# Find A Closed Formula For 1 2 0 3 T Simplex Method – Algorithm of Solving Linear Programming Problems

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## Simplex Method – Algorithm of Solving Linear Programming Problems

The simplex method is used in linear programming to optimize an objective function over a number of linear constraints. If you need a tabular representation of data when writing articles for EzineArticles, consider using ASCII art tables within the HTML pre tag. I have used them to illustrate simplex algorithm iterations. They can be a source of inspiration for you.

Now let’s see how to solve a linear programming model:

```Maximize: Z = 1x1 + 2x2subject to:
3x1 + 4x2 ≤ 5
6x1 + 7x2 ≥ 8
x1 ≥ 0, x2 ≥ 0********* (1) *********```

This model will be converted to canonical form (all constraints become equations). A constraint of type “≤ 0” becomes an equation by adding a non-negative variable (slack variable) and a constraint of type “≥ 0” changes into an equation by subtracting one (extra variable).

```Maximize: Z = 1x1 + 2x2 + 0x3 + 0x4subject to:
3x1 + 4x2 + 1x3 + 0x4 = 5
6x1 + 7x2 + 0x3 + -1x4 = 8x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0**************** (2)***************```

Note that max-7, -3, 0, +4, +6 = 6 = –minute+7, +3, 0, -4, -6. That’s why max(Z(x)) = –minute(-Z(x)). In other words, we get maxBy calculating (Z(x)). minute(-Z(x)) and changing the sign at the end of the algorithm.

A two-step method of constructing the initial basis required for the first iteration of the simplex algorithm requires the addition of an artificial variable for each constraint. Finally, they must all equal 0 for the bottom model to be equal to the top model. This trick allows us to get the unit vectors of Initialize the vector space and iteration for the auxiliary minimization model (minimizing the sum of artificial variables). The above maximal model leads us to the following minimal model:

```Minimize: Z = -1x1 + -2x2 + 0x3 + 0x4 + 0x5 + 0x6subject to:
3x1 + 4x2 + 1x3 + 0x4 + 1x5 + 0x6 = 5
6x1 + 7x2 + 0x3 + -1x4 + 0x5 + 1x6 = 8x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0********************** (3) **********************```

As we mentioned, after calculation, all artificial variables (“x5”, “x6”) must be 0. Since artificial variables also satisfy the non-negativity condition (“≥ 0”), we conclude that they are all equal to 0 if and only if their sum is at least 0. So, in the first step we minimize the sum of artificial variables:

```Minimize: Z' = 0x1 + 0x2 + 0x3 + 0x4 + 1x5 + 1x6subject to:
3x1 + 4x2 + 1x3 + 0x4 + 1x5 + 0x6 = 5
6x1 + 7x2 + 0x3 + -1x4 + 0x5 + 1x6 = 8x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0, x5 ≥ 0, x6 ≥ 0********************** (4) **********************```

Now we begin the first iteration of the simplex method.

```.----+----+----+----+---+----+---+---+---+----+---.
|CB *|B * |XB *|cj * 0 * 0 ** 0 * 0 * 1 * 1 * | * |
+----+----+----+----+---+----+---+---+---+----+---+
| ** | ** | ** | ** |a1 |a2↓ |a3 |a4 |a5 |a6 *|θi |
|1 * |a5 *|5 * | ** |3 *|4 * |1 *|0 *|1 *|0 * |5/4|
|1 * |a6← |8 * | ** |6 *|7 * |0 *|-1 |0 *|1 * |8/7|
+----+----+----+----+---+----+---+---+---+----+---+
|Z'1 = 13 **** |zj * 9 * 11 * 1 * -1* 1 * 1 * | * |
+--------------+----+---+----+---+---+---+----| * |
| ************ |Δj * 9 * 11 * 1 * -1* 0 * 0 * | * |
'--------------+----+---+----+---+---+---+----+---'******************** Tableau 1 ********************```

Legend:

```• cj: Objective function (Z') coefficients, j = 1...6• aj: Column vectors of the constraints, j = 1...6• B = a5, a6: Standard basis of R² vector space• XB = (5, 8): Column of the basic feasible solution(the elements of the solutionX = (x1, x2, x3, x4, x5, x6) = (0, 0, 0, 0, 5, 8)corresponding to vectors outside the basis B are all equal to 0)• CB: Coefficients of vectors from basis B of objective function (Z')• Z'1 = ((CB)^T)*(XB) = (1, 1) * ((5, 8)^T)
Z'1 = 1 * 5 + 1 * 8 = 13
(CB)^T: Transpose of column vector CB• zj = ((CB)^T)*(aj), j = 1...6• Δj = zj - cj, j = 1...6
Since not all Δj: 9, 11, 1, -1, 0, 0 elements are ≤ 0,
we have not yet achieved the minimum of the objective function Z'.
Δ2 = 11 = max9, 11, 1, -1, 0, 0
Δ2 ⇒ a2↓• θi = XBi / a2↓i, i ∈ 5, 6  and a2↓i > 0
θi = --, if a2↓i ≤ 0
θ6 = 8/7 = min5/4, 8/7
θ6 ⇒ a6←```

The new vector “a2↓” replaces the old vector “a6←” in base B which is leaving base B. The pattern element where intersects the entering vector (“a2↓”) and the leaving vector (“a6←”). known as the pivot element ( 7 ). We must split all elements in the pivot row 8, 6, 70, -1, 0, 1 to get a by the pivot element +1 At the pivot element position 8/7, 6/7, +1, 0, -1/7, 0, 1/7. The remaining elements of the pivot column become 0’s.

Now we need to remove all the coefficients in the pivot column except the pivot element. This is done by simple Gaussian operations. For example to remove a pivot column element in some row “k”, we proceed as follows:

• (new row “k”) = (row “k”) – (pivot column coefficient in row “k”) * (pivot row)

• (3/7, -3/7, 0, 1, 4/7, 1, -4/7) = (5, 3, 4, 1, 0, 1, 0) – 4 * (8/7, 6/7, 1, 0, -1/7, 0, 1/7)

A second iteration is about to begin.

```.----+----+----+----+------+---+----+------+---+------+---.
|CB *|B * |XB *|cj * 0 **** 0 * 0 ** 0 **** 1 * 1 *** | * |
+----+----+----+----+------+---+----+------+---+------+---+
| ** | ** | ** | ** |a1 ** |a2 |a3↓ |a4 ** |a5 |a6 ** |θi |
|1 * |a5← |3/7 | ** |-3/7 *|0 *|1 * |4/7 * |1 *|-4/7 *|3/7|
|0 * |a2 *|8/7 | ** |6/7 * |1 *|0 * |-1/7 *|0 *|1/7 * |-- |
+----+----+----+----+------+---+----+------+---+------+---+
|Z'2 = 3/7 *** |zj * -3/7 * 0 * 1 ** 4/7 ** 1 * -4/7 *| * |
+--------------+----+------+---+----+------+---+------| * |
| ************ |Δj * -3/7 * 0 * 1 ** 4/7 ** 0 * -11/7 | * |
'--------------+----+------+---+----+------+---+------+---'*********************** Tableau 2 ************************```

Since all Δj: -3/7, 0, 1, 4/7, 0, -11/7 elements are not ≤ 0, we still have not obtained the minimum of the objective function Z’.

```.----+----+----+----+------+---+---+------+----+------+---.
|CB *|B * |XB *|cj * 0 **** 0 * 0 * 0 **** 1 ** 1 *** | * |
+----+----+----+----+------+---+---+------+----+------+---+
| ** | ** | ** | ** |a1 ** |a2 |a3 |a4 ** |a5 *|a6 ** |θi |
|0 * |a3 *|3/7 | ** |-3/7 *|0 *|1 *|4/7 * |1 * |-4/7 *| * |
|0 * |a2 *|8/7 | ** |6/7 * |1 *|0 *|-1/7 *|0 * |1/7 * | * |
+----+----+----+----+------+---+---+------+----+------+---+
|Z'3 = 0 ***** |zj * 0 **** 0 * 0 * 0 **** 0 ** 0 *** | * |
+--------------+----+------+---+---+------+----+------| * |
| ************ |Δj * 0 **** 0 * 0 * 0 **** -1 * -1 ** | * |
'--------------+----+------+---+---+------+----+------+---'************************ Tableau 3 ************************```

Since all Δj: 0, 0, 0, 0, -1, -1 elements are ≤ 0, the objective function Z’ has reached its minimum. The minimum sum of the artificial variables in the final table results in Z’3 = 0. So, below are the solutions in the minimal model.

```Minimize: Z = -1x1 + -2x2 + 0x3 + 0x4subject to:
3x1 + 4x2 + 1x3 + 0x4 = 5
6x1 + 7x2 + 0x3 + -1x4 = 8x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0***************** (5)****************```

It is useful to reuse data. In the second step, we will complete another table with the “aj”, “XB”, “B” columns of the previous table and the coefficients of the new objective function “Z” of the above model. We will reconstruct the other calculations and then resume the iterations:

```.----+----+----+----+-------+----+---+------+----.
|CB *|B * |XB *|cj * -1 **** -2 * 0 * 0 *** | ** |
+----+----+----+----+-------+----+---+------+----+
| ** | ** | ** | ** |a1 *** |a2 *|a3 |a4↓ * |θi *|
|0 * |a3← |3/7 | ** |-3/7 * |0 * |1 *|4/7 * |3/4 |
|-2 *|a2 *|8/7 | ** |6/7 ** |1 * |0 *|-1/7 *| -- |
+----+----+----+----+-------+----+---+------+----+
|Z1 = -16/7 ** |zj * -12/7 * -2 * 0 * 2/7 * | ** |
+--------------+----+-------+----+---+------| ** |
| ************ |Δj * -5/7 ** 0 ** 0 * 2/7 * | ** |
.--------------+----+-------+----+---+------+----.******************** Tableau 4 ******************```

Since all Δj: -5/7, 0, 0, 2/7 elements are not ≤ 0, we still have not obtained the minimum of the objective function Z.

```.----+----+----+----+------+----+------+----+---.
|CB *|B * |XB *|cj * -1 ** -2 ** 0 **** 0 * | * |
+----+----+----+----+------+----+------+----+---+
| ** | ** | ** | ** |a1 ** |a2 *|a3 ** |a4 *|θi |
|0 * |a4 *|3/4 | ** |-3/4 *|0 * |7/4 * |1 * | * |
|-2 *|a2 *|5/4 | ** |3/4 * |1 * |1/4 * |0 * | * |
+----+----+----+----+------+----+------+----+---+
|Z2 = -5/2 *** |zj * -3/2 * -2 * -1/2 * 0 * | * |
+--------------+----+------+----+------+----| * |
| ************ |Δj * -1/2 * 0 ** -1/2 * 0 * | * |
.--------------+----+------+----+------+----+---.******************* Tableau 5 *******************```

Since all Δj: -1/2, 0, -1/2, 0 elements are ≤ 0, the objective function Z has reached its minimum (Z2 = -5/2). Therefore, the maximum of the initial model is (1) Z = -Z2 = -(0 * 3/4 ​​+ (-2) * 5/4) = 5/2. The optimal solution is (x1, x2) = (0, 5/4).

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