Consider the following LP: Maximize z = -3x1

Consider the following LP: Maximize z = -3x1 - 2x2 Subject to -x1 + x2 ? 1 6x1 + 4x2 ? 24 x2 ? 2 x1, x2 ? 0 (a) Convert the LP given above to the standard form. Determine all the basic feasible solutions (bfs) of the problem. Give the values of both basic and non-basic variables in each bfs. (b) Identify the adjacent basic feasible solutions of each extreme point of the feasible region. (c) Using the graphical solution technique, solve the problem. (d) Which constraint(s) is(are) active (binding) in the optimal solution? Which constraint(s) is(are) redundant? (e) Using the big-M simplex method, find the optimal solution. (f) Using the method of transformation technique, write the dual of the problem. (g) Write directly the dual of the original problem, without using the method of transformation.

Transcript

00:03 In this problem we have given our maximum z which is equals to minus x1 plus 2 x q plus 3 x4.

00:13 So the sub 2 x1 minus x2 plus 2 x3 plus s1 is equal to 8 and we have 4 x1 plus 2x2 plus 7 x3 plus 9xy plus s2 is equal to 8 and we have 4 x1 plus 2x2 plus 7 x3 plus 9 x y plus s 2 is 2 is 2 is 2.

00:34 Is equal to 30 also we have 2x1 plus 3 x q plus 7 x 4 plus s 3 which is equal to 20 okay and we have a 3x1 plus x2 minus 3x3 plus 4 x4 which is equal to 1 also now the trick is here that are x1 x2 x3 and x4 is greater than 3.

01:05 0 okay so this will be minus x1 2x cube plus 3 x cube plus 0 s plus os 2 s 2 plus os 3 okay and we'll have to find the values of s 1 s 2 s 3 also here okay but before that we have our x1 minus x2 plus 2 x2 plus 2 x cube plus 0 x4 plus s 1 plus 0 s 2 plus 0 s 2 which is equal to 8 and we have is in the second equation we have 4x1 2x2 7x cube 9x4 plus os 1 now for the s 2 o s 2 e which is equal to 30 and for the 3 case we have 2x1 plus o x2 here and 3x cube plus 7x raised to power 4 os 1 o s 1 plus o x 2 and now for the 3 which will be equal to 20 that means here are 3x1 plus x2 minus 3 x q plus 4x4 plus os 1 plus os 1 plus os 2 is equal to what it is equal to 1 this is what we have already given in the question okay now what we need to do is we need to form initial table and follow it by iteration okay now let's make an initial table for this problem.

02:41 So for the initial table, remember the formula, form x, z, which is equal to cj minus z j which is this also.

03:02 Okay.

03:02 Now let's come.

03:04 We have our c b of i.

03:07 We have our c of j.

03:08 Right? let me make a table here.

03:20 It should be in this form.

03:25 Right? here will be our solutions and here will be our ratios.

03:31 And this will be the complete table then okay so the line will be here also between this so let's now start that is our b to be right minus 1 0 plus 2 plus 3 0 and 0 this is our for the solutions and this is for the ratios okay now see we have three situations 0, 0 and 0.

04:06 That means our s1, our s2 here and our s3 is here.

04:11 Okay.

04:12 And in this case we have our x1 this, x2 we have this, x3, x4 and this is our s1, s2 and s3.

04:22 Okay.

04:23 So from here we have is 1, 4 and this is we have 2.

04:28 And in this case we have minus 1, 2 and 0, right? and in this case we have minus 1, 2 and 0, right? and in this this case we have two seven three in this case we have in the export we have a zero nine seven okay and in this case we have is one zero zero zero zero one here zero and again zero one okay so from here this to this this this we will get eight from adding all of these we will get here what thirty and from this we will get twenty and from this we will get 20 and the ratio of this will begin 8 by 0 is equals to what infinity no this is wrong we can't take it it will be 30 the 30 by 9 will be what it will be 3 .33 this we can take the solutions as and we have 20 so it will become 7 by 20 which is equals to 30 0 .35 this is also can be taken okay now in this case we have our z j and we will find our cj minus z j j j also as for the given problem so it will become let's see so from here we will get the equation the series we have is this one right and we have is the series x4 0 to 9 and 7 this is 0 97 right so that means we have this is we have 0 0 0 0 0 0 0 0 0 0 0 this is we have 0 0 0 0 0 0 0 0 0 0 0 this is and 0.

06:19 From here we will get 0 0 0.

06:22 This will be 3 to 0 and minus 1.

06:26 Right...

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