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We have to solve the linear programming problem.
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Maximum of z is equal to twice x1 plus 5 times of x2.
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Given to us an x1 plus x2 is less than or equal to 3.
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Then x1 less than or equal to 2.
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X2 less than or equal to 2.
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Then x1 comma x2 are greater than or equal to 0.
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This is provided to us.
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We have to find x1, x2.
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Here for the first equation subject to constants, if we take x1 is 0, then it implies that x2 will be equal to 3.
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And if we are taking x2 is equal to 0, then x1 will be equal to 3.
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And from second we get that x1 is equal to 2.
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And from third we get x2 is equal to 2.
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Now by graphical method we form a graph here.
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Let this be y and x axis respectively.
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After that this is x1 axis and this is x2 axis.
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We will plot a point.
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Here is o.
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Then we will plot.
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So this is the required graph.
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So x1 is equal to 2 is a line which is parallel to x2 axis.
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Then x2 is equal to 2 is nothing but a line parallel to x1 axis.
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And this is a line passing from 2 .03 and 3 .0.
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So this is x1 plus x2 less than or equal to 3.
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And this is the feasible reason.
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So this is the shaded reason is known as a feasible reason.
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Now at these points, this as i am circling that points, at that points we will find the value of maximum of z.
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So first point is o, 0, 0.
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At this point maximum of z is equal to 0.
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Then at the point a which is equal to 2, 0.
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At that point our z is equal to 4.
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At the point b which is equal to 2, 1 that z is equal to 9.
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Then at the point c which is equal to 1, 2.
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At this point z is equal to 12.
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Then at the point d where point is 0, 2.
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At this z is given by 10.
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Therefore maximum of z is nothing but equal to 12.
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And it occurs at the point c which is equal to c of 1, 2.
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So this is the maximum point.
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Therefore this 12 is said to be the optimal value and this 1, 2 is point optimal solution.
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Now by simplex method we have to solve it.
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So by simplex method we will solve and first we will convert the given problem in the standard method.
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So maximum of z will be equal to twice x1 plus 5 x2.
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First we will write the constraints x1 plus x2.
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It was less than or equal to 3.
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Therefore we will add the slack variable s1 then it will be equal to 5.
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Then x1 plus s2 will be the slack variable.
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It is equal to 2.
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Then x2 plus s3 is the slack variable.
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It is equal to 2.
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Then it is 0 s1 plus 0 s2 plus 0 s3.
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So this is the maximum of z function.
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Where s1, s2, s3 are the slack variables.
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They are greater than or equal to 0.
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Also x1, x2 and x3 are greater than or equal to 0.
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Now first iteration is initial for giving the initial basic physical solution.
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So initial basic physical solution for this we will put x1, x2 is equal to 0.
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So we will get here s1 is equal to 3, s2 is equal to 2 and s3 is equal to 2.
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Now we will form the first iteration here.
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So first we will write here the cb.
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Then xb, after that x1, x2 variables and then s1, then s2 and s3.
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After that b and last is the theta.
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Then this will be the required table here...