Question

4.6-1.* Consider the following problem. Maximize Z= 2x1 + 3x2, subject to x1 + 2x2 54 x1 + x2 = 3 and X120, X2 0. DI (a) Solve this problem graphically. (b) Using the Big M method, construct the complete first simplex tableau for the simplex method and identify the corresponding initial (artificial) BF solution. Also identify the initial entering basic variable and the leaving basic variable. I (c) Continue from part (b) to work through the simplex method step by step to solve the problem.

Answer 1

is Green Linear programming probleme - Max z = Quit 342 Elulb.fect to 1 +242 S4 hli + x2 = 3 x 7,0 42,0 @ (i) Treating 1st constraint as quality, we have se f286 = 4 if m=0 12 = 2 is x 20, y = 4 so we sketch 10141 Ha lalo in equelity, we have Treating and constraint as 24 faly = 0 r=0 x2=3 Th Hy so 14=3 so we sketen 24 -

心 2

Now we check the realue of objective function - at the corner point of the feasible region. 1. At (211) Max Z= 2x2 + 3x1= 7 2. At (3,0) - Maxz = 3x2 fox 3 = 6 - so optimal value of a 187 at 41=2 lg =1

- ) 18t we convert the brobleme to canonical form. iRtcom constraint is of a type, so we add a slack reanias and is an exuality, 80 we add an astificial sarias - For the big-M methol, LPP becomes a an artificial variable Al. ---- - Maxz = dritare torsi-MAL - nit 202 + s = 4 Altay +Al=3 Hol 22, Su Al2O - Simplex table is - ý 2 - co XB a o 4 1 3 na Ⓡ e si 1 -M A 0 Min Ratio > si 2-cj -M-2 -M-3 0 0 Megatine minimum to be enter & and јя и кҒ Ути v- is -N-3 minimum ration у o s leave for and column, in red column a bogig , 0 o continuing Relneul = some part Rilold) Relneul 1 - Recold) - Ri(new) B 366 CB 3 U -M XB 2 n a lei 11alo -MI! 0 si A la 0 -1/2 I M430 ron Ratio 4 I Zj-cj

2 for the 18t columnis. basis and Negative minimum y-ci is - - &t column Minimum ratio in in and now so an enters the 4 lеало— Reneul = 2 Recold) Rilnewl = Ricoid) - Recnew) B 29 -. ca 3 q ng 1 2 a 0 3 na 1 0 0 . si 1 -1 1 Zj-6 0 Мl Кош 4 —ј 770 Но афета | Blue и ig shuud X2 , , = Max z = 2x241x3 = 7 cit
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