Pratique du Shadowing: Lec-6 Simplex Method | Maximization Problem | Mathematical Example Solution | Operations Research - Apprendre l'anglais à l'oral avec la vidéo

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Welcome to Academic Game Tutorials.

In this Operations Research Lecture Series, the topic for today's lecture is simplex method.

In the previous lecture we have seen how to convert a linear programming problem from general form to standard form.

So, here we have a linear programming problem and in today's lecture we will solve this question using simplex method.

So, this linear programming

Problem is given to us in general form.

To solve this using simplex method we have to convert it into standard form at first.

Now, at step number 1, by introducing slack variables s1, s2 and s3, we will convert the problem into standard form.

As we can see here, this is a maximization problem.

So, we don't have to change a thing in the objective function.

So, we will write the objective function as it is written here.

The objective function is, maximize, z equals to 3x1, plus 2x2, plus 5x3.

Now, let's write the constraints.

Here we have the constraints.

As we can see here, the decision variables x1, x2, and x3 are all greater than or equal to 0.

So, we don't have to change the decision variables in the constraints.

If there were any unrestricted decision variables then we would have to rewrite the constraint in equations.

Now, we will write the constraints as equations by introducing slack variables.

So, for the first equation, we will write subject to x1 plus 2x2 plus x3,

then leave some space and write equals to 430.

For the second equation, we will write 3x1 plus 2x3, then leave some space and write equals to 460.

For the third equation, we will write x1 plus 4x2 equals to 420.

Now, on the first case, here we can see less than or equal to sine,

that means, this left hand side is smaller than the right hand side, and to make both the sides equal,

we have to add slack variable to the left hand side or the smaller side,

so, on the left hand side we added the slack variable plus s1.

If there was a greater than or equal to sign, then we would have subtracted a surplus variable from the left hand side.

Similarly, on this case also we can see less than or equal to sign.

So, on the left hand side or the smaller side we added the slack variable, plus s2.

Now, on this third case also, here we can see less than or equal to sign.

So, on the left hand side we added the slack variable, plus s3.

Now, since we have introduced the slack variables s1, s2 and s3 in our equations, we also have to write s1,

s2 and s3 in the objective function over here.

So in the objective function we add 0 s1 plus 0 s2 plus 0 s3 as we know

that the cost of slack variable is 0, we have written 0 before them.

And, below the constraint equations we will write,

x1, x2, x3, s1, s2, s3 greater than or equal to 0.

So, we have converted the problem from general form to standard form at first.

Now, before solving this problem by simplex method, we need to remember some rules or steps for simplex method

which is also known as simplex algorithm this is nothing

but the steps to solve the problem using simplex method as

we can see here the first step was to convert the

given general linear programming problem into a standard linear programming problem form

so we have already converted the problem from general form to standard form using these four steps

we have explained these four steps of converting from general form to standard form in our previous lecture.

So, our first step is done.

Now, the second step in the simplex method is, write the values of the initial basic feasible solution.

So, for the second step we will find the initial basic feasible solution.

If we see here, we have six decision variables now.

The decision variables x1, x2 and x3 were given in the problem

and we introduced the other three variables s1 s2

and s3 the value of the decision variables

that were given in the problem will be 0 so we will write x1 equals to x2 equals to x3

equals to 0 then we have to find out the values of s1 s2 and s3

now in these constraints we can see if the values of x1 x2

and x3 is 0

then from the first one the value of s1 will be

430 from the second one the value of s2 will be 460

and from the third one the value of s3 will be 420.

So, our second step is done.

Now, the third step in the simplex method is, write the standard form LPP into matrix form.

So, in step 3, we have to write the standard linear programming problem into matrix form,

which is, A into x equals to B. So, A into x equals to B is the matrix form

and we have to write the standard linear programming problem into this A into X equals to B form.

Now if we take a look at the constraints here, the cost of X1 is 1,

the cost of X2 is 2 as we can see the cost is written in front of X2.

Similarly, the cost of X1 over here is 3.

So, in the matrix form, the letter A represents the cost of these six variables.

Again, the letter X in the matrix form represents these 6 variables.

And the letter B in the matrix form represents the values we have here in the right hand side.

So, let's write this standard form LPP into matrix form.

In the headings, at first here we have written all the variables X1, X2, X3, S1, S2 and S3.

Now, we will put the cost or the values of these variables from this standard LPP.

Here we can see that, for the first constraint equation, the value of x1 is 1, the value of x2 is 2,

the value of x3 is 1, and the value of s1 is 1.

So, we put the values 1, 2, 1 and 1 over here.

Since, the first constraint equation does not have S2 and S3, so, the value of S2 and S3 is 0.

Similarly, for the second constraint equation, the value of X1 is 3, there is no X2 here.

So, value of X2 is 0, the value of X3 is 2, there is no S1 here.

So, value of s1 is 0, the value of s2 is 1, and, since there is no s3 here, so, value of s3 is 0.

And, for the third equation, the value of x1 is 1, the value of x2 is 4, there is no x3, so,

the value of x3 is 0, there is no s1, and s2 here, so, value of s1, and s2 is also 0,

and, the value of s3 is 1.

So, we have written the values of A in the matrix form.

Now, for the values of X in the matrix form, we will write all the variables over here as the value of X.

So, we have written all the variables X1, X2, X3, S1, S2, and S3 as the value of X.

And, finally on the right hand side,

we will write these values from the equations 430, 460 and 420.

These are the values of B in the matrix form.

So, in this third step, we have written the standard LPP into the matrix form which is AX equals to B.

Now, in step number 4, we have to construct initial simplex table.

So, let's see how to make a simplex table.

Here, we can see an initial simplex table.

You have to make an initial simplex table exactly like this one.

And remember to leave some space on the right hand side of the table.

We will use this space later.

Now, the first row on the table is for cj.

To fill up this row for cj, we will take a look at the objective function.

Here, in the objective function we have 6 variables, x1, x2, x3, s1, s2, and s3.

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And in front of each variable there is the cost of these variables, 3, 2, 5, 0, 0, and 0.

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We will write the cost of these variables from objective function, to fill up the row for cj over here.

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So we have written the costs 3 for x1 2 for x2 5 for x3 0 for s1 s2 and s3

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So this row has been filled

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Now we will fill up the columns the column B over here is the column for basic variables

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When we will make the first initial table

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The slack or surplus variables that we introduced into the problem earlier will be written in this column B.

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So, here we can see that we have three slack variables.

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So, in the column B we will write the three slack variables, S1, S2 and S3.

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And, the column CB over here is for the cost of these basic variables in column B.

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So, the cost of S1 is zero, the cost of S2 and S3 are also zero.

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Since this is the initial simplex table, the basic variables we write in this column B are all slack variables.

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And we already know that, the cost of slack and surplus variables is 0.

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So, the value of CB will be 0 at first for all variables in the initial simplex table.

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Now, in the column XB over here, we will write the values from the right hand side of the matrix form,

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430, 460, and 420.

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are the values on the right hand side.

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If you haven't made the matrix yet, you can take the values from right hand side of these equations also.

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So in the column xp we write the values from right hand side of the matrix 430,

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460 and 420.

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Now we have to fill up this region.

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We will put the values of the variables for these rows directly from the matrix over here.

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In the first row we have 1, 2, 1, 1, 0, and 0.

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In the second row we have 3, 0, 2, 0, 1, and 0.

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And, in the last row we have 1, 4, 0, 0, 0, and 1.

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The initial simplex table is now ready.

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So, in this fourth step, we have created the initial simplex table.

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Now, the step number 5 is, calculate the values of Zj minus Cj and check the basic feasible solution for optimality.

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So in this fifth step, we will at first calculate the values of Zj minus Cj using this formula and following these steps.

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So, let's find the value of Zj minus Cj at first.

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Now in the initial simplex table, we have a row down here.

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This is the row for Zj minus Cj.

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For the calculation of Zj minus Cj,

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the general formula is Zj minus Cj equals to CB into Xj minus Cj.

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So, this is a general formula and based on this general formula, for the calculation of Z1 minus C1,

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the formula will be Z1 minus C1 equals to CB into X1 minus C1.

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Here, since this is the initial simplex table, the basic variables we write in this column b are all slack variables.

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And we already know that, the cost of slack or surplus variables is 0.

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So the value of CB will be 0 at first for all variables in the initial simplex table.

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Now, we have CB into x1.

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The value of CB is 0, 0, and 0.

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And the value of x1 is 1, 3, and 1.

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So the value of CB into X1 will be 0 into 1 plus 0 into 3 plus 0 into 1.

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Then give the minus sign.

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And in the CJ row, we can see the value of C1 is 3.

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So we write 3 here.

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And finally, the value of Z1 minus C1 will be minus 3.

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So in this row, here we will write minus 3.

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Now, in the same way we will find out value of Z2 minus C2.

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As we said earlier, since this is the initial simplex table, the basic variables we write in this column B are all slack variables.

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So the value of CB will be zero at first for all variables.

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So in the initial simplex table, the value of ZJ minus CJ usually depends on CJ only,

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and the value of ZJ minus CJ will be minus CJ.

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So, we will find out value of Z2 minus C2.

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The formula will be CB into X2 minus C2.

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Here we will have 0 into 2 plus 0 into 0 plus 0 into 4 minus the value of C2 is 2.

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And finally, the value of Z2 minus C2 will be minus 2.

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So, in this row, here we will write, minus 2.

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Similarly, we can calculate the value for Z3 minus C3 will be, minus 5.

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The value for Z4 minus C4 will be, 0.

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And these two values will be also 0.

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So, we have found out the values for ZJ minus CJ using this formula.

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Now, after finding the values, the first condition is, if all the values of ZJ minus CJ are greater than or equal to zero,

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then the optimal solution will be obtained.

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So, if the values of ZJ minus CJ that we found out are either greater than zero or equal to zero,

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then we will obtain the optimal solution.

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Now, let's check over here, the values of ZJ minus CJ are not greater than or equal to zero, rather we have negative values here,

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minus 3, minus 2, minus 5, so, all the values are not positive or equal to zero, thus the first condition is not fulfilled,

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and we will not have the optimal solution here.

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So, let's move to the second condition, the second condition says,

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if at least one value of ZJ minus CJ is negative then indicate by an arrow

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and that column will be called key column.

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If we see in the table we have three negative values of ZJ minus CJ

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if there was only one negative value then we would indicate

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that negative value by an arrow and that column would be called key column.

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Since we have more than one negative value the second condition is also not applicable we will move to the third condition.

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The third Third condition says, if more than one or multiple values of ZJ minus CJ are negative, then choose the most negative value of them,

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then, indicate that most negative value by an arrow, and that column will be called key column.

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If we see in the table, we have three negative values of ZJ minus CJ, minus 3, minus 2, minus 5,

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the most negative value among these three values is, minus 5, we will now indicate this most negative value by an arrow

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and this corresponding column containing this arrow indicated value will be called key column.

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So this column X3 is the corresponding column to this value which we indicated with an arrow.

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This column of X3 with the values 1, 2 and 0 will be called the key column.

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And X3 will be called incoming vector.

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So in step number 5, we have found out the values of ZJ minus CJ.

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Then, based on these three conditions we found out the key column and the incoming vector.

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Now, in step number 6, we have to calculate the minimum ratio.

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Here we have the formula for calculating minimum ratio equals to Xp divided by Ck.

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Here, Ck is the key column and the value of Ck has to be greater than zero.

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So, let's find out the minimum ratios.

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Now, on the right hand side of the initial simplex table we had left some space.

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We will use that space for calculation of minimum ratio.

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When we made this initial simplex table, we did not make these column for minimum ratio at first,

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because after finding these values of ZJ minus CJ, if we found out that all the values are greater than or equal to 0,

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Then we would not have to calculate minimum ratio and we would not make the minimum ratio column.

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But here we have negative values of ZJ minus CJ.

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So we need to calculate the minimum ratio.

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Thus we made the minimum ratio column now.

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So the formula of minimum ratio is XP divided by key column.

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Here the key column is this X3 column.

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So we write the formula minimum ratio equals to XP divided by X3.

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Here, x3 is the key column and the value of x3 has to be always greater than 0.

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If the value of CK is equal to 0 or less than 0, then we will not find the minimum ratio for that row.

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Now for the first row, value of XP is 430 and the value of x3 is 1.

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So, the minimum ratio will be 430 divided by 1 equals to 430.

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For the second row, value of xp is 460 and the value of x3 is 2.

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So, the minimum ratio will be 460 divided by 2 equals to 230.

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And, for the last row, value of xp is 420 and the value of x3 is 0.

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Now, since the value of x3 or the key column is 0 in this case, we will not calculate the minimum ratio for this row

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if we divided 420 by 0 it would ultimately give an undefined result

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so we will not calculate the minimum ratio for this row

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because x3 has to be greater than 0 now among 430 and 230

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The minimum value is 230.

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So we will indicate this minimum value 230 by using an arrow like this.

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Now we made the first arrow here

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and this column X3 is the corresponding column to this value which we indicated with the first arrow.

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This column of X3 is called the key column.

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And X3 is called incoming vector.

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Now, in the case of minimum ratio, when we got the minimum value from minimum ratio,

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230, we also indicated this minimum value with an arrow, and this row is the corresponding row to this value which we indicated with this arrow.

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So this corresponding row is called the key row.

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And the variable in front of the key row is S2.

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So this variable S2 is called outgoing vector.

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And, this common element between this key row and key column is called key element.

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Here, 2 is the common element between key row and key column, so, 2 is the key element.

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So, in step number 6, we have found out the minimum ratios, then, based on the minimum value, we found out the key row,

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the outgoing vector, and the key element.

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Now, in step number 7, we have to construct the new simplex table by entering incoming vector.

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And after that we will repeat step number 5 and 6.

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So we have to repeat the step number 5 and step number 6 again after constructing the new simplex table.

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So, moving forward, here we have made a new simplex table.

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In the heading we have written first iteration.

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So this is the first iteration simplex table we have constructed this new simplex table exactly like the first table.

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When we filled up the values for the initial simplex table, we had easily looked at the matrix,

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and filled the values from the matrix to the initial simplex table without much calculation.

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But, in this new simplex table, we have to calculate each value to fill up the table.

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The value of Cj row will remain same, because the cost of the objective function has remained same.

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So, we will copy and write the value of cj from the table number 1 to this new table.

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Here we will write 3, 2, 5, 0, 0, and 0 as the cost will remain same.

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In the basic variable, in the first table we had s1, s2, and s3.

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Here, s2 is the outgoing vector and in place of s2, the incoming vector is x3.

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So, in place of S2, we will write X3.

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Here we have written the basic variables, S1, X3, and S3.

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So we have replaced S2 by X3.

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Now we will write the cost of these variables here.

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So, the cost of S1 is 0, the cost of X3 is 5, and the cost of S3 is 0.

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Now starting from XP, we have to fill up the rest of this table by doing some calculations

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and we will do the calculations by looking at table number 1.

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Now, 2 is the key element here.

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This key element has very important role in doing the calculations.

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Now, this key element is in second row.

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So we will divide all the values of the second row by this key element.

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In other words, we will divide all elements of the key row by the key element.

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We will divide 460 by 2, divide 3 by 2, similarly the other elements also.

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So, when we divide 460 by 2, in the new table we will write 230 in the place of 460,

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we are replacing the second row in the old table, so we will write the new values in the second row of the new table.

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So here we will have 3 divided by 2, then 0 divided by 2, which gives us 0, 2 divided by 2,

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which gives us 1, then we will have 0, 1 divided by 2, and 0.

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So, why are we doing this?

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If you observe carefully, in table number 1, the basic variables were S1, S2, and S3.

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If you observe the values of S1 column here, we have 1, 0, 0.

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In the S2 column here, we have 0, 1, 0.

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And in the S3 column here, we have 0, 0, 1.

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So, this forms an identity matrix.

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The basic variables are written serially as S1, S2 and S3 and if the S1,

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S2 and S3 columns are arranged in the same serial, we get an identity matrix like this.

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Now, in the table 2 also, the basic variables are S1, X3 and S3.

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If the S1, X3 and S3 columns are arranged in the same serial, we should get an identity matrix.

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The first basic variable is S1.

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In the first table we saw, the value of the first basic variable s1 column was 1, 0, 0.

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So, in new table, in the first basic variable s1 column we will write 1, 0, 0.

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Now, the second basic variable is x3.

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In the first table we saw, the value of the second basic variable s2 column was 0, 1, 0.

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So, in new table, in the second basic variable x3 column we will write 0, 1, 0.

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And, the third basic variable is s3.

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In the first table we saw, the value of the third basic variable s3 column was 0, 0, 1.

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So, in new table, in the third basic variable s3 column we will write 0, 0, 1.

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Now, if we arrange the columns of S1, X3 and S3 serially, we will get an identity matrix form like this one.

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In this lecture we will see two methods of forming this table,

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we are learning one method now and if we require a new simplex table for finding the values,

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then in the second iteration table we will see another method of forming this table,

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you can solve using any one method that seems easier to you.

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So, in this first method, we have formed the table and filled some portion of it,

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but we don't know the values to fill up these empty spaces.

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So, to find out these remaining values we have to look back in the first table.

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Let's find out the value for this position at first, in the table number 1, in this position we have 430.

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So, we go to table number 1, then we will directly cross multiply 430 with the key element.

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Here the key element is 2.

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So here we write 430 multiplied by 2.

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We have to find this position in new table.

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So we directly multiplied 430 by key element 2.

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And then we can see that here we have 460.

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And here we have 1.

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If we multiply 460 and 1,

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it will form across diagonally we multiplied 430 with 2 and we will multiply 460 with 1 over here.

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So it will be a cross multiplication.

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So in calculation we write 430 multiplied with 2 minus 460 into 1.

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Then we will divide this whole thing by the key element.

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Here the key element is 2.

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So we divide the whole expression by 2 and finally we will get a value of 200.

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We were finding this value for this position.

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So we will write 200 in this position.

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Now, we will find out the value for this position.

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In the table number 1, in this position we have 420.

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We will directly cross multiply 420 with the key element.

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Here the key element is 2.

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So, we will multiply 420 with 2.

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And then, we can see that here we have 460.

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And here we have 0.

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If we multiply 460 and 0, it will form a cross diagonally.

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We multiplied 420 with 2.

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And we will multiply 460 with 0 over here.

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So it will be a cross multiplication.

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So, in calculation we write, 420 multiplied with 2, minus, 460 into 0.

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Then we will divide this whole thing by the key element.

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Here the key element is 2.

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So we divide the whole expression by 2.

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And finally we will get a value of 420.

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We were finding this value for this position.

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So we will write 420 in this position.

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Similarly, we will find out the value for this position.

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In the first table, in this position we have 1.

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We will directly cross multiply 1 with the key element 2.

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Remember that, we need to find this position in new table.

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So we will multiply this value with key element first.

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Then we will multiply the other two corner values of the imaginary rectangle that makes a perfect cross with this one.

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So, we will cross multiply 1 with 2.

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We can see that here we have 3.

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And here we have 1.

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If we multiply 3 and 1, it will form a cross diagonally.

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So, in calculation we write 1 multiplied with 2 minus 3 into 1.

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Then we will divide this whole thing by the key element, 2.

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And finally we will get a value of minus 1 by 2.

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We were finding this value for this position.

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So we will write minus 1 by 2 in this position.

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Similarly, we will find out the value for this position.

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In the first table, in this position we have 1.

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We will directly cross multiply 1 with the key element 2.

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We can see that here we have 3 and here we have 0.

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If we multiply 3 and 0, it will form across diagonally.

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So in calculation we write 1 multiplied with 2 minus 3 into 0.

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Then we will divide this whole thing by the key element, 2.

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And finally we will get a value of 1.

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We were finding this value for this position.

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So we will write 1 in this position.

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Similarly, we will find out the value for this position.

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In the first table, in this position we have 2.

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We will directly cross multiply 2 with the key element 2.

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We can see that here we have 0.

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And here we have 1.

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If we multiply 0 and 1, it will form a cross diagonally.

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So, in calculation we write 2 multiplied with 2 minus 0 into 1.

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Then we will divide this whole thing by the key element 2.

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And finally we will get a value of 2.

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We were finding this value for this position.

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So we will write 2 in this position.

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Now, we will find out the value for this position.

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In the first table, in this position we have 4.

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We will directly cross multiply 4 with the key element 2.

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We can see that here we have 0 and here we have 0.

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So, we will multiply 0 and 0.

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It will form across diagonally.

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So, in calculation we write 4 multiplied with 2 minus 0 into 0.

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Then we will divide this whole thing by the key element 2.

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And finally we will get a value of 4.

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We were finding this value for this position.

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So we will write 4 in this position.

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We have two more empty spaces left.

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We will find out the value for this position now.

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In the first table, in this position we have 0.

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We will directly cross multiply 0 with the key element 2.

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We can see that here we have 1 and here we have 1.

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So, we will multiply 1 and 1.

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It will form a cross diagonally.

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So, in calculation we write 0 multiplied with 2 minus 1 into 1.

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Then we will divide this whole thing by the key element 2

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and finally we will get a value of minus 1 by 2.

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We were finding this value for this position.

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So we will write minus 1 by 2 in this position.

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So this is the last empty space in this table.

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We will find out the value for this position now.

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In the first table in this position we have 0.

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We will directly cross multiply 0 with the key element 2.

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We can see that here we have 0 and here we have 1.

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So, we will multiply 0 and 1.

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It will form across diagonally.

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So, in calculation we write 0 multiplied with 2 minus 0 into 1.

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Then we will divide this whole thing by the key element 2 and the value is 0.

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We were finding this value for this position.

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So, we will write 0 in this position.

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So, we have prepared the simplex table for the first iteration.

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We have filled all the values in this first iteration table.

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Now, we will find out the values for ZJ minus CJ and fill up this last row.

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We will find out the values for ZJ minus CJ using this formula just like we did for the initial table.

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The general formula is ZJ minus CJ equals to CB into XJ minus CJ.

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For the calculation of Z1 minus C1,

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the formula will be Z1 minus C1 equals to CB into X1 minus C1.

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In the initial table, all the values of CB were 0, but in this first iteration table, the values of CB are 0, 5, and 0.

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Here we have to multiply CB with X1,

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so 0 into minus 1 by 2, which gives us 0 plus 5 into 3 by 2 plus 0 into 1.

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Now here we write 0 into minus 1 by 2 plus 5 into 3 by 2 plus 0 into 1.

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Then give the minus sign from formula and in the CJ row we can see the value of C1 is 3.

402

So we write 3 here and finally the value of Z1 minus C1 will be 9 by 2.

403

So, in this row, here we will write 9 divided by 2.

404

Now, in the same way we will find out value of Z2 minus C2.

405

The formula will be CB into X2 minus C2.

406

Here we will have 0 into 2, which gives us 0, plus 5 into 0,

407

which gives us 0, plus 0 into 4, which gives us 0.

408

So, the value of CB into X2 will be 0, minus, the value of C2 is, 2.

409

And finally, the value of Z2 minus C2 will be, minus 2.

410

So, in this row, here we will write, minus 2.

411

Now, in the same way we will find out value of Z3 minus C3.

412

First multiply CB into X3.

413

Here we will have, 0 into 0, which gives us 0, plus,

414

5 into 1 which gives us 5 plus 0 into 0 which gives us 0.

415

So, the value of CB into X3 will be 5 minus the value of C3 is 5.

416

And finally, the value of Z3 minus C3 will be 5 minus 5 equals to 0.

417

So, in this row, here we will write 0.

418

Similarly for Z4 minus C4, here we will have 0 into 1, which gives us 0, plus 5 into 0,

419

which gives us 0, plus 0 into 0, which gives us 0.

420

So the value will be 0, minus the value of C4 is 0, and finally the value of Z4 minus C4 will be 0.

421

So in this row, here we will write 0.

422

For Z5 minus C5, here we will have 0 into minus 1 by 2, which gives us 0, plus 5 into 1 by 2,

423

which gives us 5 by 2, plus 0 into 0, which gives us 0.

424

So the value will be 5 by 2 as the other values are 0, minus the value of C5 is 0.

425

And finally, the value of Z5 minus C5 will be 5 by 2.

426

So, in this row, here we will write 5 by 2.

427

And, for Z6 minus C6, here we will have 0 into 0, which gives us 0, plus 5 into 0,

428

which gives us 0, plus 0 into 1, which gives us 0.

429

So, the value will be 0, minus, the value of C6 is 0.

430

And finally, the value of Z6 minus C6 will be 0.

431

So, in this row, here we will write, 0.

432

Now, we have found out the values for ZJ minus CJ and filled up this row.

433

Here we can see a negative value of ZJ minus CJ, which is, minus 2.

434

We will indicate this negative value of minus 2 by an arrow, and this column will be the key column.

435

The variable X2 will be the incoming vector.

436

Now, on this side we will make a column for the calculation of minimum ratio.

437

The formula for minimum ratio is xp divided by the key column

438

and the value of key column has to be greater than 0.

439

Here, the key column is x2.

440

So, we will divide the values of column xp by the values of key column x2 for calculating minimum ratio.

441

For the first row, xp is 200 and value of x2 is 2.

442

So, we divide 200 by 2.

443

The minimum ratio will be 100.

444

For the second row, xp is 230 and value of x2 is 0.

445

Since the value of the key column is 0 in this row, we will not calculate the minimum ratio for this row

446

because we said earlier that the value of key column has to be greater than 0.

447

It cannot be 0 or less than 0.

448

Now, for the third row, xp is 420 and value of x2 is 4.

449

So, we divide 420 by 4, the minimum ratio will be 105.

450

Now between these two values 105 and 100, the minimum value is 100.

451

We will indicate this minimum value of 100 by an arrow and this corresponding row will be the key row.

452

This variable s1 will be the outgoing vector.

453

And this common element between this key row and key column is called key element.

454

Here, 2 is the common element between key row and key column, so 2 is the key element.

455

So in this first iteration simplex table, this is the key column, x2 is the incoming vector, this is the key row,

456

s1 is the outgoing vector and this common element 2 is the key element.

457

Now, we will make the second iteration table.

458

So here we have the simplex table for second iteration.

459

The value of cj row will remain same because the cost of the objective function has remained same.

460

So, we will copy and write the value of cj from the first iteration table to this new table.

461

Here we will write 3, 2, 5, 0, 0 and 0.

462

In the basic variable column, in the first iteration table we had s1, x3 and s3.

463

Here, S1 is the outgoing vector and in place of S1, the incoming vector is X2.

464

So in place of S1, we will write X2.

465

Here we have written the basic variables X2, X3 and S3.

466

So we have replaced S1 by X2.

467

Now we will write the cost of these variables here.

468

So, the cost of X2 is 2, the cost of X3 is 5, and the cost of S3 is 0.

469

Now, we have to fill up the rest of this table by doing the calculations.

470

In the first iteration table, we saw one method of finding the values.

471

In the first method we cross multiplied the old values to get new values.

472

Now in this second iteration table, we will see another method of calculating the values to fill up the table.

473

In this second method we will find out the values by doing some row operations and column operations options,

474

you can either use the first method or this second method that we will see now, whichever seems easier to you.

475

So, for the first row, we will divide all the values of the key row by the key element.

476

So we divide 200 by the key element 2 which gives us 100,

477

divide minus 1 over 2 by the key element 2 which gives us minus 1 by 4,

478

then divide key element 2 by 2 which gives us 1.

479

Here we will have 0, then 1 by 2, minus 1 by 4, and 0.

480

So, why did we divide this whole row by 2?

481

Since, this is the key row, here we have the key element.

482

When we make the new table, the key element has to be changed and written as 1.

483

Now if we divide the key element 2 by 2, then in the second iteration table we will get 1 in the same position.

484

Again, since we cannot divide just one element from the row,

485

So we divided the whole key row by 2 and got this new row in the second iteration table.

486

As we can see here, the key element 2 has been converted into 1 in this new table.

487

So, we have shown the key element of this key column as 1 in this new table.

488

Now, we have to show the other two elements of the key column as 0.

489

Now, this second value of the key column is already 0.

490

So we don't have to change anything in this row to make it 0 row,

491

we will write this row in the new table as it is written here.

492

So, here we write 230, 3 by 2, 0, 1, 0, 1 by 2, and 0.

493

So, the first element of the key column has been converted to 1.

494

The next element is 0.

495

Now, the third element of the key column is, 4.

496

We have to do some row operation to convert 4 to 0 in this new table.

497

So, the row operation for For this third row will be R3 dash equals to I3 minus 4 R1.

498

We will use R1 to solve this operation.

499

Because R1 was the key row in the previous table, we have to get 0 in the position of 4 here.

500

So, in this row operation, we will multiply this new first row by 4 and then subtract from the old column's third row.

501

Let's see in rough.

502

To get the new row 3, our 3 dash, first we will write the old row 3,

503

421, 400, 00, and 1.

504

So, this is our 3.

505

Then we multiply this new row 1, by 4, and write here, so, 4 into 100, gives us,

506

400, 4 into minus 1 by 4, gives us, minus 1.

507

Then we have 4, 0, 2, minus 1, and 0.

508

So, this is 4 R1.

509

Now, subtract 4 R1 from R3.

510

We get 20, 2, 0, 0, minus 2, 1, and 1.

511

So, this is R3 dash, or the new R3 row.

512

Now, we will put this values of new R3 row.

513

20, 2, 0, 0, minus 2, 1, and 1.

514

Now, if we look carefully, in this table, the basic variables are x2, x3, and s3.

515

If the x2, x3, and s3 columns are arranged in the same serial, we should get an identity matrix as we have seen earlier.

516

Here, the first basic variable is x2, in the x2 column we have 1, 0, 0.

517

Now, the second basic variable is x3, in the x3 column we have 0, 1, 0.

518

And the third basic variable is s3, in the s3 column we have 0, 0, 1.

519

Now if we arrange the columns of x2, x3 and s3 serially, we will get an identity matrix form like this one.

520

Now we will find the values for the zj minus cj row.

521

We have seen how to calculate ZJ minus CJ in the previous two tables, so let's calculate them quickly.

522

For the calculation of Z1 minus C1, here we have to multiply CB with X1.

523

So, 2 into minus 1 by 4 plus 5 into 3 by 2 plus 0 into 2.

524

Then we will solve it.

525

And in the CJ row, we can see the value of C1 is 3.

526

So, we will subtract 3 from the value of CB into X1.

527

And finally, the value of Z1 minus C1 will be 4.

528

So, in this row, here we will write 4.

529

Now, in the same way we will find out value of Z2 minus C2.

530

First we have to multiply CB with X2.

531

So, 2 into 1 plus 5 into 0 plus 0 into 0.

532

Then we will solve it.

533

And in the CJ row, we can see the value of C2 is 2.

534

So, we will subtract 2 from the value of CB into X2.

535

And finally, the value of Z2 minus C2 will be 0.

536

So, in this row, here we will write 0.

537

If you want you can write the detailed calculation in rough and check the values.

538

Since we have seen the detailed calculations for Zj minus Cj in the previous tables, we are solving quickly now.

539

So, in the same way we will find out value of Z3 minus C3, first multiply Cb into x3.

540

Here we will have 2 into 0 plus 5 into 1 plus 0 into 0.

541

So, the value of CB into X3 will be 5 minus the value of C3 is 5.

542

And finally, the value of Z3 minus C3 will be 5 minus 5 equals to 0.

543

So, in this row, here we will write 0.

544

Similarly for Z4 minus C4, here we will have 2 into 1 by 2

545

which gives us 1 plus 5 into 0 which gives us 0 plus 0 into minus 2 which gives us 0.

546

So the value will be 1 minus the value of C4 is 0

547

and finally the value of Z4 minus C4 will be 1.

548

So in this row here we will write 1.

549

For Z5 minus C5, here we will have 2 into minus 1 by 4 plus

550

5 into 1 by 2 plus 0 into 1 minus the value of C5 is 0.

551

And finally, the value of Z5 minus C5 will be 2.

552

So, in this row, here we will write 2.

553

And, for Z6 minus C6, here we will have 2 into 0, plus 5 into 0, plus 0 into 1,

554

minus, the value of C6 is, 0.

555

And finally, the value of Z6 minus C6 will be 0.

556

So, in this row, here we will write, 0.

557

So, we have found out the values for Zj minus Cj, and filled up this row.

558

Now, if we look at the values of ZJ minus CJ, all these values are either greater than 0,

559

or equal to 0, there is no negative value of ZJ minus CJ here,

560

since, all the values of ZJ minus CJ are greater than or equal to 0, that is, they are all positive values,

561

so, we can say that our solution has reached the optimal stage.

562

Finally we will write, since, all values of ZJ minus CJ are greater than or equal to 0, the solution is optimum.

563

Then we will write the values of X1, X2, and X3 from the basic variables.

564

Here we can see that, value of X2 is 100, value of X3 is 230, and there is no X1.

565

So we write x2 equals to 100, x3 equals to 230 and x1 equals to 0.

566

Now in our objective function we had maximize z.

567

So we will calculate max z here.

568

The formula of max z will be cb into xp.

569

Now we will multiply values of cb with xp and then we will add them.

570

In CB column here we have 2, 5 and 0 and in the xp column we have 100, 230 and 20.

571

So, we will write CB into xp equals to 2 into 100 plus 5 into 230 plus 0 into 20.

572

The answer will be 1350.

573

Thus, the optimum value of max z is 1350.

574

So, in this lecture we have solved a maximization problem by simplex method.

575

In the next lecture we will solve a minimization problem by simplex method.

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Vocabulaire et conseils d’expression pour cette leçon

Cette leçon d’expression orale s’appuie sur la vidéo « Lec-6 Simplex Method ». Les mots qui reviennent le plus souvent : value, minus, table, row, key. Cette vidéo contient 575 phrases et 6973 mots à répéter en shadowing. La partie parlée dure 49:14. Le locuteur parle à un rythme régulier d’environ 142 mots par minute, confortable pour le shadowing. Seuls 80 % des mots font partie des 3 000 mots les plus courants en anglais, le vocabulaire est donc exigeant.

Vocabulaire clé de cette vidéo

15 mots de la vidéo à retenir, avec leur prononciation et leur sens :

Mot	Prononciation	Sens
minus	/ˈmaɪ.nəs/	moins
column nom	/ˈkɒləm/	colonne
variable adjectif	[ˈvæɹ.i.ə.bl̩]	variable
element nom	/ˈɛləmənt/	élément
divide verbe	/dɪˈvaɪd/	diviser, fendre
ratio nom	/ˈɹeɪ.ʃoʊ/	rapport, ratio
calculation nom	/ˌkælkjuˈleɪʃn̩/	calcul
formula nom	/ˈfɔː.mjʊ.lə/	formule
calculate verbe	/ˈkælkjʊleɪt/	calculer
slack adjectif	/slæk/	lâche
vector nom	/ˈvɛktɚ/	vecteur
arrow nom	/ˈæɹəʊ/	flèche
equation nom	/ɪˈkweɪ.ʒən/	équation
indicate verbe	/ˈɪndɪˌkeɪt/	indiquer, signaler
solve verbe	/sɒlv/	résoudre, régler

Les verbes à particule que vous entendrez

Mot	Sens
find out verbe	découvrir, savoir
fill up verbe	remplir

La grammaire de cette vidéo

Les structures que le locuteur utilise le plus, avec les mots exacts de la vidéo :

Structure	Dans la vidéo
Present perfect have/has + participe passé — une action passée qui compte encore maintenant	have seen · have introduced · have written
Voix passive be + participe passé — l’accent est mis sur ce qui arrive, pas sur qui le fait	is given · is written · is also known

Prononciation à surveiller

Le locuteur utilise 5 contractions et formes réduites, comme don't, haven't. Prononcez-les sous leur forme courte, telles que vous les entendez.

Les sons « sh » et « zh »: ratio /ˈɹeɪ.ʃoʊ/, calculation /ˌkælkjuˈleɪʃn̩/, equation /ɪˈkweɪ.ʒən/, dash /ˈdæʃ/, iteration /ˌɪt.əˈɹeɪ.ʃən/
Mots longs — placez bien l’accent: similarly /ˈsɪm.ə.lə.li/, applicable /əˈplɪk.ə.bəl/, imaginary /ɪˈmæd͡ʒɪˌn(ɛ)ɹi/, iteration /ˌɪt.əˈɹeɪ.ʃən/, diagonally /daɪˈæɡ.nə.li/

Les sons difficiles pour les francophones :

/tʃ/ et /dʒ/ — à ne pas adoucir en « ch » et « j »: objective /ɒbˈd͡ʒɛk.tɪv/, lecture /ˈlɛk.t͡ʃə/, arrange /əˈɹeɪnd͡ʒ/, imaginary /ɪˈmæd͡ʒɪˌn(ɛ)ɹi/, whichever /wɪt͡ʃˈɛvɚ/
/r/ anglais — langue recourbée, sans frotter la gorge: variable [ˈvæɹ.i.ə.bl̩], ratio /ˈɹeɪ.ʃoʊ/, matrix /ˈmeɪ.tɹɪks/, arrow /ˈæɹəʊ/, linear /ˈlɪn.i.əɹ/
/ŋ/ final — sans prononcer de /g/ après: incoming /ˈɪnˌkʌmɪŋ/, outgoing /aʊtˈɡəʊɪŋ/, heading /ˈhɛdɪŋ/

Comment s’entraîner avec cette vidéo

Écoutez la vidéo en entier une fois sans parler et notez les mots que vous ne connaissez pas.
Répétez phrase par phrase à vitesse normale, en reprenant chacune jusqu’à ce que votre rythme corresponde à celui du locuteur.
Enregistrez-vous et comparez avec l’original, en faisant attention à des mots comme minus, column, variable.

Qu'est-ce que la technique du Shadowing ?

Le Shadowing est une technique d'apprentissage des langues fondée sur la science, développée à l'origine pour la formation des interprètes professionnels. Le principe est simple mais puissant : vous écoutez de l'anglais natif et le répétez immédiatement à voix haute — comme une ombre suivant le locuteur avec un décalage de 1 à 2 secondes. Les recherches montrent une amélioration significative de la précision de la prononciation, de l'intonation, du rythme, des liaisons, de la compréhension orale et de la fluidité.

Technique du shadowing : lire le guide complet étape par étape →

Pratique du Shadowing: Lec-6 Simplex Method | Maximization Problem | Mathematical Example Solution | Operations Research - Apprendre l'anglais à l'oral avec la vidéo

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Vocabulaire et conseils d’expression pour cette leçon

Vocabulaire clé de cette vidéo

Les verbes à particule que vous entendrez

La grammaire de cette vidéo

Prononciation à surveiller

Comment s’entraîner avec cette vidéo

Qu'est-ce que la technique du Shadowing ?