Figure 1. Inventory Level vs. Time - EOQ Problem

Transcription

1 IEOR 54 Sprng, 009 rof Leahman otes on Eonom Lot Shedulng and Eonom Rotaton Cyles he Eonom Order Quantty (EOQ) Consder an nventory tem n solaton wth demand rate, holdng ost h per unt per unt tme, and replenshment ost A (ndependent of quantty replenshed) What s the optmal replenshment lot sze Q? Fgure plots nventory level vs tme for ths problem for the ase,000 and Q s taken to be 00 ote the sawtooth pattern, whereby nventory s replenshed by the order quantty (00 n ths ase), and then steadly falls to 0, whereupon the yle repeats tself Fgure Inventory Level vs me - EOQ roblem Inventory Level me We formulate a funton expressng the total ost rate as a funton of Q he relevant osts are the nventory holdng ost and replenshment ost ote that replenshment orders wll be plaed one every Q/ unts of tme he nventory level has a maxmum of Q and falls lnearly to 0; thus the average nventory level s Q/ he total ost rate s therefore expressed as RC ( Q) A / Q + hq / We take a dervatve to fnd the best value of Q:

2 * RC / Q A / Q + h / Q A / h he optmal tme between replenshments s gven by * * Q / A/ h Fgure plots RC (Q) vs Q for the ase of h 05,,000 and A 0 he best value of Q s where the orderng ost rate and the holdng ost rate are equalzed; ths ours at Q 00 n ths ase ote that the total ost rate urve s very flat around the best order sze Fgure otal Cost Rate vs Order Quantty otal Cost Rate A/Q hq/ RC(Q) Order Quantty, Q he Eonom Manufaturng Quantty (EMQ) ow suppose that replenshment s not nstantaneous but ours at a fnte produton rate > ow what s the optmal replenshment quantty Q and optmal tme between replenshments? Whle replenshment s underway, we add to nventory at rate but nventory s onsumed by demands at rate he net rate of hange n nventory durng replenshment s Fgure 3 plots nventory versus tme for ths problem n the ase,000, 4,000 and Q s taken to be 00 We stll have a sawtooth pattern for the

3 nventory yle, but now replenshment ours gradually rather than all at one, so the trangles are no longer rght trangles urng replenshment, nventory aumulates at rate After a quantty Q has been produed, nventory delnes at rate untl the next replenshment bath s started Fgure 3 Inventory Level vs me - EMQ roblem Inventory Level me If we replensh a quantty Q, the duraton of the replenshment s Q/ and the heght of the nventory at the ompleton of the replenshment s therefore [Q/]( ) Q( /) he average nventory level s half of ths heght (why?) he total ost rate s therefore expressed as RC ( Q) A + hq Q akng a dervatve and solvng for Q, we fnd Q * A h Fgure 4 plots RC (Q) vs Q for the ase h 05, A 0,,000, 4,000 hs s the same data as n Fgure exept we have added a fnte replenshment rate 4,000 For ths ase, the optmal bath sze s 30, e, about 5% bgger than the optmal order 3

4 sze n the ase of an nfnte replenshment rate As before, the total ost rate urve s very flat n the vnty of the best pont Fgure 4 otal Cost Rate vs Bath Sze otal Cost Rate A/Q h(-/)q/ RC(Q) Bath Sze, Q he optmal tme between replenshments n ths ase s expressed as opt Q * / A h ote that we ould have formulated ths problem wth the tme between replenshments as the deson varable: A CR( ) + h, and takng a dervatve of ths expresson wth respet to and solvng for results n opt A h 4

5 When we onsder replenshment of multple tems, t wll be more onvenent to work wth ths varable rather than wth the order quantty varable hs tme between replenshments varable also s ommonly referred to as the yle length 3 Replenshment Lead me and Reorder ont Suppose there s a delay tme L between the tme of request for replenshment and the tme replenshment starts to arrve (fnte produton rate ase) or the tme the replenshment arrves (EOQ) ase hen the replenshment order must be plaed before the nventory s exhausted If the demand ours exatly at the ontnuous rate, then the replenshment order may be made when the nventory falls to the level s L hs level s termed the reorder pont ow suppose the demand s unertan wth mean rate and suppose the standard devaton of demand durng the lead tme s σ L Suppose we wsh the probablty of a stok-out to be very low It s ustomary to set the reorder pont to be s L + kσ L where k s alled the safety fator For example, f k 3 and f demand durng the lead tme has a normal dstrbuton, then the probablty of stok-out durng the lead tme s about one n one thousand 4 Changeover tme Suppose there s a fxed delay between the tme a produtonreplenshment s requested and the tme the replenshment of nventory begns (hs s an example of an nventory replenshment lead tme as defned above) hs delay s sometmes termed the hangeover tme or the setup tme he exstene of a setup tme enfores a onstrant on the tme between replenshments: there must be tme to make the setup and omplete produton before the replenshment yle expres, e, +, or, equvalently, hat s, the mnmum yle length that s feasble s mn he lowest-ost, feasble yle length n ths ase s 5

6 * Max { }, mn opt 5 Smple Common Rotaton Cyle for Replenshment of Multple Items ow suppose dfferent nventory tems must be replenshed usng the same mahne Only one tem may be replenshed at a tme here s a hangeover tme when the mahne s swthed to produe tem after produng some other tem he demand rate for tem s, the produton rate s, the hangeover ost s A, and the holdng ost s h One way to handle ths stuaton s establsh a ommon rotaton yle of length for all tems hat s, the tems are set up n sequene and a quantty lastng tme unts s produed of eah tem After a tme has elapsed, the yle starts over agan Obvously, must be long enough to make all the hangeovers and produe a yle quantty of eah tem, e, +, or mn Consderng the relevant osts, the total ost rate s A CR ( ) + and the unonstraned optmal yle length s gven by h opt A h he best feasble yle length s therefore gven by * Max {, } mn opt 6

7 6 Common Rotaton Cyle wth Item Multplers Consder the same stuaton as above but suppose some of the tems have very low demand rates relatve to the others Forng all tems onto the same replenshment yle may seem wasteful Instead, t mght be more reasonable that a low demand tem should not be produed every yle but nstead be produed one every k yles Here, the parameter k s alled the multpler At least one tem wll have k, e, t wll be produed every yle, and ts yle length s termed the fundamental yle length All other tems are replenshed n an nteger multple of he total ost rate n ths ase s expressed as RC( k, ) A k + h k As smple as ths generalzaton of the ommon rotaton yle problem s to state, there s no polynomal algorthm for determnng optmal values of the k and Fortunately, several pratal heurst algorthms have been developed that generate near-optmal solutons for ths problem One suh algorthm, known as the oll and Whybark roedure (named after ts authors), works as follows: Step For eah tem, ompute ts optmal yle length n solaton, e, A h Step Set the fundamental yle length to be Mn { } Step 3 For eah, set k to the nteger round-up or nteger round-down of /, whhever mnmzes RC A k + h k Step 4 Consderng all possble ombnatons of tems that would be produed n the same yle, ompute the mnmum feasble yle length, e, mn Max yles j j j k 7

8 8 Step 5 Re-ompute the fundamental yle length usng the k s from Step 3 and the mn from Step 4 as, mn h k k A Max Step 6 If the value of s unhanged, termnate the algorthm Otherwse, return to Step 3 to re-ompute the multplers