year the average annual cost of holding and setup

EOQ With Finite Production Rate — Example Solution: What is the maximum level of the on-hand inventory of the EPROM?

Quantity Discount Models Two kinds of quantity discount: 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

Quantity Discount Models –all units discount Trash bag company’s price schedule:

Quantity Discount Models –all units discount Procedure: Starting from the lowest price interval and determine the largest realizable EOQ value. Compare the value of average annual cost at the largest realizable EOQ and at all the price breakpoints that are greater than the largest realizable EOQ. The optimal one is the one with lowest average annual cost.

Quantity Discount Models –all units discount --example Trash bag company’s price schedule: For c=0.28, Q*=414 X c=0.29, Q*=406 X c=0.30, Q*=400 OK

Quantity Discount Models –all units discount --example G(400)=204 G(500)=198.1 G(1,000)=200.8 Q=500 with lowest average annual cost

Quantity Discount Models –Incremental discount Trash bag company’s price schedule: And G(Q) becomes

Quantity Discount Models –Incremental discount Procedure: Find C(Q) equation for all price intervals Substitute C(Q) into G(Q), compute the minimum values of Q for each price intervals Determine which minima computed from previous step are realizable, compute the average annual costs at the realizable EOQ values and pick the lowest one.

Quantity Discount Models –Incremental discount --example Trash bag company’s price schedule: 1. 2.

Quantity Discount Models –Incremental discount --example 2.

Quantity Discount Models –Incremental discount --example 3. Compare G0 and G1

Quantity Discount Models --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: EOQ:

Resource Constrained Multi-Product Systems When the condition that is met, the solution procedure is straightforward. EOQ

Resource Constrained Multi-Product Systems 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. j = demand rate for product j Pj= production rate for product j hj = holding cost per unit per unit time for product j Kj= cost of setup the production facility for product j

EOQ Models for Production Planning The goal is to determine the optimal procedure for producing n products on the machine to minimize the cost of holding and setups, and to guarantee that no stock-outs occur during the production cycle. For the problem to be feasible we must have that

EOQ Models for Production Planning We also assume that rotation cycle policy is used. That is, in each cycle, there is exactly one setup for each product, and the products are produced in the same sequence in each production cycle. Let T be the cycle time, and during time T, exactly one cycle of each product are produced.

EOQ Models for Production Planning So the lot size for product j during time T is And the average annual cost for product j is For all products

EOQ Models for Production Planning Since  So The goal is to find optimal cycle time to minimize G(T)

EOQ Models for Production Planning So However, if setup time is a factor, one needs to check if having enough time for setup and production

EOQ Models for Production Planning Let sj be the setup time for product j So And So we choose the cycle time T =max(T*, Tmin)

EOQ Models for Production Planning -- Example A machine serves as a cutting machine for different products. The rotation policy is used and setup cost is proportion to the setup time. Data are followed. Products Annual Demand (units/year) Production Rate Setup time (hours) Variable costs ($/unit) A 4,520 35,800 3.2 40 B 6,600 62,600 2.5 26 C 2,340 41,000 4.4 52 D 2,600 71,000 1.8 18 E 8,800 46,800 5.1 38 F 6,200 71,200 3.1 28 G 5,200 56,000 31

EOQ Models for Production Planning -- Example The firm estimates that the setup costs amount to an average of $110 per hour, based on the cost of worker time and the cost of forced machine idle time during setups. Holding costs are based on a 22 percent annual interest rate charge. Please find the optimal cycle time for those products.

EOQ Models for Production Planning -- Example Solution: Verify if is valid. Compute the setup costs and modified holding costs Setup cost K1=$110*3.2=$352, … etc. Modified holding cost

EOQ Models for Production Planning -- Example Setup costs(Kj) Modified Holding Costs 352 7.69 275 5.12 484 10.79 198 3.81 561 6.79 341 5.62 6.19 Total=2,695

EOQ Models for Production Planning -- Example Solution: The sum of setup cost is 2,695 The sum of the products of the modified holding costs and the annual demands is 230,458.4 So

The Power of the two policy The solution could be hard to find in complex cases even the demand is deterministic. Approximation is required in these cases The i8dea is to choose the best replenishment interval from a set of possible interval proportional to the powers of two. Basic EOQ

The Power of the two policy The optimal time between placement of orders, T* is: It is possible that thye optimal order intervals are unrealistic, for example, 3.3999weeks.

The Power of the two policy Assuming that the order interval is the multiple of base time, TL To find the optimal solution under the constraint, one might simply compare the costs between the two closest multiples of TL to T* and pick the one with the least cost. That is, find k for which

The Power of the two policy Now, a further restriction is added, the order intervals must be of the form of 2kTL k0. What is the worst cost error we will incur relative to that of the optimal reorder interval T* ? As k increases, the error is getting larger. How about cost error? No, it would not! Hint: cost function is our concern.

The Power of the two policy Cost function is relatively insensitive respective to T Average annual cost Q=T

The Power of the two policy Case of powers of the two So That is,

The Power of the two policy Rearranging terms So, we can prove that