CagebWhVg ba gb HcXeTg baf FTaTZX Xag 6S. Linear Programming BTafbb D uwv$ >Xcg) by FTaTZX Xag CaYbe Tg ba LlfgX f RNLM HF HiXei Xj Class Overview (Ch. ) X Mgmt of Quality/ Six Sigma Quality (Ch. 9, 1) Demand Mgmt Forecasting (Ch 3) Operations, Productivity, and Strategy (Ch. 1, 2) Queueing/ Simulation (Ch. 18) Aggregated Planning (Ch. 13) Project Management (Ch. 17) Supply Chain Management (Ch 11) Inventory Management (Ch. 12) Strategic Capacity Planning (Ch. 5, 5S) Location Planning and Analysis (Ch. 8) MRP & ERP (Ch 14) Process Selection/ Facility Layout; LP (Ch. 6, 6S) X X X X X JIT & Lean Mfg System (Ch. 15) Term Project, - MbWTlqf Hhg ax o P Tg f EI: o Bbj gb Ybe h TgX FbWX $: o Bbj gb fb ix: n =b chg az gbb f o FL(?kVX )Lb ixe o E awb be E azb$ o Bbj gb Tcc l: P Tg f EI E axte IebZeT az$ : o FTg X Tg VT gxv a dhx Algorithm$ n Gbg Vb chgxe cebzet az o ; bvtgxf gxw exfbhevxf gb TV XiX Ta bu XVg ix 목적을추구하기위해제약된자원을어떻게할당하는가하는문제 $ o I baxxexw Ul George Dantzig a Pbe W PTe CC n >XiX bcxw jbe TU X fb hg ba VT XW Simplex Method a,4/2. /
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XTf U X KXZ ba HU XVg ix havg ba 12 1 8 6 4 2 Feasible Region 1 2 3 4 5 6 7 8 12 1 8 6 4 2 7*X 1 + 5*X 2 = 41 Iso-profit line 1 2 3 4 5 6 7 8 4,+?kgeX X Ib agf Hcg T Lb hg ba 12 1 8 6 4 2 Iso-profit line Possible Corner Point Solution 1 2 3 4 5 6 7 8 12 1 8 6 4 2 Iso-profit line X1 = 3 X2 = 4 Possible Corner Point Solution Optimal solution 1 2 3 4 5 6 7 8,,,-
L c Xk ; Zbe g Nf az > Vg batel$ FTk mx 2Q, Q - /Q,.Q - -/+ X XVgeba Vf VbafgeT ag -Q,,Q -,++ TffX U l VbafgeT ag Q, + TaW Q - + abaaxztg ix VbafgeT agf Basic variables /Q,.Q - Q. 8 -/+ 89 Q. 8 -/+ ( /Q, (.Q - -Q,,Q - Q / 8,++ 89 Q / 8,++ ( -Q, (,Q - S 8 2Q, Q - 89 FTk S 8 2Q, Q - Q, Q - Q. Q / + Nonbasic variables L c Xk ; Zbe g Nf az > Vg batel$ Q. 8 -/+ ( /Q, (.Q - Q / 8,++ ( -Q, (,Q - S 8 2Q, Q - Q, 8+ Q - 8+ Q. 8-/+ Q / 8,++ S8+$ Deciding Entering Variable ( 투입변수결정 ) Bbj hv S VTa avextfxw j Xa Q, be Q - TeX avextfxw: aw Q fhv g Tg FTkn2 o 89 Q, BXaVX?agXe az OTe TU X f Q, ) Deciding Leaving Variable( 이탈변수결정 ), Bbj hv Q, VTa avextfxw abg gb i b TgX abaaxztg ix VbafgeT ag: aw Q fhv g Tg F an-/+*/,++*-o 89 Q / Q / 8,++ ( -Q, (,Q - 8 9 Q, 8 + p Q - *- p Q / *-,$ Calculation,?agXe,$ gb IebU X,.,/ L c Xk ; Zbe g Nf az > Vg batel$ L c Xk ; Zbe g Nf az > Vg batel$ Q, 8 + p Q - *- p Q / *- Q. 8 -/+ p / + p Q - *- p Q / *- $ (.Q - S 8 2 + p Q - *- p Q / *- $ Q - ((((((((((((((((((((((((((((((((((((((((((((((( Q, 8 + p Q - *- p Q / *- Q. 8 /+ p Q - -Q / S 8.+.Q - *- p 2Q / *- Q, 8+ Q - 8+ Q. 8/+ Q / 8+ S8.+$ ((((((((((((((((((((((((((((((((((((((((((((((( Deciding Entering Variable aw Q fhv g Tg FTkn.*- (2*-o 89 Q - BXaVX?agXe az OTe TU X f Q - ) Deciding Leaving Variable, aw Q fhv g Tg F an+*,*-$ /+*,$o 89 Q. Q. 8 /+ p Q - -Q / 8 9 Q - 8 /+ p Q. -Q /,$ Calculation, Q - 8 /+ p Q. -Q / Q, 8 + p r /+(Q. -Q / $ p Q / *- S 8.+.*- /+(Q. -Q / $ (2Q / *- (((((((((((((((((((((((((((((((((((((( Q - 8 /+ p Q. -Q / Q, 8.+ Q. *- (.Q / *- S 8 /,+ p.q. *- ( Q / *- Q, 8.+ Q - 8/+ Q. 8+ Q / 8+ S8/,+$ ((((((((((((((((((((((((((((((((((((((((( Deciding Entering Variable aw Q fhv g Tg FTkn(.*- (,*-o 7 + Gb bex cebix Xag! Gb cbff U X?agXe az OTe TU X M X VheeXag fb hg ba f bcg T! Z * = 41, X 1* = 3, X 2* = 4?agXe,$ gb IebU X,,1
L c Xk ; Zbe g MTU X be Tg$ L c Xk ; Zbe g Maximize 7X 1 + 5X 2 s.t. 4X 1 + 3X 2 24 2X 1 + 1X 2 1 X 1, X 2 Min -7X 1-5X 2 s.t. 4X 1 + 3X 2 + X3 = 24 2X 1 + 1X 2 + X4 =1 X 1, X 2, X 3, X 4 Z 1 7 5 X 3 4 3 1 24 X 4 2 1 1 1 Current Solution: X 3 = 24, X 4 = 1 Z =,2 Z 1 7 5 X 3 4 3 1 24 X 4 2 1 1 1 Step1: Find Entering Variable among non-basic variable Since Max {7,5}, X 1 is Entering Variable Step2: Find Leaving Variable among basic variable Since Min {24/4=6,1/2=5}, X 4 is Leaving Variable Step3: Pivoting with X 1,3 I ibg az j g Q, I ibg az j g Q - Z 1 7 3/2 5-7/2-35 X 3 4 31 1-2 24 X 41 21 1/2 1 1/2 1 1 5 Z 1 3/2-3/2-7/2-1/2-35 -41 X 32 1 1-2 4 X 1 1 1/2-1/2 1/2 3/2 5 3 New Solution: X 1 = 5, X 3 = 4 Z = -35 Step1: Find Entering Variable among non-basic variable Since Max {3/2}, X 2 is Entering Variable Step2: Find Leaving Variable among basic variable Since Min {4/1=4,5/(1/2)=1}, X 3 is Leaving Variable,4 New Solution: X 1 = 3, X 2 = 4 Z = -41 Step1: Find Entering Variable among non-basic variable But, since all negative (-3/2, -1/2), this solution is optimal -+
Lb hg ba LXTeV az ITg 12 1 8 6 4 2 Feasible Region 1 2 3 4 5 6 7 8 L c Xk ; Zbe g o LgXc+5MTU Xg be h Tg ba o LgXc,5 aw?agxe az OTe TU X Q $ T baz GbaUTf V OTe TU Xf n Q fhv g Tg FTk S (= 9 + n CY g XeX f ab VTaW WTgX g X VheeXag f bcg T fb hg ba o LgXc-5 aw EXTi az OTe TU X T baz VheeXag <Tf V OTe TU Xf n Q e fhv g Tg F a nu ( *l 5 l 9 +o p F a h KTg b MXfg n CY l + Hcg T Lb hg ba f haubhawxw o I ibg az j g Q TaW KXcXTg LgXc, -, -- NaUbhaWXW =TfX NaUbhaWXW =TfX o FTk Q,.Q - f)g) Q, ( -Q - / (Q, Q -. Q, Q - + X 2 -X 1 + X 2 = 3 Z 1 1 3 X 3 1-2 1 4 X 4-1 1 1 3 X 1-2X 2 = 4 Z 1 4-3 -9 X 3-1 1 2 1 X 1 -. X 2-1 1 1 3 -/
NaUbhaWXW =TfX Z 1 4-3 -9 X 3-1 1 2 1 X 2-1 1 1 3 ; gxeatg ix Lb hg ba =TfX o F a (-Q, ( /Q - f)g) Q, -Q - Q. 8 / (Q, Q - Q/ 8. Q, Q - Q. Q / + X 2 Since all Y ik, unbounded solution X 1 - -1 ; gxeatg ix =TfX Z 1 2 4 X 3 1 2 1 4 X 4-1 1 1 1 ; gxeatg ix =TfX Z 1 6-4 -4 X 3 3 1-2 2 X 2-1 1 1 1 Z 1 6-4 -4 X 3 3 1-2 2 X 2-1 1 1 1-2 Z 1-2 -8 X 1 1 1/3-2/3 2/3 X 2 1 1/3 1/3 5/3-3
; gxeatg ix =TfX Z 1-2 -8 X 1 1 1/3-2/3 2/3 X 2 1 1/3 1/3 5/3 FbWX az?kt c Xf( IebWhVg F k Product Wiring Drilling Inspection Unit Profit XJ21 XM897 TR29 BR788.5 1.5 1.5 1. 3 1 2 3 2 4 1 2.5 1..5.5 $9 $12 $15 $11 Z 1-2 -8 X 1 1 1 4 X 4 3 1 1 5-4 Department Capacity (hr) Product Wiring Drilling Inspection 1,5 2,35 2,6 1,2 XJ21 XM897 TR29 BR788 Minimum Production Level 15 1 3 4.+ IebWhVg F k be h Tg ba FTk 4Q,,-Q -,Q.,,Q / f)g) +)Q,,)Q -,)Q.,Q /,++.Q,,Q - -Q..Q / -.+ -Q, /Q -,Q. -Q / -1++ +)Q,,Q - +)Q. +)Q /,-++ Q,,+ Q -,++ Q..++ Q / /++ Q, Q - Q. Q / + IebWhVg ba LV XWh az?kt c X Month Mfg Cost Selling Price July August September October November December $ 6 $ 6 $ 5 $ 6 $ 7 - - $ 8 $ 6 $ 7 $ 8 $ 9 Production Lead time: 1 month Maximum Sales for each month: 3 units Maximum Capacity of Warehouse: 1 units Variables: X 1, X 2, X 3, X 4, X 5, X 6 : number of units manufactured from July to Dec. Y 1, Y 2, Y 3, Y 4, Y 5, Y 6 : number of units sold from July to Dec.., Objective Function: Max 8Y 2 +6Y 3 +7Y 4 +8Y 5 +9Y 6 - (6X 1 +6X 2 +5X 3 +6X 4 +7X 5 ).-
IebWhVg ba LV XWh X be h Tg ba Max 8Y 2 +6Y 3 +7Y 4 +8Y 5 +9Y 6 -(6X 1 +6X 2 +5X 3 +6X 4 +7X 5 ) s.t I 1 = X 1 I 2 = I 1 +X 2 -Y 2 I 3 = I 2 +X 3 -Y 3 I 4 = I 3 +X 4 -Y 4 I 5 = I 4 +X 5 -Y 5 I 6 = I 5 +X 6 -Y 6 I i 1, for all i I 6 = Y i 3, for all i X i, Y i, I i, for all i Inventory Constraints: Inventory at end of this month = Inventory at end of prev. month + Current month s production This month s Sales > Xg IebU X o o M XeX TeX g exx ZeT af Ybe =Tj6 Q R TaW S bhe i gt af ; < = > a ZeT a, Z Vitamin Grain X Grain Y Grain Z A B C D 3 g/1kg 2 g 1 g 6 g 2 g 3 g g Na g Vbfgf Ybe ZeT af6 +)+- Q$ +)+/ R$ +)+- S$ F a h exdh ex Xagf cxe T VTj5 bixe 1/Z i gt a ;$ bixe 3+Z i gt a <$ bixe,1z i gt a =$ bixe,-3z i gt a >$ AeT a S VTa abg Uhl ab bex g Ta 3+ Z Bbj hv ZeT af f bh W UX UbhZ g gb a mx g X gbgt Vbfg: 8 g 4 g 1 g 2 g 4 g.../ > Xg IebU X be h Tg ba o o o >XV f ba OTe TU Xf5 Q, 8 Z by ZeT a Q Q- 8 Z by ZeT a R Q. 8 Z by ZeT a S HU XVg ix havg ba F a mx S 8 +)+-Q, +)+/Q - +)+-Q. =bafget agf O gt a ; VbafgeT af5.q, -Q - /Q. 1/ O gt a < VbafgeT af5 -Q,.Q -,Q. 3+ O gt a = VbafgeT af5,q, +Q - -Q.,1 O gt a > VbafgeT af5 1Q, 3Q - /Q.,-3 AeT a S VbafgeT ag5 Q. 3+ GbaaXZTg ix =bafget ag5 Q, Q - Q. + BP o KXi Xj ; XkT c Xf TaW fb ixw cebu X f Ul TaWf TaW j g FL(?kVX..1
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