Chapter 4. Models for Known Demand

Chapter 4 Models for Known Demand Introduction EOQ analysis is based on a number of assumptions. In the next two chapters we describe some models where these assumptions are removed. This chapter keeps the condition that all variables take known values (giving deterministic models) while the next chapter allows some uncertainty (giving probabilistic models). We concentrate on the most widely used models, and use these to illustrate some general principles. 1

Variable Costs Unit costs vary with the quantity ordered Reordering cost might also vary with order size, particularly if it includes quality control inspections, or some other processing of deliveries. The transport cost also varies with the amount delivered, and can have quite complex patterns. The holding cost can vary with order size. It is common for all of the costs to vary, often in quite complicated ways. We will, therefore, illustrate the general approach with problems where the unit cost decreases in discrete steps with increasing order quantity. Price Discounts from Suppliers UC 1 UC UC 3 Unit Cost Unit Cost Order Quantity Lower Limit Order Quantity Upper Limit UC 1 0 Q a UC Q a Q b UC 3 Q b Q c UC 4 Q c Q d UC 4 UC 5 Q d Q e UC 5 Q a Q b Q c Q d Order Quantity

Total Cost Curve using UC 1 Total Cost Order Quantity Price Discounts from Suppliers: Valid Range of the Total Cost Curve using UC 1 Total Cost Valid Range of Curve Invalid Range of Curve Q a Order Quantity 3

Price Discounts from Suppliers Total Cost Q a Q b Q c Q d Order Quantity Price Discounts from Suppliers: Finding the Lowest Valid Cost EOQ: Q o = RC D HC If the holding cost can be expressed as a proportion, I, of the unit cost, and for each unit cost UC i, the minimum point of the cost curve comes with Qo i Q oi = RC D I UC i For each curve with unit cost UC i this minimum is either valid or invalid. 4

Price Discounts from Suppliers Total Cost Valid Minimum Invalid Minimum Q a Q b Q c Q d Order Quantity Price Discounts from Suppliers: Finding the Lowest Valid Cost Every set of cost curves will have at least one valid minimum, and a variable number of invalid minima. The valid total cost curve always rises to the left of a valid minimum. This means that when we search for an overall minimum cost it is either at the valid minimum or somewhere to the right of it. There are only two possible positions for the overall minimum cost: it is either at a valid minimum, or else at a cost break point. 5

Price Discounts from Suppliers: Finding the Lowest Valid Cost Worked Example 1: Price Discounts from Suppliers The annual demand for an item is,000 units, each order costs 10 and the annual holding cost is 40% of unit cost. The unit cost depends on the quantity ordered as follows: 1 for order quantities less than 500 0.80 for quantities between 500 and 999 0.60 for quantities of 1,000 or more. What is the optimal order size? 6

Worked Example 1: Price Discounts from Suppliers RC D TC = UC D + Q D = 000 units/yr RC = 10 per order HC= (40%)(UC) UC= 1.00 (Q<500); 0.80 (500 Q < 999) 0.60 (Q 1000) + HC Q Worked Example 1: Price Discounts from Suppliers D= 000 unit/yr RC= 10.00 per order I= 40% of UC HC= I UC UC= 1.00 Q<500 0.80 500<=Q<999 0.60 1000<=Q Q>=1000 500<=Q<999 Q<500 UC 0.60 0.80 1.00 Qo 408.5 353.55 316.3 Invalid Invalid Valid Order size 1000 500 316.3 TC 1,340.00 1,70.00,16.49 7

Worked Example : Price Discounts from Suppliers During a 50-week year, IH James and Partners notice that demand for one of its products is more or less constant at 10 units a week. The cost of placing an order, including delivery, is around 150. The company aims for 0% annual return on assets. The supplier of the item quotes a basic price of 50 a unit, with discounts of 10% on orders of 50 units or more, 15% on orders of 150 units or more and 0% on orders of 500 units or more. What is the optimal order size for the item? Worked Example : Price Discounts from Suppliers D= 500 unit/yr RC= 150.00 per order I= 0% of UC HC= I UC UC= 50.00 Q<50 5.00 50<=Q<150 1.50 150<=Q<500 00.00 500<=Q 500<=Q 150<=Q<500 50<=Q<150 Q<50 UC 00.00 1.50 5.00 50.00 Qo 61.4 59.41 57.74 54.77 Invalid Invalid Valid Invalid Order size 500 150 57.74 50 TC 110,150.00 109,937.50 115,098.08 17,750.00 8

Rising Delivery Cost: Finding the Lowest Valid Cost This method for considering unit costs that fall in discrete steps can be extended to any problem where there is a discrete change in cost. For example, the cost of delivery and therefore the reorder cost rises in steps as more vehicles are needed. Worked Example 3: Rising Delivery Cost Kwok Cheng Ho makes a range of high quality garden ornaments. On an average day he uses 4 tons of fine grain sand. This sand costs 0 a ton to buy, and 1.90 a ton to store for a day. Deliveries are made by modified lorries that carry up to 15 tons, and each delivery of a load or part load costs 00. Find the cheapest way to ensure continuous supplies of sand. 9

Worked Example 3: Rising Delivery Cost D= 4 unit/day RC= 00.00 per lorries HC= 1.90 per day UC= 0.00 Lorries 1 3 4 Q<=15 15<Q<=30 30<Q<=45 45<Q<=60 Order Cost 00.00 400.00 600.00 800.00 Qo 9.0 41.04 50.6 58.04 Invalid Invalid Invalid Valid Order size 15 30 45 58.04 TC 147.58 161.83 176.08 190.7 Finite Replenishment Rate Stock from Production If the rate of production is less than demand, each unit arriving is immediately transferred to a customer and there are no stocks. Goods only accumulate when the production rate is greater than demand. The stock level rises at a rate that is the difference between production and demand. If we call the rate of production P, stocks will build up at a rate P D. This increase will continue as long as production continues. This means that we have to make a decision at some point to stop production of this item. The purpose of this analysis is to find the best time to stop production, which is equivalent to finding an optimal batch size. 10

Finite Replenishment Rate Produce at rate P per unit time Transfer to stock Stock increases at (P D) per unit time Meet demand Meet demand of D per unit time Finite Replenishment Rate We still consider a single item, with demand that is known, constant and continuous, with costs that are known and constant, and no shortages allowed. We have replenishment at a rate P and demand at a rate D, with stock growing at a rate P D. After some time, PT, we decide to stop production. Then, stock is used to meet demand and declines at a rate D. After some further time, DT, all stock has been used and we must start production again. We assume there is an optimal value for PT (corresponding to an optimal batch size) that we always use. 11

Finite Replenishment Rate A Stock level D (P-D) PT DT Time Finite Replenishment Rate: Optimal Batch Size The overall approach of this analysis is the same as the EOQ: find the total cost for one stock cycle, divide this total cost by the cycle length to get a cost per unit time, minimize the cost per unit time. Consider one cycle of the modified saw-tooth pattern. If we make a batch of size Q, the maximum stock level with instantaneous replenishment would also be Q. With a finite replenishment rate, though, stock never reaches this level, as units are continuously being removed to meet demand. The maximum stock level is lower than Q and occurs at the point where production stops. 1

Finite Replenishment Rate Q A Stock level (P-D) D PT T DT Time Finite Replenishment Rate: Optimal Batch Size Looking at the productive part of the cycle we have: A = (P D) PT We also know that total production during the period is: Q = P PT or PT = Q P Substituting for PT into the equation for A gives: A = Q (P D) P 13

Finite Replenishment Rate: Optimal Batch Size 1. Find the total cost of one stock cycle. Total cost per cycle = unit cost component + reorder cost component + holding cost component HC A T = UC Q + RC + HC Q T = UC Q + RC + (P D) P Finite Replenishment Rate: Optimal Batch Size. Divide this total cost by the cycle length to get a cost per unit time. Total cost per unit time Since Q = D T = UC Q T + RC T + HC Q (P D) P TC = UC D + RC D Q + HC Q (P D) P 14

Finite Replenishment Rate: Optimal Batch Size 3. Minimize this cost per unit time. d(tc) d(q) = RC D Q + HC (P D) P = 0 Solve for Q o (Optimal Batch Size), Q o = RC D HC P (P D) Finite Replenishment Rate: Optimal Batch Size Order quantity Q o = Finite Replenishment Rate RC D HC P (P D) Q o = EOQ RC D HC Cycle length T o = RC HC D P (P D) T o = RC HC D Variable cost VC o = RC HC D (P D)/P = HC Q o (P D) P VC o = RC HC D = HC Q o Total cost TC o = UC D + VC o TC o = UC D + VC o Production time PT o = Q o P 15

Worked Example 4: Finite Replenishment Rate Demand for an item is constant at 1,800 units a year. The item can be made at a constant rate of 3,500 units a year. Unit cost is 50, batch set-up cost is 650, and holding cost is 30% of value a year. What is the optimal batch size for the item? If production set-up time is weeks, when should this be started? Worked Example 5: Finite Replenishment Rate Forecasters at Saloman Curtis have estimated the demand for an item that they import from Taiwan to average 0 units a month. They pay 1,000 for each unit. Last year the Purchasing and Inward Transport Department arranged the delivery of,000 orders and had running costs of 5,000,000. The Accounts Section quotes the annual holding costs as 0% of unit cost for capital and opportunity loss, 5% for storage space, 3% for deterioration and obsolescence and % for insurance. All other costs associated with stocking the item are combined into a fixed annual cost of 4,000. Calculate the economic order quantity for the item, the time between orders and the corresponding total cost. By making the item at a rate of 40 units a month Saloman Curtis could avoid the fixed cost of 4,000 a year, reduce the unit cost to 900 and have a batch set-up cost of 1,000. Would it be better for the company to make the item itself rather than buy it in? 16

Worked Example 5: Finite Replenishment Rate Purchase Production Qo 63.5 59.63 PTo 0.145999 To 0.6353138 0.48451997 VCo 18,973.67 8,049.84 TCo 58,973.67 4,049.84 Fixed 4,000.00 -- Total 8,973.67 4,049.84 Alternatives for Customers when their Demand cannot be met Customer demands an item The item is out of stock Customer waits for the next replenishmentgiving back-orders Customer moves to another suppliergiving lost sales Customer keeps all future business with the supplier Customer transfers some future business to another supplier Customer transfers some future business to another supplier Customer transfers all future business to another supplier 17

Back-orders A back-order occurs when a customer demands an item that is out of stock, and then waits to receive the item from the next delivery from suppliers. Back-ordering is more likely when the unit cost is high, there is a wide range of items, it would be too expensive to hold stocks of the complete range, lead times from suppliers are reasonably short, there is limited competition, and customers are prepared to wait. In extreme cases an organization keeps no stock at all and meets all demand by back orders. This gives make to order rather than make for stock operations. Back-orders We will look at the more common case where some stock is kept, but not enough to cover all demand. The key question is, how much of the demand should be met from stock and how much from back-orders? There is always some cost associated with back-orders, including administration, loss of goodwill, some loss of future orders, emergency orders, expediting, and so on. This cost is likely to rise with increasing delay. We can, then, define a shortage cost, SC, which is time-dependent and is a cost per unit per unit time delayed. 18

Stock Cycle with Back-orders Stock level Q Q-S T T 1 S Time T Back-orders 1. Find the total cost of one stock cycle. Total cost per cycle = unit cost component + reorder cost component + holding cost component + shortge cost component = UC Q + RC + HC (Q S) T 1 Where T 1 = (Q S) D and T = S D; then + SC S T Total cost / cycle = UC Q + RC + HC (Q S) D + SC S D 19

Back-orders. Divide this total cost by the cycle length to get a cost per unit time. Total cost / cycle Since Q = D T = UC Q + RC + HC (Q S) D + SC S D TC = UC D + RC D Q + HC (Q S) Q + SC S Q Back-orders 3. Minimize this cost per unit time. d(tc) d(q) = RC D Q + HC HC S SC S Q Q = 0 and d(tc) HC S = HC + SC S d(s) Q Q Solving these simultaneous equations, = 0 0

Back-orders Q o = RC D (HC + SC) HC SC In addition, S o = T 1 = Q o S o D RC HC D SC (HC + SC) ; T = S o D ; T = T 1 + T Worked Example 6: Back-orders Demand for an item is constant at 100 units a month. Unit cost is 50, reorder cost is 50, holding cost is 5% of value a year, shortage cost for back-orders is 40% of value a year. Find an optimal inventory policy for the item. 1

Lost Sales An initial stock of Q runs out after a time Q/D, and all subsequent demand is lost until the next replenishment arrives. There is an unsatisfied demand of D T Q. When shortages are allowed, the aim of minimizing cost is no longer the same as maximizing revenue. If shortages are allowed we might, in some circumstances, minimize costs by holding no stock at all but this would certainly not maximize revenue. In this analysis we will maximize the net revenue, which is defined as the gross revenue minus costs. For this we have to define SP as the selling price per unit. Lost Sales We also need to look at the cost of lost sales, which has two parts. First, there is a loss of profit, which is a notional cost that we can define as SP UC per unit of sales lost. Second, there is a direct cost, which includes loss of goodwill, remedial action, emergency procedures, and so on. We will define this as DC per unit of sales lost.

Stock Level with Lost Sale Q Stock level D Q/D Time T Lost Sales 1. Find the total cost of one stock cycle. Total cost per cycle = unit cost component + reorder cost component + holding cost component + lost sales cost component = UC Q + RC + HC Q Q + DC (D T Q) D then Net Revenue = SP Q UC Q RC + HC Q D + DC (D T Q) 3

Lost Sales. Divide the net revenue by the cycle length to get the net revenue per unit time. R = 1 [Q DC + SP UC T HC Q RC D DC D T] Let LC = cost of each unit of lost sales including loss of profits = DC + SP UC; and Z = proportion of demand satisfied = Q/(D T) R = Z [D LC RC D Q HC Q DC D] Lost Sales 3. Minimize this cost per unit time. d(r) Z RC D = d(q) Q Z HC Solving this equation, = 0 Q o = RC D HC R o = Z D LC Q o HC = Z D LC RC HC D 4

Lost Sales If D LC > If D LC < If D LC < RC HC D; set Z = 1 no shortage RC HC D; set Z = 0 Don t stock RC HC D; Z = any value Worked Example 7: Lost Sales Find the best ordering policy for three items with the following costs. Item D RC HC DC SP UC 1 50 150 80 0 110 90 100 400 00 10 00 170 3 50 500 400 30 350 30 5

Worked Example 7: Lost Sales Item1 Item Item3 D= 50 100 50 RC= 150.00 400.00 500.00 HC= 80.00 00.00 400.00 DC= 0.00 10.00 30.00 SP= 110.00 00.00 350.00 UC= 90.00 170.00 30.00 LC= 40.00 40.00 60.00 D LC=,000.00 4,000.00 3,000.00 SQRT( RC HC D) 1,095.45 4,000.00 4,47.14 Z= 1 0.5 0 Ro= 904.55 - - Qo= 13.69306394 0 11.18033989 Do Not Stock Constraints on Stock If each inventory item is completely independent, then we can calculate an optimal order policy for each item in isolation and there is no need to consider interactions with other items. There are situations where, although demand for each item is independent, we may have to look at these interactions. When several items are ordered from the same supplier; and we can reduce delivery costs by combining orders for different items in a single delivery. When there are constraints on some operations, such as limited warehouse space or a maximum acceptable investment in stock. 6

Constraints on Stock Problems of this type can become rather complicated, so we will illustrate some broader principles by looking at problems with constraints on stock levels. These constraints could include limited storage space, a maximum acceptable investment in stock, a maximum number of deliveries that can be accepted, maximum size of delivery that can be handled, a maximum number of orders that can be placed, etc. Most problems with constraints can be tackled using the same methods. The first shows an intuitive approach to problems where there is a limited amount of storage space; the second is a more formal analysis for constrained investment. Constraints on Storage Space If an organization uses the economic order quantity for all items in an inventory, the resulting total stock might exceed the available capacity. Then we need some way of reducing the stock until it is within acceptable limits. One approach puts an additional cost on space used. Then the holding cost is in two parts: the original holding cost, HC, and an additional cost, AC, related to the storage area (or volume) used by each unit of the item. Then the total holding cost per unit per unit time becomes HC + AC Si, where Si is the amount of space occupied by one unit of item i. 7

Constraints on Storage Space The economic order quantity becomes: Q i = RC i D i HC i + AC S i If space is constrained, we can give AC a positive value and order quantities of all items are reduced from Q oi to Q i. As the average stock level is Q i /, this automatically reduces the amount of stock. The reduction necessary and the consequent value of AC depend on the severity of space constraints. A reasonable way of finding solutions is iteratively to adjust AC until the space required exactly matches available capacity. Worked Example 8: Constraints on Storage Space The Brendell Work Centre is concerned with the storage space occupied by three items that have the following characteristics. Item UC($) D Space (m 3 /unit) 1 100 500 1 00 400 3 300 00 3 The reorder cost for the items is constant at $1000, and holding cost is 0% cent of value a year. If the company wants to allocate an average space of 300 m 3 to the items, what would be the best ordering policy? How much does the constraint on space increase variable inventory cost? 8

Worked Example 8: Constraints on Storage Space Item UC D Space RC HC(1) A HC Qo Space Req. VCo 1 $ 100.00 500 1 $ 1,000.00 $ 0.00 0 $ 0.00 3.61 111.80 $ 4,47.14 $ 00.00 400 $ 1,000.00 $ 40.00 0 $ 40.00 141.4 141.4 $ 5,656.85 3 $ 300.00 00 3 $ 1,000.00 $ 60.00 0 $ 60.00 81.65 1.47 $ 4,898.98 375.70 $ 15,07.97 Goal Seek: Item UC D Space RC HC(1) A HC Q Space Req. VC 1 $ 100.00 500 1 $ 1,000.00 $ 0.00 11.3666 $ 31.37 178.55 89.8 $ 4,585.8 $ 00.00 400 $ 1,000.00 $ 40.00 11.3666 $ 6.73 11.93 11.93 $ 5,800.66 3 $ 300.00 00 3 $ 1,000.00 $ 60.00 11.3666 $ 94.10 65.0 97.80 $ 5,03.5 300.00 $ 15,409.99 Solver: Item UC D Space RC HC(1) A HC Q Space Req. VC 1 $ 100.00 500 1 $ 1,000.00 $ 0.00 11.3667 $ 31.37 178.55 89.8 $ 4,585.8 $ 00.00 400 $ 1,000.00 $ 40.00 11.3667 $ 6.73 11.93 11.93 $ 5,800.66 3 $ 300.00 00 3 $ 1,000.00 $ 60.00 11.3667 $ 94.10 65.0 97.80 $ 5,03.5 300.00 $ 15,410.00 Constraints on Average Investment in Stock Suppose an organization stocks N items and has an upper limit, UL, on the total average investment. The calculated EOQ for each item i is Q oi, but again we need some means of calculating the best, lower amount Q i which allows for the constraint. The average stock of item i is Q i / and the average investment in this item is UC Q i /. The problem then becomes one of constrained optimization minimize the total variable cost subject to the constraint of an upper limit on average investment. 9

Constraints on Average Investment in Stock Minimize VC = Subject to N i=1 N i=1 RC i D i Q i UC i Q i + HC i Q i UL Constraints on Average Investment in Stock The usual way of solving problems of this type is to eliminate the constraint by introducing a Lagrange multiplier and incorporating this into a revised objective function. In particular, we can adjust the constraint to make it less than or equal to zero, and the result is multiplied by the Lagrange multiplier, LM. This is incorporated in the revised objective function: Minimize VC = N i=1 RC i D i Q i + HC i Q i + LM N i=1 UC i Q i UL 30

Constraints on Average Investment in Stock To solve this, we differentiate with respect to both Q i and LM, and set both of these derivatives to zero to give two simultaneous equations. We can then eliminate the Lagrange multiplier to give the overall result: Q i = Q oi UL HC i N UC i i=1 VC oi When the holding cost is a fixed proportion of the unit cost, the best order quantity for each item is a fixed proportion of the EOQ. Worked Example 9: Constraints on Average Investment in Stock A company is concerned with the stock of three items with the following demand and unit cost. Item UC($) D 1 100 100 60 300 3 40 00 The items have a reorder cost of 100 and a holding cost of 0% of value a year. What is the best ordering policy if the company wants to limit average investment in these item to 4,000? 31

Worked Example 9: Constraints on Average Investment in Stock Item UC D RC HC Qo VCo Avg. Inv. 1 100.00 100 100.00 0.00 31.6 63.46 1,581.14 60.00 300 100.00 1.00 70.71 848.53,11.3 3 40.00 00 100.00 8.00 70.71 565.69 1,414.1,046.67 5,116.67 Item Qo VCo Ratio Q Avg. Inv' 1 31.6 63.46 0.781758 4.7136 1,36.07 70.71 848.53 0.781758 55.7864 1,658.36 3 70.71 565.69 0.781758 55.7864 1,105.57,046.67 4,000.00 Discrete, Variable Demand Often, an organization has to meet demand for an integer number of units every period, e.g., a car showroom, electrical retailer, furniture warehouse or computer store. In this section, we will look at a model for discrete, variable demand. If we know exactly what these demands are going to be, we can build a deterministic model and find an optimal ordering policy. The way to approach this is to assume there is some optimal number of period s demand that we should combine into a single order. 3

Discrete, Variable Demand Discrete, Variable Demand Consider one order for an item, where we buy enough to satisfy demand for the next N periods. The demand is discrete and varies, so we can define the demand in period i of a cycle as D i, and an order to cover all demand in the next N periods will be: A = N i=1 D i 33

Discrete, Variable Demand Discrete, Variable Demand The average variable cost of stocking the item per period: VC N = RC N + HC N i=1 D i Starting with N = 1, increasing N will follow the graph downward until costs reach a minimum and then start to rise. If it is cheaper to order for N + 1 periods than for N periods, we are on the left-hand side of the graph. If we keep increasing the value of N, there comes a point where it is more expensive to order for N + 1 periods than N. At this point, we have found the optimal value for N and if we increase it any further, costs will continue to rise. 34

Discrete, Variable Demand The average variable cost of stocking the item over N periods: VC N = RC N + HC N i=1 D i The average variable cost of stocking the item over N + 1 periods: VC N+1 = RC N + 1 + HC N+1 i=1 D i If VC N+1 > VC N, RC N + 1 + HC N+1 i=1 D i > RC N + HC N i=1 N N + 1 D N+1 > RC HC D i ; or Discrete, Variable Demand 35

Discrete, Variable Demand This method usually gives good results, but it does not guarantee optimal ones. It is sometimes possible for the variable cost to be higher for orders covering N + 1 periods rather than N, but then fall again for N + periods. If VC N+ > VC N, RC N + + HC N+ i=1 D i > RC N + HC N i=1 N N + [D N+1 +D N+ ] > 4 RC HC D i ; or Discrete, Variable Demand When we find a turning point and we know it is cheaper to order for N periods than for N + 1, we simply see if it is also cheaper to order for N + periods. If the inequality above is valid, we stop the process and accept the solution. If the inequality is invalid we continue the process until we find another turning point. 36

Worked Example 10: Discrete, Variable Demand K C Low calculated her total cost of placing an order for an item and having it delivered at $1,00, with a holding cost of $50 a unit a week. She had a fixed production schedule and needed the following units over the next 11 weeks. Week 1 3 4 5 6 7 8 9 10 11 Demand 4 6 9 9 6 3 4 4 6 9 Find an ordering pattern that gives a reasonable cost for the item. Worked Example 10: Discrete, Variable Demand Week 1 3 4 5 6 7 8 9 10 11 Demand 4 6 9 9 6 3 4 4 6 9 N= 1 3 1 3 4 1 3 N(N+1)D 8 36 108 18 36 36 80 8 36 108 less less greater less less less greater less less greater N(N+)(D) 5 168 5 valid Valid Valid VC 150 750 700 145 1050 1000 975 1300 800 750 37