ISTM206: Lecture 3 Class Notes


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1 IST06: Leture 3 Clss otes ikhil Bo nd John Frik Simple ethod. Outline Liner Progrmming so fr Stndrd Form Equlity Constrints Solutions, Etreme Points, nd Bses The Representtion Theorem Proof of the Optiml Bsis Theorem The Simple ethod Simple ethod: Bsi Chnge Converting the Stndrd Form to the Augmented Form Simple Emple Getting Bsi Fesile Solution Phse Approh Big method Determining Optimlity of Bsis. Liner Progrmming so fr Up till now in lss we hd overed the following spets of liner progrmming: Formulte liner progrms in vrious ontets. Trnsforming liner progrms into stndrd form. The intuition ehind the Simple ethod to find the est orner point fesile solution. Some of the mthemtil detils: Corner point fesile or si fesile solutions orrespond to set of n tive onstrints. Any set of tive onstrints orresponds to sis from the mtri A. The sis is set of linerly independent olumns.
2 .3 Stndrd Form Stndrd form for liner progrmming prolem is: imize... sujet to j 0, j.. The onise version is: imize s.t. A 0 where A is n m y n mtri: n vriles, m onstrints..4 Equlity Constrints In ses where the prolem ontins n equlity onstrint, it n e repled y two inequlity onstrints. For emple: n e repled y nd The seond onstrint n e negted in this se, yielding: Solutions, Etreme Points, nd Bses Key Ft: If Liner Progrm hs n optiml solution, then it hs n optiml etreme point solution. Bsi Fesile Solution (Corner Point Fesile: The vetor is n etreme point of the solution spe iff it is BFS or A, 0. If A is of full rnk then there is t lest one sis B of A. B is set of linerly independent olumns of A. B gives us si solution. If this is fesile then it is lled si fesile solution (BFS or orner point fesile (CPF.
3 .6 The Representtion Theorem The proof of the Optiml Bsis Theorem requires the Representtion Theorem. Representtion Theorem: Consider the set S { : A, 0}. Let V {V... V } e the etreme points of S. If S is nonempty, V is nonempty nd every fesile point S e written s: d k α i V i i where α i, α i 0 i nd d stisfies Ad 0, d 0 A simple eplntion of this theorem is tht every point in solution spe is omintion of the onve orners of tht spe. We disrd the unounded ses for simplifiittion in the following proof:.7 Proof of the Optiml Bsis Theorem Sy is finite optiml solution of liner progrm. The representtion theorem then sys: se d 0 d k α i V i i k T T α i V i i k T α i V i i It is known tht α i nd tht α i 0. Let V rg m T V i then T α i T V T V Therefore T T V Optiml sis theorem If liner progrm hs finite optiml solution then it hs n optiml si fesile solution. The Simple ethod The intuition ehind the Simple method is tht it heks the orner points nd gets etter solution t eh itertion. A simplified outline of the Simple method:. Find strting solution.. Test for optimlity. If optiml then stop. 3. Perform one itertion to new CPF (BFS solution. Bk to. 3
4 . Simple ethod: Bsis Chnge. One si vrile is repled y nother. This proess essentilly hooses whih vriles re tive nd then hnges the tive onstrints.. The optimlity test identifies nonsi vrile to enter the sis. The entering si vrile is inresed until one of the other si vriles eomes zero. The vrile tht rehes zero is identified using the minimum rtio test. Tht vrile deprts the sis.. Converting the Stndrd Form to the Augmented Form Doing this onversion llows you to pply the Simple method to the ugmented form nd is reltively strightforwrd. The gist of the ide is to dd slk vrile to dd slk vrile to reple eh inequlity An emple would e onverting from the following left set of equtions to the following right set: Stndrd Form to Augmented Form m i j i j..,.. 0,, sujet to m! 0 sujet to m! # A j j.. 0, sujet to m! A is n m y n mtri: n vriles, m onstrints 0 ], sujet to[ m! # $ % & ' ( I A A more onise representtion of the sme onversion is: Stndrd Form to Augmented Form m i j i j..,.. 0,, sujet to m! 0 sujet to m! # A j j.. 0, sujet to m! A is n m y n mtri: n vriles, m onstrints 0 ], sujet to[ m! # $ % & ' ( I A.3 Simple Emple The following emples re from the OR tutor CD. This is one itertion of the Simple method vi the lgeri method: 4
5 Algeri Simple ethod  Introdution To demonstrte the simple method, onsider the following liner progrmming model: et This is the model for Leo Coo's prolem presented in the demo, Grphil ethod. Tht demo desries how to find the optiml solution grphilly, s displyed on the right. Thus the optiml solution is,, nd. We will now desrie how the simple method (n lgeri proedure otins this solution lgerilly. 5
6 Algeri Simple ethod  Formultion et The Simple Formultion To solve this model, the simple method needs system of equtions insted of inequlities for the funtionl onstrints. The demo, Interprettion of Slk Vriles, desries how this system of equtions is otined y introduing nonnegtive slk vriles, nd. The resulting equivlent form of the model is The simple method egins y fousing on equtions ( nd ( ove. 6
7 Algeri Simple ethod  Initil Solution et The Initil Solution Consider the initil system of equtions ehiited ove. Equtions ( nd ( inlude two more vriles thn equtions. Therefore, two of the vriles (the nonsi vriles n e ritrrily ssigned vlue of zero in order to otin speifi solution (the si solution for the other two vriles (the si vriles. This si solution will e fesile if the vlue of eh si vrile is nonnegtive. The est of the si fesile solutions is known to e n optiml solution, so the simple method finds sequene of etter nd etter si fesile solutions until it finds the est one. To egin the simple method, hoose the slk vriles to e the si vriles, so nd re the nonsi vriles to set equl to zero. The vlues of nd now n e otined from the system of equtions. The resulting si fesile solution is,,, nd. Is this solution optiml? 7
8 Algeri Simple ethod  Cheking Optimlity et Cheking for Optimlity To test whether the solution,,, nd is optiml, we rewrite eqution (0 s Sine oth nd hve positive oeffiients, n e inresed y inresing either one of these vriles. Therefore, the urrent si fesile solution is not optiml, so we need to perform n itertion of the simple method to otin etter si fesile solution. This egins y hoosing the entering si vrile (the nonsi vrile hosen to eome si vrile for the net si fesile solution. Algeri Simple ethod  Entering Bsi Vrile et Seleting n Entering Bsi Vrile The entering si vrile is: Why? Agin rewrite eqution (0 s. The vlue of the entering si vrile will e inresed from 0. Sine hs the lrgest positive oeffiient, inresing will inrese t the fstest rte. So selet. This seletion rule tends to minimize the numer of itertions needed to reh n optiml solution. You'll see lter tht this prtiulr prolem is n eeption where this rule does not minimize the numer of itertions. 8
9 Algeri Simple ethod  Leving Bsi Vrile et Seleting Leving Bsi Vrile The entering si vrile is: The leving si vrile is: Why? Choose the si vrile tht rehes zero first s the entering si vrile ( is inresed (wth inrese. when. Wht if we inrese until? 9
10 Algeri Simple ethod  Leving Bsi Vrile et Seleting Leving Bsi Vrile The entering si vrile is: The leving si vrile is: Why? Choose the si vrile tht rehes zero first s the entering si vrile ( is inresed. when. Wht if we inrese until (wth inrese? when. However is now negtive, resulting in n infesile solution. Therefore, nnot e the leving si vrile. Algeri Simple ethod  Gussin Elimintion et Sling the Pivot Row In order to determine the new si fesile solution, we need to onvert the system of equtions into proper form from Gussin elimintion. The oeffiient of the entering si vrile ( in the eqution of the leving si vrile (eqution ( must e. The urrent vlue of this oeffiient is: Therefore, nothing needs to e done to this eqution. 0
11 Algeri Simple ethod  Gussin Elimintion et Eliminting from the Other Equtions et, we need to otin oeffiient of zero for the entering si vrile ( in every other eqution (equtions (0 nd (. The oeffiient of in eqution (0 is: 0 To otin oeffiient of 0 we need to: Add 0 times eqution ( to eqution (0. The oeffiient of in eqution ( is: 3 Therefore, to otin oeffiient of 0 we need to: Sutrt 3 times eqution ( from eqution (. Algeri Simple ethod  Cheking Optimlity et Cheking for Optimlity The new si fesile solution is,,, nd, whih yields. This ends itertion. Is the urrent solution optiml? o. Why? Rewrite eqution (0 s. Sine hs positive oeffiient, inresing from zero will inrese. So the urrent si fesile solution is not optiml. Here is the sme proess using Simple tleu:
12 Tulr Simple ethod  Introdution et To demonstrte the simple method in tulr form, onsider the following liner progrmming model: This is the sme prolem used to demonstrte the simple method in lgeri form (see the demo The Simple ethod  Algeri Form, whih yielded the optiml solution (, (0, 7, s shown to the right. Tulr Simple ethod  Initil Tleu et Using the Tleu After introduing slk vriles (,, et., the initil tleu is s shown ove. Choose the slk vriles to e si vriles, so nd re the nonsi vriles to e set to zero. The vlues of nd n now e otined from the righthnd side olumn of the simple tleu. The resulting si fesile solution is,,, nd.
13 Tulr Simple ethod  Entering Bsi Vrile et Seleting n Entering Bsi Vrile Ahe entering si vrile is: Why? This is the vrile tht hs the lrgest (in solute vlue negtive oeffiient in row 0 (the eqution (0 row. Tulr Simple ethod  Leving Bsi Vrile et Seleting Leving Bsi Vrile The entering si vrile is: The leving si vrile is: Why? Apply the minimum rtio test s shown ove. 3
14 Tulr Simple ethod  Gussin Elimintion et Sling the Pivot Row Although it is not needed in this se, the pivot row is normlly divided y the pivot numer. Tulr Simple ethod  Gussin Elimintion et Eliminting from the Other Rows Add 0 times row to row 0. Sutrt 3 times row from row. Why? To otin new vlue of zero for the oeffiient of the entering si vrile in every other row of the simple tleu. 4
15 Tulr Simple ethod  Gussin Elimintion et Eliminting from the Other Rows Add 0 times row to row 0. Sutrt 3 times row from row. Why? To otin new vlue of zero for the oeffiient of the entering si vrile in every other row of the simple tleu. Tulr Simple ethod  Cheking Optimlity et Cheking for Optimlity This ends itertion. Is the urrent si fesile solution optiml? o. Why? Beuse row 0 inludes t lest one negtive oeffiient..4 Getting Bsi Fesile Solution Given liner progrm where B is the sis of A: imize s.t. A 0 To get si solution rewrite the liner progrm in terms of the sis: 5
16 A [B] A B B B R m, R n m 0, B B is si fesile solution if it is fesile..5 Phse pproh Initilly, revie the onstrins of the originl prolem y introduing rtifiil vriles s needed to otin n ovious initil BF solution for the rtifiil prolem. Phse : The ojetive for this phse is to find BF solution for the rel prolem. To do this, inimize Z the sum of the rtifiil vriles, sujet to revised onstrins Phse : The ojetive of this phse is to find n optiml solution for the rel prolem. Sine the rtifiil vriles re not prt of the rel prolem, these vriles n now e dropped. Strting from the BF solution otined t the end of phse, use the simple method to solve the rel prolem. pg 35 : A 0 look for solution to : y 0 nd y nd 0 for solve phse prolem inimize e yfesiilities s.t. A Iy 0 y 0 if y 0 then you hve prolem euse you re missing term Solve phse A Iy 0 y 0 6
17 .6 Big method Solve originl Liner Progrm with ojetive plus lrge numer imize e y s.t. A Iy 0 y 0 Computtion worked out on pge 3 of OR.7 Determining Optimlity of Bsis One you hve si fesile solution in the following form: imize B B s.t. B B B B, 0 Algerilly mnipulting the progrm we n solve for B : B B B B ( Then rerrnging : B B BB ( X BB BB z 0 j (z j j j where: z 0 BB z j BB A j, j We n now see tht: m z 0 j (z j j j So to test optimlity we n hek: z j j 0 j 7
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