Dynamic Lot Size Models A Production Engineering Special presented by the ENM Faculty Fine print: This is section 7.2 and Appendix 7-A in the text. You will want to read it!

Why would demands fluctuate? Material Requirements Planning (MRP) methodology (planned order releases) optimal production lot sizes produce to meet contract orders by specified dates seasonal demands replacement parts trends in demands shifts in market place (competitor, advertising, discounts, etc. change in sales force

The Problem A known set of time-varying demands Setup or order costs independent of order or lot size (Q) Holding costs proportional to the number of time periods item is maintained in inventory What order quantities or production lot sizes will minimize order + holding cost over the planning horizon?

Methods for dealing with “lumpy” demands (D1, D2, …,Dn) A set of lumpy demands Production Smoothing Use EOQ (assumes constant demand) Simple rules fixed period demand period order quantity lot for lot reordering Heuristic rules Silver-Meal method least unit cost part period balancing (PPB) Wagner-Whitin algorithm Transportation Problem

Some Assumptions Demands, Dj, are known for periods j =1, …n Demand Dj must be satisfied in period j and available at the start of the period Replenishments arrive at the beginning of a period No quantity discounts Unit costs do not change over planning horizon no shortages permitted lead-times are known and constant entire order quantity arrives at the same time items are independent of one another carrying costs applies only to inventory carried over from one time period to another

Using EOQ Setup cost is $132 Holding costs are $.60 per item per week Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 Setup cost is $132 Holding costs are $.60 per item per week

Using EOQ Total Setup cost is $132 x 4 = $528 Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 Q 139 End Inv 97 55 23 11 124 106 92 16 117 Sum = 653 Total Setup cost is $132 x 4 = $528 Total Holding costs are $.60 x (653-117) = $321.60 Total cost = $849.60

Fixed Period Demand (D1, D2, …,Dn) Rule: Order m months worth of demands. Example: m = 3 1st order quantity: D1 + D2 + D3 2nd order quantity: D4 + D5 + D6

Fixed Period Demand Q End Inv Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 84   44 138 59 114 Q End Inv M = 2; cost = 5 x 132 + 218 x .60 = $790.80 Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 154   285 70 173 128 114 Q End Inv M = 5; cost = 2 x 132 + 699 x .60 = $683.40

Period Order Quantity (POQ) (D1, D2, …,Dn) 1. Establish an average lot size - L. 2. Determine the average demand - Davg 3. Set m = L / Davg 4. Order in period j, Dj+1 , Dj+2 , …,Dj+m Example: week 1 2 3 4 5 6 demands 19 14 21 25 18 23 Davg = 120 / 6 = 20 If L = 40 then m = 40 / 20 = 2 order: D1 + D2 , D3+ D4 , D5 + D6

Period Order Quantity (POQ) (D1, D2, …,Dn) 1 L = 150 (perhaps breakpoint for quantity discounting) 2. Davg = 43.9  44 3. Set m = L / Davg = 150 / 44 = 3.4  3 Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 116   150 135 74 138 90 Q End Inv cost =4 x 132 + 460 x .60 = $804.00

Lot for Lot (L4L) (D1, D2, …,Dn) Set order quantity equal to Dj That is, order for each period , the expected demands. Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 Q End Inv cost = 10 x 132 = $1,320 Note that holding costs are zero!

Silver-Meal Heuristic Try to minimize average cost per period costs includes ordering (set-up) (K) and holding cost (h). Assume K and h are constant over the planning horizon. assume holding costs occur at the end of the period assume quantity needed in a period is used at the beginning of the period

Silver-Meal Heuristic (D1, D2, …,Dn) Continue until cost / period starts increasing Order quantity is then D1 + D2 + … + Dm m

Silver-Meal Heuristic Example: K = $90 ; h = 1.20 per unit per month Month 1 2 3 4 5 6 Demands 20 30 23 19 32 28 Order Period Ordering cost Holding Cost Avg cost / period one month 90 90 two months 90 1.2(30) = 36 126 / 2 = 63 three months 90 1.2(30) +2.4(23)=91.2 181.2 / 3 = 60.4 Order quantity = 20 + 30 + 23 = 73 M=3 Four months 90 91.2 + 3.6 (19) = 159.6 249.6 / 4 = 62.4

Repeat for next order Example: K = $90 ; h = 1.20 per unit per month Demands 20 30 23 19 32 28 Order Period Ordering cost Holding Cost Avg cost / period one month 90 90 two months 90 1.2(32) = 38.4 128.4 / 2 = 64.2 three months 90 1.2(32) +2.4(28)=105.6 195.6 / 3 = 65.2 Order quantity = 19 + 32 = 51 M=2

Least Unit Cost Heuristic Compute average cost per unit demanded rather average cost per period.

Example: K = $90 ; h = 1.20 per unit per month Demands 20 30 23 19 32 28 Order Period Ordering cost Holding Cost Avg cost / unit one month 90 90/20 = 4.50 two months 90 1.2(30) = 36 126/ 50 = 2.52 three months 90 1.2(30) +2.4(23)=91.2 181.2 / 73 = 2.48 Order quantity = 20 + 30 + 23 = 73 M=3 Four months 90 91.2 + 3.6 (19) = 159.6 249.6 / 92 = 2.71

Part Period Balancing (PPB) Attempts to minimize the sum of the variable cost for all lots. Definition: Part period = one unit held in inventory for one period PPm = part period for m periods PP1 = 0 PP2 = D2 PP3 = D2 + 2D3 PPm = D2 + 2D3 + … + (m-1) Dm

Continued Part Period Balancing (PPB) Inventory holding cost = h (PPm) Find m so that K h(PPm) or PPm K / h Order quantity = Q = D1 + D2 + … + Dm

Example problem continued Example: K = $90 ; h = 1.20 per unit per month Month 1 2 3 4 5 6 Demands 20 30 23 19 32 28 K / h = 90 / 1.2 = 75 PP1 = 0 PP2 = 30 PP3 = (30) +2(23)= 76 stop! Q1 = 20 + 30 + 23 = 73 Starting month 4: PP1 = 0 PP2 = (32) PP3 =32 + 2 (28) = 88 stop Q2 = 19 + 32 + 28 = 79

An Old Favorite PPm  K/h = 132 / .6 = 220 pp1 = 0 pp2 = 45 pp1 = 0 Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 PPm  K/h = 132 / .6 = 220 pp1 = 0 pp2 = 42 Pp3 = 42 + (2) 32 = 106 pp4 = pp3 + 3(12) = 142 pp5 = pp4 + 4(26) = 246 Q1 = 42 + 42 + 32 + 12 + 26 = 154 pp1 = 0 pp2 = 45 pp3 = 45 + (2) 14 = 73 pp4 = 73 + 3(76) = 301 Q6 = 112 + 45 + 14 + 76 = 247 Q10 = 38 Wk1 Wk2 Wk3 Wk4 Wk5 Wk6 Wk7 Wk8 Wk9 Wk10 42 32 12 26 112 45 14 76 38 154   247 70 135 90 cost =3 x 132 + 328 x .60 = $724.20

The Wagner-Whitin Algorithm Our Feature Presentation The Wagner-Whitin Algorithm The Whole Thing! The Big Enchilada

The General Problem Qt = production or order quantity in period t It = inventory at the end of period t Ct(Qt) = cost of production in period t ht(It) = holding cost from period t to t+1

The Linear Problem Qt = production or order quantity in period t It = inventory at the end of period t Kt = fixed cost of production in period t Ct = cost of production in period t ht = holding cost per unit carried from period t to t+1

A Simpler Problem?

It-1 Qt = 0 Property 1: A replenishment only takes place when the inventory level is zero. Therefore Qk = 0, or Dk or Dk + Dk+1 or … or Dk + Dk+1 + … + Dn It-1 Qt = 0 Property 2: There is an upper limit to how far before a period j we would include its requirements, Dj in a replenishment quantity. That is, the carrying costs become so high that it is less expensive to have a second replenishment occur.

A Wagner-Whitin Example The Maka Parte Company makes parts for General Motors automobiles. One part they make is a vulcanized tri-solenoid distributor. A primary component used in the manufacture of this distributor is a silicon computer chip. This chip is purchased from a vendor - Outspeak Corp. Ordering costs are $70 and holding costs are $ .5 per item per month. Demands for the next six months based upon an exponential smoothing model with seasonal effects are: Nov Dec Jan Feb Mar Apr 120 80 94 78 86 110

A Wagner-Whitin Example n = 6 (Apr) Q6 = 110; f = $70 Order Cost = $70 Holding cost = .5 n = 5 (Mar/Apr) Q5 = 86, 196 f5(86) = 70 + 70 =140 f5(196) = 70 + .5 (110) = 125 Nov Dec Jan Feb Mar Apr 120 80 94 78 86 110 n = 4 (Feb/Mar/Apr) Q4 = 78, 164, 274 f4(78) = 70 + 125 = 195 f4(164) = 70 + .5 (86) + 70 = 183 f4(274) = 70 + .5(86) + 1.00 (110) = 223

A Wagner-Whitin Example Order Cost = $70 Holding cost = .5 Nov Dec Jan Feb Mar Apr 120 80 94 78 86 110 n = 3 (Jan/Feb/Mar/Apr) Q3 = 94, 172, 258, 368 f3(94) = 70 + 183 = 253 f3(172) = 70 + .5 (78) + 125 = 234 f3(258) = 70 + .5(78) + 1.00 (86) + 70 = 265 f3(368) = 70 + .5(78) + 1.00 (86) + 1.5(110) = 360

A Wagner-Whitin Example Order Cost = $70 Holding cost = .5 Nov Dec Jan Feb Mar Apr 120 80 94 78 86 110 n = 2 (Dec/Jan/Feb/Mar/Apr) Q2 = 80, 174, 252, 338, 448 f2(80) = 70 + 234 = 304 f2(174) = 70 + .5 (94) + 183 = 300 f2(252) = 70 + .5(94) + 1.00 (78) + 125 = 320 f2(338) = 70 + .5(94) + 1.00 (78) + 1.5(86) + 70 = 394 f2(448) = 70 + .5(94) + 1.00 (78) + 1.5(86) + 2(110) = 544

A Wagner-Whitin Example Order Cost = $70 Holding cost = .5 n = 1 (Nov/Dec/Jan/Feb/Mar/Apr) Q1 = 120, 200, 294, 372, 458, 568 f1(120) = 70 + 300 = 370 f1(200) = 70 + .5 (80) + 234 = 344 f1(294) = 70 + .5(80) + 1.00 (94) + 183 = 387 f1(372) = 70 + .5(80) + 1.00 (94) + 1.5(78) + 125 = 446 f1(458) = 70 + .5(80) + 1.00 (94) + 1.5(78) + 2(86)+70= 563 f1(568) = 70 + .5(80) + 1.00 (94) + 1.5(78) + 2(86) + 2.5(110) = 768 Nov Dec Jan Feb Mar Apr 120 80 94 78 86 110 Q1 = 200; Q3 = 172; Q5 = 196 ; Cost = $344

Capacity Constraints

Why isn’t Wagner-Whitin used more frequently? relatively complex needs a well-defined ending point all information out to end point needed even to compute initial quantity within MRP systems using rolling schedules, the solution will keep changing the assumption that replenishments can be made only at discrete intervals computational requirements

No Fixed Cost

The Transportation Problem xtj = number of units produced in month t to satisfy month j demands

The Example h = $2 per unit per month

What can we conclude from all of this? Most heuristics outperform EOQ the Silver-Meal heuristic incurs an average cost penalty relative to Wagner-Whitin of less than 1 percent. Significant costs penalties using Silver-Meal will incur if demand pattern drops rapidly over several periods when there are a large number of periods having no demand

Can we have some really neat homework problems? Huh? Text: Chapter 7: problems 13, 14, 17, 18, 19, 22

Safety Stock When demand or lead-time is random (or both), then safety stock may be established as a “hedge” against uncertain demands. For the deterministic case: R = D L For the stochastic case: R = LTDavg + s where LTD = a random variable, the lead-time demand, LTDavg = average lead-time demand and s is the safety stock.

Safety Stock based on Fill Rate Shortage probability Pr{LTD > R) = p LTD R LTDavg s Fill rate criterion: set s = z STD where STD = standard deviation of the lead-time demand distribution then R = LTDavg + z STD

But I need to know when demands are lumpy, don’t I? Compute the variability coefficient, v = variance of demand per period square of average demand per period If V < .25 , use EOQ with Davg else use a DLS method