Managing Cycle Inventories. Matching Supply and Demand

Transcription

1 Managng Cycle Inventores Matchng Supply and Demand 1

2 Outlne Why to hold cycle nventores? Economes of scale to reduce fxed costs per unt. Jont fxed costs for multple products Long term quantty dscounts Short term quantty dscounts: Promotons 2

3 Role of Inventory n the Supply Chan Overstockng: Amount avalable exceeds demand Lqudaton, Obsolescence, Holdng Understockng: Demand exceeds amount avalable Lost margn and future sales Goal: Matchng supply and demand 3

4 Batch or Lot sze Batch = Lot = quantty of products bought / produced together But not smultaneously, snce producton can not be smultaneous Q: Lot sze. R: Demand per tme. Consder sales at a Jean s retaler wth demand of 10 jeans per day and an order sze of 100 jeans. Q=100. R=10/day. Inventory Q R 0 Order Q/R Order Cycle Order Tme 4

5 Demand affected by vsblty Demand s hgher when the nventory s hgher and s smaller when the nventory s smaller. When I am buyng coffee, t s often not fresh. Why? Fresh coffee s consumed fast but stale coffee s not. Or because: Inventory Coffee becomes stale Store owner does not prepare new coffee Expects that coffee wll fnsh n the next 2 hours Hours I arrve at the coffee shop 5

6 Batch or Lot sze Cycle nventory=average nventory held durng the cycle =Q/2=50 jean pars Average flow tme Remember Lttle s law =(Average nventory)/(average flow rate)=(q/2)/r=5 days Long flow tmes make a company vulnerable to product / technology changes Lower cycle nventory decreases workng (operatng) captal needs and space requrements for nventory Then, why not to set Q as low as possble? 6

7 Why to order n lots? Fxed orderng cost: S Increase the lot sze to decrease the fxed orderng cost per unt Materal cost per unt: C Holdng cost: Cost of carryng 1 unt n the nventory: H H=h.C h: carryng $1 n the nventory > nterest rate Lot sze s chosen by tradng off holdng costs aganst fxed orderng costs (and sometmes materal costs). Where to shop from: Fxed cost (drvng) Materal cost Convenence store low HIGH Sam s club HIGH low 7

8 Economc Order Quantty - EOQ Annual TC = carryng + cost Annual orderng cost + Purchasng cost Q 2 hc R TC = + Q S + CR Total cost s smple functon of the lot sze Q. Note that we can drop the last term, t s not affected by the choce of Q. 8

9 Cost Mnmzaton Goal Annual Cost The Total-Cost Curve s U-Shaped Q R TC = hc + S + CR 2 Q Holdng costs Orderng Costs Q (optmal order quantty) Order Quantty (Q) 9

10 Dervng the EOQ Usng calculus, we take the dervatve of the total cost functon and set the dervatve equal to zero and solve for Q. Total cost curve s convex.e. curvature s upward so we obtan the mnmzer. EOQ 2RS EOQ 2S = T = = n= R = hc R RhC EOQ RhC 2S T: Reorder nterval length = EOQ/R. n: Orderng frequency: number of orders per unt tme = R/EOQ. The total cost curve reaches ts mnmum where the nventory carryng and orderng costs are equal. Total cost( Q = EOQ ) = 2RShC 10

11 EOQ example Demand, R = 12,000 computers per year. Unt cost, C = $500 Holdng cost, h = 0.2. Fxed cost, S = $4,000/order. Fnd EOQ, Cycle Inventory, Average Flow Tme, Optmal Reorder Interval and Optmal Orderng Frequency. Q = , say 980 computers Cycle nventory = Q/2 = 490 unts Average Flow Tme = Q/(2R) = 0.49 month Optmal Reorder nterval, T = year = 0.98 month Optmal orderng frequency, n=12.24 orders per year. 11

12 Key Ponts from Batchng In decdng the optmal lot sze the trade off s between setup (order) cost and holdng cost. If demand ncreases by a factor of 4, t s optmal to ncrease batch sze by a factor of 2 and produce (order) twce as often. Cycle nventory (n unts) doubles. Cycle nventory (n days of demand) halves. If lot sze s to be reduced, one has to reduce fxed order cost. To reduce lot sze by a factor of 2, order cost has to be reduced by a factor of 4. Ths s what JIT strves to do. 12

13 Strateges for reducng fxed costs In producton Standardzaton / dedcated Smplfcaton Set up out of the producton lne In delvery Thrd party logstcs Aggregatng multple products n a sngle order» Temporal, geographc aggregaton Varous truck szes, dffcult to manage 13

14 Example: Lot Szng wth Multple Products Demand per year R L = 12,000; R M = 1,200; R H = 120 Common transportaton cost per delvery, S = $4,000 Product specfc order cost per product n each delvery s L = $1,000; s M = $1,000; s H = $1,000 Holdng cost, h = 0.2 Unt cost C L = $1,000; C M = $1,000; C H = $1,000 14

15 Delvery Optons No Aggregaton: Each product ordered separately Complete Aggregaton: All products delvered on each truck Talored Aggregaton: Selected subsets of products on each truck 15

16 No Aggregaton: Order each product ndependently Ltepro Medpro Heavypro Demand per year 12,000 1, Fxed cost / order $5,000 $5,000 $5,000 Optmal order sze 1, Order frequency 11.0 / year 3.5 / year 1.1 / year Annual cost $109,544 $34,642 $0,954 Total cost = $155,140 16

17 Complete Aggregaton: Order all products jontly Total orderng cost S*=S+s L +s M +s H = $7,000 n: common orderng frequency Annual orderng cost = n S* Total holdng cost: Total cost: TC ( n) n * = L Rn hc R M n hc R H n hc L + M * h = S n + n R C + R C + 2 R C ( ) ( + + ) h R C R C R C 2 S L L M M H H L L M M H H * H 17

18 Complete Aggregaton: Order all products jontly Ltepro Medpro Heavypro Demand per year 12,000 1, Order frequency 9.75/year 9.75/year 9.75/year Optmal order sze 1, Annual holdng cost $61,512 $6,151 $615 Annual order cost = 9.75 $7,000 = $68,250 Annual total cost = $136,528 Orderng hgh and low volume tems at the same frequency cannot be a good dea. 18

19 Talored Aggregaton: Orderng Selected Subsets Example Orders may look lke (L,M); (L,H); (L,M); (L,H). Most frequently ordered product: L M and H are ordered n every other delvery. We can assocate fxed order cost S wth product L because t s ordered every tme there s an order. Products other than L are assocated only wth ther ncremental order costs (s values). An Algorthm: Step 1: Identfy most frequently ordered product Step 2: Identfy frequency of other products as a relatve multple Step 3: Recalculate orderng frequency of most frequently ordered product Step 4: Identfy orderng frequency of all products 19

20 Talored Aggregaton: Orderng Selected Subsets s the generc ndex for products, s L, M or H. Step 1: Fnd most frequently ordered tem: n = hc R 2( S + s ) The frequency of the most frequently ordered tem wll be modfed later. Ths s an approxmate computaton. Step 2: Relatve order frequency of other tems, m n hc R n = m = s 2 n m are relatve order frequences, they must be ntegers. n = max{ n } 20

21 Talored Aggregaton: Orderng Selected Subsets Step 3: Recompute the frequency of the most frequently ordered tem. Ths tem s ordered n every order whereas others are ordered n every m orders. The average fxed orderng cost s: n * = S + s m Annual orderng cost Annual holdng cost 2 S + R m hc s m = s = n( S + ) m R 2n / m hc dfferent than (10.8) on p.263 of Chopra 21

22 Talored Aggregaton: Orderng Selected Subsets Step 4: Recompute the orderng frequency n of other products: n n = m Total Annual orderng cost: ns+n H s H +n M s M +n L s L Total Holdng cost: R L n hc R M R L + hc M + 2 2n 2n L M H H hc H 22

23 Step 1: n L = Talored Aggregaton: Orderng Selected Subsets hclrl 2( S + s ) L = 11, n = 3.5, n = 1.1 n = max{ n } = M H 11 Step 2: hcm RM n nm = nh mm = mh sm = 7.7, = 2.4; = 2, = 5 2 nm Item L s ordered most frequently. Every other L order contans one M order. Every 5 L orders contan one H order. At ths step we only now relatve frequences, not the actual frequences. 23

24 Step 3: Step 4: Talored Aggregaton: Orderng Selected Subsets Total orderng cost: n M ns+n H s H +n M s M +n L s L =11.47(4000)+11.47(1000)+5.73(1000)+2.29(1000) Total holdng cost = n m * M = 5.73 R L n h C R M R L + h C M n 2 n L n * = M h C R m s 2 ( S + ) m = H H h C = ( ) (. ) 2 ( ) (. ) 2 ( ) (. ) n H H = n m * H =

25 Talored Aggregaton: Order selected subsets Ltepro Medpro Heavypro Demand per year 12,000 1, Order frequency 11.47/year 5.73/year 2.29/year Optmal order sze Annual holdng cost $52,810 $10,470 $2,630 Annual order cost = $65,370 Total annual cost = $130,650 25

26 Lessons From Aggregaton Informaton technology can decrease product specfc orderng costs. Aggregaton allows frm to lower lot sze wthout ncreasng cost Order frequences wthout aggregaton and wth talored aggregaton» (11; 3.5; 1.1) vs. (11.47; 5.73; 2.29) Complete aggregaton s effectve f product specfc fxed cost s a small fracton of jont fxed cost Talored aggregaton s effectve f product specfc fxed cost s large fracton of jont fxed cost 26

27 Quantty Dscounts Lot sze based All unts Margnal unt Volume based How should buyer react? What are approprate dscountng schemes? 27

28 All-Unt Quantty Dscounts Cost/Unt Total Materal Cost $3 $2.96 $2.92 5,000 10,000 q 1 q 2 Order Quantty 5,000 10,000 Order Quantty 28

29 All-Unt Quantty Dscounts Fnd EOQ for prce n range q to q +1 If q EOQ < q +1,» Canddate n ths range s EOQ, evaluate cost of orderng EOQ If EOQ < q,» Canddate n ths range s q, evaluate cost of orderng q If EOQ q +1,» Canddate n ths range s q +1, evaluate cost of orderng q +1 Fnd mnmum cost over all canddates 29

30 Fndng Q wth all unts dscount Total Cost Q = 1 2RS hc 1 Q = 2 2RS hc 2 Q = 3 2RS hc 3 Quantty 30

31 Fndng Q wth all unts dscount Total Cost Q = 1 2RS hc 1 2 Q = 3 2RS hc 3 Q = 2 2RS hc 2 Quantty 31

32 Fndng Q wth all unts dscount Total Cost 2 1 Quantty 32

33 Margnal Unt Quantty Dscounts Cost/Unt Total Materal Cost c 0 c 1 c 2 $3 $2.96 $2.92 V 2 V 1 5,000 10,000 q 1 q 2 Order Quantty 5,000 10,000 Order Quantty 33

34 Margnal Unt Quantty Dscounts V = C ost of buyng exactly q. V = 0. V = c ( q q ) + c ( q q ) c ( q q ) If q Q q + 1 A nnual order cost = A nnual holdng cost = T otal cost( Q ) Q, E O Q R Q S A nnual m ateral cost = For range, h 2 R Q V Q q c ( V + ( Q q ) c ) 0 ( + ( ) ) R = ( ) Q S c h R Q V q c = 2 = ( + ) 2 R S V q c hc 0 34

35 Margnal Unt Quantty Dscounts V = C ost of buyng exactly q. V = 0. V = c ( q q ) + c ( q q ) c ( q q ) If q Q q + 1 A nnual order cost = A nnual holdng cost = T otal cost( Q ) Q, E O Q R Q S A nnual m ateral cost = For range, h 2 R Q V Q q c ( V + ( Q q ) c ) 0 ( + ( ) ) R = ( ) Q S c h R Q V q c = 2 = ( + ) 2 R S V q c hc 0 35

36 Margnal-Unt Quantty Dscounts Fnd EOQ for prce n range q to q +1 If q EOQ < q +1,» Canddate n ths range s EOQ, evaluate cost of orderng EOQ If EOQ < q,» Canddate n ths range s q, evaluate cost of orderng q If EOQ q +1,» Canddate n ths range s q +1, evaluate cost of orderng q +1 Fnd mnmum cost over all canddates 36

37 Margnal Unt Quantty Dscounts Total cost EOQ 1 EOQ 3 q 1 q 2 Lot sze Compare ths total cost graph wth that of all unt quantty dscounts. Here the cost graph s contnuous whereas that of all unt quantty dscounts has breaks. 37

38 Margnal Unt Quantty Dscounts Total cost EOQ 1 EOQ 2 EOQ 3 q 1 q 2 Lot sze 38

39 Why Quantty Dscounts? The lot sze that mnmzes retalers cost does not necessarly mnmze suppler and retaler s cost together. Coordnaton n the supply chan Wll suppler and retaler be wllng to operate wth the same order szes, frequences, prces, etc.? How to ensure ths wllngness? Va contracts. Quantty dscounts gven by a suppler to a retaler can motvate the retaler to order as the suppler wshes. 39

40 Coordnaton for Commodty Products: Suppler and Retaler Coordnaton Consder a suppler S and retaler R par R = 120,000 bottles/year S R = $100, h R = 0.2, C R = $3 S S = $250, h S = 0.2, C S = $2 Retaler s optmal lot sze = 6,324 bottles Retaler s annual orderng and holdng cost = $3,795; If Suppler uses the retaler s lot sze, Suppler s annual orderng and holdng cost = $6,009 Total annual supply chan cost = $9,804 40

41 Coordnaton for Commodty Products What can the suppler do to decrease supply chan costs? Combne the suppler and the retaler Coordnated lot sze: 9,165= 2R( SS + SR) h( C + C ) Retaler cost = $4,059; Suppler cost = $5,106; Supply chan cost = $9,165. $639 less than wthout coordnaton. S R ChooseQ R RetalerCost(Q Coordnaton Savngs by MnmzeRetalerCost(Q), then R ) + SupplerCost(Q = R ) MnmzeRetalerSupplerCost (Q) { RetalerCost(Q ) + SupplerCost(Q )} R Q R - MnmzeRetalerSupplerCost (Q) Q 41

42 Coordnaton va Prcng by the Suppler Effectve prcng schemes All unt quantty dscount» $3 for lots below 9,165» $ for lots of 9,165 or more What s suppler s and retaler s cost wth the all unt quantty dscount scheme? Not the same as before. Who gets the savngs due to coordnaton? Pass some fxed cost to retaler (enough that the retaler rases order sze from 6,324 to 9,165) 42

43 Quantty Dscounts for a Frm wth Market Power (Prce dependent demand) No nventory related costs Demand curve 360,000-60,000p Retaler dscounts to manpulate the demand Retaler chooses the market prce p, manufacturer chooses the sales prce C R to the retaler. Manufacturng cost C M =$2/unt Manufacturer Manufacturer s Prce, C R demand Retaler Market Prce, p demand 43

44 Quantty Dscounts for a Frm wth Market Power Retaler proft=(p-c R )(360,000-60,000p) Manufacturer proft=(c R -C M ) (360,000-60,000p) Note C M =$2 If each optmzes ts own proft: Manufacturer assumes that p= C R Sets C R =$4 to maxmze (C R -2) (360,000-60,000C R ) Retaler takes C R =$4 Sets p=$5 to maxmze (p-4)(360,000-60,000p) Q=60,000. Manufacturer and retaler profts are $120K and $60K respectvely. Total SC proft s $180K. Observe that f p=$4, total SC profts are (4-2)120K=$240K. How to capture =$60K? 44

45 Two Part Tarffs and Volume Dscounts Desgn a two-part tarff that acheves the coordnated soluton. Desgn a volume dscount scheme that acheves the coordnated soluton. Impact of nventory costs Pass on some fxed costs wth above prcng 45

46 Two part tarff to capture all the profts Manufacturer sells each unt at $2 but adds a fxed charge of $180K. Retaler proft=(p-2)(360,000-60,000p)-180,00 Retaler sets p=$4 and obtans a proft of $60K Q=120,000 Manufacturer makes money only from the fxed charge whch s $180K. Total proft s $240K. Manufacturer makes $60K more. Retaler s proft does not change. Does the retaler complan? Splt of profts depend on barganng power Sgnalng strength Other alternatve buyers and sellers Prevous hstory of negotatons; credblty (of threats) Mechansm for conflct resoluton: teratve or at once 46

47 All unts dscount to capture all profts Suppler apples all unt quantty dscount: If 0<Q<120,000, C R =$4 Else C R =$3.5 If Q<120,000, we already worked out that p=$5 and Q=60,000. And the total proft s $180,000. If Q>=120,000, the retaler chooses p=$4.75 whch yelds Q=75,000 and s outsde the range. Then Q=120,000 and p=$4. Retaler proft=(4-3.5)120,000=60,000 Manufacturer proft=(3.5-2)120,000=180,000 Total SC profts are agan $240K. Manufacturer dscounts to manpulate the market demand va retaler s prcng. 47

48 Lessons From Dscountng Schemes Lot sze based dscounts ncrease lot sze and cycle nventory n the supply chan Lot sze based dscounts are justfed to acheve coordnaton for commodty products Volume based dscounts are more effectve n general especally n keepng cycle nventory low End of the horzon panc to get the dscount: Hockey stck phenomenon Volume based dscounts are better over rollng horzon 48

49 Short Term Dscountng Why? To ncrease sales, Ford To push nventory down the SC, Campbell To compete, Peps Leads to a hgh lot sze and cycle nventory because of strong forward buyng 49

50 Weekly Shpments of Chcken Noodle Soup. Forward Buyng Shpments Consumpton Dscountng 50

51 Short Term Promotons Promoton happens only once, Optmal promoton order quantty Q d s a multple of EOQ Quantty Q d EOQ Tme 51

52 Short Term Dscountng C: Normal unt cost d: Short term dscount R: Annual demand h: Cost of holdng $1 per year Q d : Short term (once) order quantty Q d d R C EOQ = + ( C - d ) h C - d Forward buy = Q d - Q* 52

53 Short Term Dscounts: Forward buyng. Ex 10.8 on p.280 Normal order sze, EOQ = 6,324 bottles Normal cost, C = $3 per bottle Dscount per tube, d = $0.15 Annual demand, R = 120,000 Holdng cost, h = 0.2 Q d =38,236 Forward buy =38,236-6,324=31,912 Forward buy s fve tmes the EOQ, ths s a lot of nventory! 53

54 Suppler s Promoton passed through to consumers Demand curve at retaler: 300,000-60,000p Normal suppler prce, C R = $3.00 Retaler proft=(p-3)(300,000-60,000p) Optmal retal prce = $4.00 Customer demand = 60,000 Suppler s promoton dscount = $0.15, C R = $2.85 Retaler proft=(p-2.85)(300,000-60,000p) Optmal retal prce = $3.925 Customer demand = 64,500 Retaler only passes through half the promoton dscount and demand ncreases by only 7.5% 54

55 Avodng Problems wth Promotons Goal s to dscourage retaler from forward buyng n the supply chan Counter measures Sell-through: Scan based promotons» Retaler gets the dscount for the tems sold durng the promoton Customer coupons; Dscounts avalable when the retaler returns the coupons to the suppler. The coupons are handed out to consumers by the suppler. Retaler realzes the dscounts only after the consumer s purchase. 55

56 Strategc Levers to Reduce Lot Szes Wthout Hurtng Costs Cycle Inventory Reducton Reduce transfer and producton lot szes» Aggregate the fxed costs across multple products, supply ponts, or delvery ponts E.g. Talored aggregaton Are quantty dscounts consstent wth manufacturng and logstcs operatons?» Volume dscounts on rollng horzon» Two-part tarff Are trade promotons essental?» Base on sell-thru (to consumer) rather than sell-n (to retaler) 56

57 Inventory Cost Estmaton Holdng cost Cost of captal Spolage cost, semconductor product lose 2% of ther value every week they stay n the nventory Occupancy cost Orderng cost Buyer tme Transportaton cost Recevng/handlng cost Handlng s generally Orderng cost rather than Holdng cost 57

58 Summary EOQ costs and quantty Talored aggregaton to reduce fxed costs Prce dscountng to coordnate the supply chan Short term promotons 58