00:01
Hello students today we will discuss about this question in this question we are given that the we need to iterate the the simplex tabula for this problem using the simple simple x method until optimal and then we need to find the optimal solution that is equals to question mark so here we are given maximize of 4x1 plus 2x2 plus 0 as x1 plus 0 s 2 plus ma1 plus ma2 plus ma3 that will be subjected to 3x1 plus 2x2 less than or equals to 8 minus 4x1 plus 3x2 is greater than or equals to minus 7 7 x1 plus 14 x2 is equal to 14 and x1 x2 that is greater than or equals to then here we can say that the first iteration, the first iteration that will be here b, cb, xb, x1, x2, s1, s2a1.
01:20
So therefore here we can write, here it is s1, s2a1, and here z is x equals to minus 14m.
01:28
So here 0 ,0 minus m, 8, 7, 14, zj, 3 minus, 3, minus, minus 4 ,7, minus 7m, x2 that will be 2 minus 3, 14, minus 14m, s1 that will be 1 ,000, s2 that will be 0 -1 -0 -0, and a1 that is 0 -0 -1 minus m.
01:57
Then similarly, for the second and the third iteration, that will be given is the, we will write the third iteration, that will be here, s1 s2 x1 z is equal to 8 here it is c b x b x1 x2 s 1 s 2 so therefore here c b c b 2 15 2 z j and here z j minus c j 0 140 minus 4 minus 4 5 2 8 6 then s 1 1 0 0 -0 -0 and 0 -1 -0...