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LPP Book

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658 views23 pages

LPP Book

Uploaded by

Sujin Prajapati
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Ox+4y < 108 x+y $24 x20 y20 210 rie Solution of a Linear Programming Problem with wo Decision Variables ‘Step1. For an appli Jblem, summarize relevant material in table the problem: and write a finear objective sing linear’ inequalities and/or the coordinates of each co ing the value of the objective function corner point. Steps. Det Step6. For terms of the original problem. vaivear programming problem is one hati concerned With TOA a esta eau (maximum or minimum valu) of 2 Tinea olecve ion ofthe form me K+ CaR + = + CaKe ‘where the decision varia constraints in the form of ‘addition, the decision v constraints x; 20, 1= 1, 2 fproblem constraints and the nonnegative reese region for the problem. Any point inthe Jroduge the optimal value ofthe objective funtion over the feasible Tejon is called an optimal solution. ‘Theorem: 1 Fundamental Theorem of Linear Programming-Version 1 ‘Theorem: 2 Existence of Optimal Solutions Problem constraints ‘Nonnegative constraints xj. X22 0 Nonnegative constraints requires 1 labor-hour from the cut from the assembly department. Each expedi inequality constraints erapli feasible region for product , how many tents of eac! day to maximize the total day tents can be sold)? Labor-hours per tent jon model Cutting 2 department Assembly | 3 4 84 department Profit per tent —_ [30 212 Value of P Rs.0 Rs. 1480 (optimal Rs.1400. can be determined from the {xy = [2 and X= 10, then P= 50 x 12 + 80x 10 0 per day, which is maximum. point (20, 6) is called an optimal solution to the problem, becau ‘maximizes the objective function and is in the feasible regi ‘general, it appears that a maximum profit occurs at one of the points. Exercise [5.2| Solve the linear programming problems graphically: 1.” Maximize P = 5x; + 5x3 & Subject to fs: pases 30 It 2x +x 510] 20 we Be cee aaa nel subject to 20) Avs: Ming ily 4424 Be Rank. 3. Maximize P= 3om +o. subject 10 2 260 dare ) a 25 2a and x, satisfy the system variables are nonnegative. the computer use, the method is ru 1g hundreds and even thousand sy and arbitrary values. Certain subject and some use solutions, are related. (0 the simplex. method boundary lines ofthe feasible regio details ofthe process. %. SE rctt : panes L AK + aK: +o +Aphy Sb DE ‘With nonnegative constraints, s he jote: Also note that the Bg hn ne coefficients of the objective function can be i a FIGURE How are basic solutions to sysiem (2) determined”, Sse @) jnvolves four variables and two equations i ‘bles into two groups, called the basic var as follows: Basic variables are selected estrction that there be as many basic variables as ‘The remaining variables are nonbasic variables, ‘ince system (2) has two equations, we can select any of the four sine ras asic variables. The remaining two variables Of then Natpasic variables. A solution found by setting, the 10 rnonbasic nonbles equal to 0 and solving for the basic variables. ‘asic lack variable. To convert a system of problem there are equations, ENED 3x1 + 4x, $84] Into a system of equations, we add variables s, and s, to the left f 2 10 the sides of (1) to obtain HA2yty — =32 ay eAvinere elas ow vation, qNote that sting two variables equal to 0in a sch of two : ‘tations, with two. variables, whch as exactly one solution, ‘nfinitely many solutions, or no solution.) ‘The variabl a i ‘5; and S) are called slack variables because each sate citference (aks up the slack) between the eft an right Of the inequalities in (1). Note that if the decision variables x, qi | Example: 1 nd two basi Satins for system (2) by fiat selecting 2408 $1 asic variables, and then by selecting X, amd.S2- A 216 bi 27 ON Programming each basi solution found in part (A witha inte Pan tthe evened) unary Une ft ee _ Migr 2 and inate which boundary Tins prodce DB wi at tas ta a ps yn part (i) are Given a system. si lem (ch sytem wllalvays have and Figure 2 ive us all the answers. We have ions in Table I, It is convenient to organize terms ofthe zero values of the nonbasic variables Table 1 Intersection Iniewseating aes a variables are called nonbasic variables Digi saaiaccatcakoe way e ‘A solution found by setting the nonbasic variables equal to 0 and X00) 50 Solving for the basic variables is called a basie solution, If basic Solution fas no negative values, its «basic feasible solution, il 1 Si ‘The following theorem, which is equivalent to the fundamental theorem (Theorem | in the preceding section) i stated without proof; Qin AG: 0 20 AIG | eta val xy +2n, =22 9 2 +0 0 pow *~? 3x,+4r,=84f NO BO gta EGA B=? . ‘Theorem: 1 xj +24 =32 LT “AR PI PROBLEM - VERSION 2 If the optimal value of the objective function in a linear programming problem exists, then that value must occur at one (or more) of the basic feasible solutions. Now we can understand why the concepts of basic and nonbasic variables and basic solutions and basic feasible solutions are so important - these concepts are central to the process of finding ‘optimal solutions to linear programming problems. Bir Oigedys 0 Coko. =P ax +4,-esf YS 3x, 44x =84] 2 6 9 0 BQD6) oes Exercise |5.3) 1, Listed in the table below are all the basic solutions forthe system Attar =32 Sytdey t= ay For each basic solution, identify the nonbasic variables and the basic ‘variables. Then classify each basi solution as feasible or not feasible. Feasible? Yes Yes No No Advanced Engineering Mathematics IL sar Programming models by have been included in the Linear Programming model. In order to solve an Linear Programming, ‘model, we must put it into standard form. Remember to make the right hhand side 20. The presentation developed here emphasizes concept important that we become proficient in constructing the m« Linear Programming Problem using the simplex method. Initial System: Given, Maximize P = 50x, + 80x; Objective function Subject to x, +2x, $32 Problem constraints... (D Ep | at ane 1.422 O nonnegative constraints Introducing slack variables , ands, , we have the following system of problem constraint equations: stint 32 Injtdyy +8, = 84] Xp X3,81582 20 |As part of the simplex method we add the objective function equation P=50x, +80x, in the form —S0x, ~80x, +P =0 to system (2) to create what is called the initial system: 42m +54 =33 ae a) =505,-80%, +P=0 21 ‘The objective function variable P is always selected as a basic variable and is never selected as a nonbasic variable. Note that a basic solution of system (3) is also a basic solution of system (2) after P is deleted, ion of system (3) is a basic feasible solution of system is called a basic solution of system (3) can contain a negative number, the value of P, the objective function variable, imental Theorem of Linear Programming ~Version3 f the optimal value of the objective function in linear programming problem exists, then that value must occur at one (or more) of the basic feasible solutions of the initial system. ‘nonbasic variables. These numbers do not change during the simplex process. Step: 2) Selecting Basic Variables. A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has in the column of ic variable. (This procedure always selects P as a basic variable, since the P column never changes during the simplex process.) ‘Step: 3} Selecting Nonbasic Variables. After the basic variables are selected in step 2, the remaining variables are selected as the nonbasic variables, (The tableau columns under the nonbasic variables usually contain more than one nonzero element.) Selecting the Pivot Element (p-element) ‘Step: 1) Locate the most negative indicator in the bottom row of the tableau to the left of the P column (the negative number with the 222 s( wbyolute value), The cokinn containing ‘column (p-colurn), we element (n the pivot column above the ‘element in the last column. The on, and we stop. pivot element is the element in the intersection of the Performing a Pivot Operation 'A pivot operation, or pivoting, consists of performing row operations as follows: Step: 1) Multiply the pivot row by Ul to transform the pivot element int 1, omit this step.) Step: 2) Add multiples of the pivot row to other rows inthe tableau to transform all other nonzero elements inthe p-column into O's foe: Rows are not to be interchanged while performing a pivot isnot a | already) is for the p-row to be multiplied by the reciprocal of the p-element.) Performing a p-operation has the following effects: 223 i ‘suuyensuoo anmedauuon, fers tz +! sruensuoo ueygont | 15+" lors tx + 'xz ‘wonduny (gO & OF + 'xOE = d PIN qurensuos aanedouuon, 3K" iets eee yuyensuoo w : supensu00 wOFGOHE YO te ez 7 sonouny aansolgg “%x+Ix0C= A Sze ors e+ stan poxpou xojduts a) Butsn waqqoxd Susur oxd eau | fyrs] 9S19FOxa Z ogres endo ou 5 2 222 da am 0g “more © oft 0 ayqean amt mh “nee __ Fae a So My ge unjoo-d am | SUBUBE OG FNS twuinyoo-d Of. Lit c0ws0 e apo to € 6 Oso Onan: Loven -sojqeuon Suntx9 pu usa say aup Asap pure “neon Xai 9X) AL hx ee ag N= + GE NE szz suoysAs yen axp 21m pues pu Is sajqetsea Noes Somposy Wop MOS ° ‘sjuyennsuos aanuouuon. enmee zis txe+ tx suupesysuo> wqa 3 A ot (eee, er wopounjaspoofgg E+ x9 =q oz WROTN :powpout xopdumsayp Bsn 9405 o=#s'0= = pur dors om [osrr olo 1c o tly s]o BE lo ee sjuaws[9 oad mou # Hupuy Aq ss09030 wadar am “mou sey amp ut neou w [INS St aI9q) BUNS fosct|1 0 or 0 ora alo 1 z 0 a1] o 0 t ic ite 1 oz an br 14 aa 0 a oz 4 +p xunew ayy io 2z da vy jo osodsuen ayn , 7 Enews amp wo 37 dS oe og = Sx + Txt Hop tee ea eter z+ x4 xoy sagt 4X7 s ov patans Scop+ ZL-+ OF = 9 azIUTUEIN fauayqoud ep oxp mx0 qordarexg enna) siutensuoo waygord 2 avs wojgosd wonezyuupxear & WI0} 0} , Y JO SOH OP 3SN. vy yo asodsuen ayy" y xUIEUL edais zdas +1 dais ‘ya wwoyqoud wojwayuayuy ¥ UOAIC) :wiojqoud tonic 04) Jo WHEW A no wy 10 vod) Ov) OK, sHUyHINSLOD can edna sent 99/809 [I] 9% NON, 2 0470 NUON HH] GOHA HA ORHAN Hon wz hay +S WI =LO=*0R= SO= NOP=N OSHS ee ‘3s1x9 uonnjos eumdo oN 0 eee som x pmee= Iie Q9C=axNE O="xpueg=hwosi=amey -Z cr I x1O8=d CW, yee be Te FE Sie te he pre. an Og Salta oy > em Gth “eee y Exege te0t + On (0) SEH Mee be HF pert + HE + ayy = JRO. noo poche Et benew te inyoad szzuvew 1 patueyd 2q pinoys doro yora yo saise Kuen mov 3 dor> uo aio Jad (002$ pur “a dors uo aine sad goes “y dox> uo 2320 30d QQT$ Jo moxd oyeu ue sue) axp JT “9}qEIEAR SKepYOM O9T JO WUE e uv asamp pur *ftonnvadsor ‘2198 Jad s44py. “a 'V sdox5 ‘p22s wo sods 2q weo oge"e$ Jo whew ¥ “fonncedsox ‘BF od -O¢$ pu “0Z$ ‘Os 9809 D “a *V sdox> 40} poos auL, ssdoto san wou ve wed ot upd pu use) 2308 OQ] & sumo sou) V ‘swuensuoo axnefouuoy 9 2 #x 1x “ae $5 byte S =he2~ — suyensuoo wargorg {715 Exe + bx on nofgns Gre ten iy 73 tre be ny anal hz + = g aE ‘suressuoo aaneouu0y, ozeettx 95% re stupensuos waiqorg ) ¢5 tx tx of yoafqns es t+ eg wonouny annsafgo — txe + Ixz = g aztunxeyy eae ential es Faruneagong aoa thyz - os- eae + tks hee TONE TENG SH TOY WTS TENT surayqoud renp,axp uy saqqeurea sours aun se woygoxd yeuyB0 oun poner beg ‘Ty Sojqeumea up asm jim am “sare] zwapD aU1090q o t sy Ms 9 iz on pafqns PK yz + Kg =a S7IUeN ‘wajqosd penp amp aieig. +g dag d= Kiz+'Kos sp = *KE+ MS 91 = aM oz wx Wz xE+Ix OS = tg + ixz cucpes ©} ;0fqns, sh 91 =9 oan Ten amp Buyzuapeons R, @R+RR, WO 1 5 2 0} 10 Plo 0 13 4 1 [400 ‘Since all the indicators in the bottom row are nonnegative, the solution to the dual problem is yi =3, atP = 420 Examining the bottom row final simplex tableau, we see the al solution to the minimization problem that we obtained always can be obtained from the botiom row of the final simplex able forthe dual problem, Advanced Engineering Mathomatles IML Example: 3 Solve the following minimization problem by maximizing the dual: Minimize C= Sxy + 10x: < + = 2y2 No e the dual problem: 5.__ Linear Pro 5 rive Yo R+Ri OR) ak od <1'o 2Ry + Ry > Ry pivot column. the pivot column above the 1. vot row. We stop the pivot Problem has no optima the orginal minimization Solve the following minimization problems by maximizing the dual i ize = 21m +502 a ann: 2 12 YS ttte 217 8 E ruts 20 2, Minimize C= 9x; + 2k: Yo tue axon 2 12 = x? > wan uP 215 FBS Bt. 29,4928? 5) oO 43, eh 2 20. ize C= 11x, + 4x2 4) Ve subject to 2xj +x. 28 HI $3 24 fb 7 xX 20 vw %rO Min C: do Maye BYRoH, mek plants to the distribution outlets. What is the minimum cost? Advanced Engineer Minimize C = 7x) + 9x2 ‘subject to. eeipruaed oho =n txr 26 Nee ; ayn 24 aS romeo 7% 6, Minimize €= 3x, + 9x2 ee ar MinCs2y Dyn 28 2f wt We uf rT xi#2u 28 7 Ky Bh oe Manufacturing- production scheduling. A food processing company produces regular and deluxe. ice cream at three plants. Pet hour of in Cedarburg produces 20 gallons of regular ice and. luces 20 gallons of regular ice jest Bend pt gallons of deluxe ice cream. It costs $70 per hour to operate the Cedarburg plant, $75 pet hour to operate the Grafton plant, and $90 per hour to operate the West Bend plant. (A) The company needs at least 300 gallons of regular ice cream and at least 200 gallons of deluxe ice cream each day. How'many hours per day should each plant be scheduled to oper the required amounts of ice cream and production? What is the minimum production ‘A Transportation Problem. A computer manufacturing company has two assembly plants, plant A and plant B, and two distribution outlets, outlet I and outlet Il, Plant A can assemble at most 700 computers a month, and plant B can assemble at most 900 computers a month. Ciutlet I must have at least 500 computers a month and Outlet IT must have 1000 computers a month, Transportation costs for shippit ‘computer from each plat are as follows; $6 from plant 5 from $4 from plant B to outlet ‘$8 from plant B to outlet I Find a shipping schedule that will rminimize the total cost of shipping the computers from the assembly goa oand x)= 13 Athin’s OF wont .6. Maximization and Minimization with Problem Constraints: The Big M Method: We introduce the big M method through a simple maximization problem with mixed problem constraints. The key parts of the method are summarized as follows: Initial Simplex Tableau Requirements: Fora system tableau to be considered an inital simplex wbleau, must satisfy the following two requirements: ‘The requisite number of basic variables must be selectable by the process described in 5-4. That is, a variable can be selected as a basic variable only if it corresponds to a column in the ee ic solution found by setting the nonbasic variables equal asibe. ‘rag Me n@Pnod— Insrodscing Sack, Surplus, and Variables to Form the Modified Problem ‘Step Li Many problem constraints have negative constants on the ‘both sides by «1 to obtain a const ‘The Big M Method - Solving the Problem ‘Step I: Form the preliminary simplex tableau for the modified If any artificial variables are nonzero in the solution to the modified problem, then the origi problem has no optimal solution. 237 Advanced Engineering Mathematics IL We now ple the key steps of the bie M method and use them to solve pide anh xian fa lxg = azqumpE Seqqeuea o1seq uou ayy Sumas Xq puny ures ayy ul ou axe Kayp pure suumnjoo 2194p cup so) "Fels. axe soygeuren otseq omy exp 295 a4 NOU SI WOK a palletes oer) zs . 1 al ro] O} fe "OW KE +S + sy wojgoud potgpow oy fay — Pe Nex + 8x4 T= -uonouny aanefgo oun 01 Sepy pure Tey “Heyy — pe ars “AqTeuL Sxpt fx hz T= tess Sige Expt se Weete- G4 tet suv ayy Suneuruma £q nae payypous ayn sop nvayges xoqduns Aseurun ie ee o=d+ N+ ‘xoq 2qp Uy parers s9jna 041 1 Sups0202 ne lee ts— sajquuzea jeroypme pur ‘snjdins “yous ayn aonponuy am “IxON, Sz Stte—le or= UUDULL 26 01 g- a8ueyD o1 |- Aq qurEsISUOD puodes 2xH 4 c= sag— tat x= LS Snel ovpotqns | Sep 4 ag + ag = g ox UNO Supuureszord svouy Suysonoy 3Mp “onqweou are Q1+A- be ¥ ( ° ove pur OHH ey @8se] os J owiMssy -suonesado ioAId UIaq MON sip ‘o = 'e sous W O1+W: OF OG a4 Is & & & ‘uumyoo 'y omy ut py SreUIUM}S MON, fino ome ; root? oot tort ae fatale) ds wis &% w & ate ae na [clin +o o ela oF a+ ‘w+ Sxor+ Sxoe + Ixog zw] oO © rF oO tf r]® ea i: een r]o = ZF 9 QIx ge fos ete ate gle L Trt ah au xk suraygoud payypou 2x) ares a4 94 AOS OY,“ XEN? =O WIN PU oz% ors aches hz soquauiny] “suoHBIedo 104i Tm S1OpSRU, "Four ar nw|gE x3] a (onnedou aq ued g “aquawa) P| -M30-ma0-MH0M 0 0 1 om | CDR, +R, > R, (M-10)R, +R; > Ry a ee Os One 1p USO eal [20 -osmi2s 0M Ri R, XR m8 ufo @ 0 2 ma Os lo P| 20 -0sm2s50 (O5R, +R, Ry (0.5M~25)R, +R3 +R Rio ol ‘The bottom row has no hegative the modified problem is P=—100atx)39,%2=2,5=4,51=0,a=0,5)=0. A n=O icators, so the optimal solution for Since ay % 0, deleting ay produces the maximization problem jig M method to solve the following ize and Minimize P = 2x) = Xp 212 =6 20 Minimize C = ~Sxy ~ 12x2 + 16x3 subject 10 xy 42x +x; < 10 en = + Paget 40 6 wa t2x; <8 2xj+%- 2x = 0 20 ‘Manufacturing resource allocation: ‘An elefltonics company manufactures two types of add-on memory modules for microcomputers, a 16k module and a 64k module, Each 16k module requires 10 minutes for assembly and 2 minutes for testing, Each 64k module requires 15 minutes for assembly and 4 ‘minutes for testing. The company makes a profit of $18 on each 16k ‘module and $30 on each 64k module. The assembly department can ‘work a maximum of 2,200 minutes per day, and the testing ‘ean work a maximum of 500 minutes a day. In order to feurrent orders, the company must produce at least $0 of the I per day. How many units of each module should the ‘company manufacture each day in order to maximize the daily profit? ‘What isthe maximum profi?” goo 243, y sve re Lt + te gee tie Qs dt - Wg word v= e+ tet c= wees + + or = se + i 4c) matt wan teh . . ‘urmgo am ‘Burppe pue 410} (2) uy suonenbs pup pur puoses aap Furajog :wonouny 2m “powgaut mau Spf wonuny a4n29°90 3 paysydoaoe am ‘pounaus Wi 31g atp uy (1) oF Yornjos ayqiseay 218% 24) STF9p UeD am aK “OraZ o} fonba Te puy pur sojqeuea jeune x a = —_ Lave x TUFMOHOH ap UrEIGO osedwios 10y powour smAWOST SH

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