Continue to find better basic feasible solutions, improving the objective function values. When a particular basic feasible solution cannot be

improved further, it becomes an optimal solution, and the simplex method terminates.

We begin by considering the details of step 2 of the simplex method by assuming that the basic solution given by system (S1) is feasible. Thus, we have an initial basic feasible solution in canonical form as follows:

Basic: x_i⫽ b_i⭓0 for i _⫽ 1, 2, . . . , m Nonbasic: x_j⫽ 0 for j _⫽ m_⫹1, . . . , n

The set of basic variables will be called a basis and will be denoted xB. Let the objective function coefficients of the basic variables be denoted cB. For the initial basis,

xB_⫽ (x , . . . , x )1 m and cB _⫽ (c , c , . . . , c )1 2 m

Since the nonbasic variables are zero, the value of the objective function Z corresponding to the initial basic feasible solution is given by

Z_⫽c xB B _⫽c b1 1<sub>⫹


⫹</sub>c bm m

Given this, the simplex method checks whether it is possible to find a better basic feasible solution with a higher value of Z. This is done by first exam-ining whether the present solution is optimal. In case it is not optimal, the simplex method obtains an adjacent basic feasible solution with a larger value of Z (or at least as large).

Definition

An adjacent basic feasible solution differs from the present basic feasible solution in exactly one basic variable.

To obtain an adjacent basic feasible solution, the simplex method makes one of the basic variables a nonbasic variable and brings in a nonbasic variable as a basic variable in its place. The problem is to select the appropriate basic and nonbasic variables such that an exchange between them will give the maximum improvement to the objective function.

In any basic feasible solution, the basic variables can assume positive val-ues while the nonbasic variables are always held at zero. Hence, making a

interested in examining adjacent basic feasible solutions only, the values of the other nonbasic variables will continue to remain zero and system (S1) can be rewritten as follows:

x_{1 ⫹}a x_{1s s ⫽} b₁

⯗

xr _⫹ a xrs s _⫽ br (S2)

⯗ xm_⫹ a xms s_⫽ bm

From system (S2), we obtain the new solution when xsincreases from 0 to 1:

xi_⫽ bi_⫺ ais for i_⫽ 1, . . . , m xs_⫽ 1

x_j⫽ 0 for j_⫽ m_⫹ 1, . . . , n and j_⫽ s The new value of the objective function is

m

Z_⫽i

冘

⫽1 c (b^{i i}_⫺ a )^is _⫹c^s

Hence, the net change in the value of Z per unit increase in xs, denotedc ,s is cs _⫽new value of Z _⫺ old value of Z

m m

⫽

冘

i⫽1 c (b^{i i}_⫺ a )^is _⫹ c^s _⫺

冘

i⫽1 c b^{i i}

m

⫽c^s _⫺i

冘

⫽1 c a^{i is} (4.5)

where cs is called the relative profit of the nonbasic variable xs, as opposed to its actual profit cs in the objective function. If cs _⬎ 0, then the objective

function Z can be increased further by making xs a basic variable. Equation (4.5), which gives the formula for calculating the relative profits, is known as the inner product rule.

Inner Product Rule

The relative profit coefficient of a nonbasic variable xj, denoted byc ,j is given by

cj_⫽cj_⫺c PB j

where cB corresponds to the profit coefficients of the basic variables and Pj

corresponds to the jth column in the canonical system of the basis under consideration.

Condition of Optimality

In a maximization problem, a basic feasible solution is optimal if the relative profits of its nonbasic variables are all negative or zero.

It is clear that when all cj _⭐ 0 for nonbasic variables, then every adjacent basic feasible solution has an objective function value lower than the present solution. This implies that we have a local maximum at this point. Since the objective function Z is linear, a local maximum automatically becomes the global maximum.

For the sake of illustration, let us assume thatcs_⫽ maxcj_⬎ 0, and hence the initial basic feasible solution is not optimal. The positive relative profit of xs implies that Z will be increased by cs units for every unit increase of the nonbasic variable xs. Naturally, we would like to increase xsas much as possible so as to get the largest increase in the objective function. But as xs

increases, the values of the basic variables change, and their new values are obtained from system (S2) as

xi_⫽bi _⫺a xis s for i _⫽ 1, . . . , m (4.6) Ifa_is⬍ 0, then x_iincreases as x_s is increased, and if a_is⫽ 0, the value of x_i does not change. However, ifa_is⬎ 0, x_idecreases as x_sis increased and may turn negative (making the solution infeasible) if x_s is increased indefinitely.

Hence, the maximum increase in x_sis given by the following rule:

bi

max x^s _⫽min^ais⬎0 冋 册^ais (4.7) Let

b_r

xⁱ _⫽bⁱ_⫺ a^is冉 冊^ars for all i br

xs _⫽ (4.8)

ars

for all other nonbasic variables xj _⫽0 j _⫽ m_⫹1, . . . , n and j _⫽ s

Since a unit increase in xs increases the objective function Z by cs units, the total increase in Z is given by

b_r (c )^s冉 冊^ars _⬎0

Equation (4.7), which determines the basic variables to leave the basis, is known as the minimum-ratio rule in the simplex algorithm. Once again, the simplex algorithm checks whether the basic feasible solution given by (4.8) is optimal by calculating the relative profits for all the nonbasic variables, and the cycle is repeated until the optimality conditions are reached.

Let us now illustrate the basic steps of the simplex algorithm with an example.

Example 4.6

Let us use the simplex method to solve the following problem:

Maximize Z _⫽3x_{1 ⫹} 2 x₂ Subject to _⫺x_{1 ⫹}2 x_2⭐ 4

3x_{1 ⫹}2 x_2⭐ 14 x_{1 ⫺} x_2⭐ 3 x_1⭓ 0 x_{2 ⭓}0

Converting the problem to standard form by the addition of slack variables, we get

Tableau 1 basic variables. The various steps of the simplex method can be carried out in a compact manner by using a tableau form to represent the constraints and the objective function. In addition, the various calculations (e.g., inner product rule, minimum-ratio rule, pivot operation) can be made mechanical. The use of the tableau form has made the simple method more efficient and convenient for computer implementation.

The tableau presents the problem in a detached coefficient form. Tableau 1 gives the initial basic feasible solution for Example 4.6. Note that only the coefficients of the variables are written for each constraint. The objective function coefficients cj are written above their respective xj’s. Basis denotes the basic variables of the initial tableau, namely x₃ for constraint 1, x₄ for constraint 2, and x₅for constraint 3. The parameter cBdenotes their respective cj values. From Tableau 1, we obtain the initial basic feasible solution im-mediately as x_{3 ⫽} 4, x_{4 ⫽} 14, x_{5 ⫽} 3, and x_{1 ⫽} x_{2 ⫽} 0. The value of the objective function Z is given by the inner product of the vectors cB and con-stants:

4 Z_⫽ (0, 0, 0) 14

冢 冣

³ _⫽ 0

To decide which basic variable is going to be replaced, we apply the minimum-ratio rule [Eq. (4.7)] by calculating the ratios for each constraint that has a positive coefficient under the x₁ column as follows:

Row Number Basic Variable Ratio

1 x_{3 ⬁}

2 x₄ ––¹⁴₃

3 x₅ –³₁

Note that no ratio is formed for the first row or is set to_⬁, since the coefficient for x₁in row 1 is negative. This means that x₁can be increased to any amount without making x₃ negative. On the other hand, the ratio ––¹⁴₃ for the second row implies that x₄ will become zero when x₁ increases to ––¹⁴₃. Similarly, as x₁ increases to 3, x₅turns zero. The minimum ratio is 3, or as x₁ is increased from 0 to 3, the basic variable x₅ will turn zero first and will be replaced by x₁. Row 3 is called the pivot row, and the coefficient of x₁ in the pivot row is called the pivot element (circled in Tableau 1). We obtain the new basic variable x₃, x₄, and x₁and the new canonical system (Tableau 2) by performing a pivot operation as follows:

1. Add the pivot row (row 3) to the first row to eliminate x₁.

2. Multiply the pivot row by_⫺3 and add it to the second row to eliminate x₁.

To check whether the basic feasible solution given by Tableau 2 is optimal, we must calculate the new relative-profit coefficients(c row). We can do this by applying the inner product rule as before. On the other hand, it is also possible to calculate the new c row through the pivot operation. Since x₁ is the new basic variable, its relative-profit coefficient in Tableau 2 should be zero. To achieve this, we multiply the third row of Tableau 1 (the pivot row) by _⫺3 and add it to the c row. This will automatically give the new c row for Tableau 2!

The movement from Tableau 1 to Tableau 2 is illustrated pictorially in Figure 4.5. The feasible region of Example 4.6 is denoted ABCDE. The dotted lines correspond to the objective function lines at Z_⫽3 and Z_⫽ 9. Note that the basic feasible solution represented by Tableau 1 corresponds to the corner

Tableau 2

cj

3 2 0 0 0

cB Basis x₁ x₂ x₃ x₄ x₅ Constants

0 x₃ 0 1 1 0 1 7

0 x₄ 0 ⑤ 0 1 _⫺3 5

3 x₁ 1 _⫺1 0 0 1 3

row

c 0 5 0 0 _⫺3 Z_⫽9

Figure 4.5. Feasible region for Example 4.6.

point A. Thec row in Tableau 1 indicates that either x₁or x₂may be made a basic variable to increase Z. From Figure 4.5 it is clear that either x₁ or x₂ can be increased to increase Z. Having decided to increase x₁, it is clear that we cannot increase x₁beyond three units (point B) in order to remain in the feasible region. This was essentially the minimum-ratio value that was ob-tained by the simplex method. Thus, Tableau 2 is simply corner point B in Figure 4.5.

Tableau 3

Alternate Optima. In Tableau 3, the nonbasic variable x₅has a relative profit of zero. This means that no increase in x₅ will produce any change in the objective function value. In other words x₅ can be made a basic variable and the resulting basic feasible solution will also have Z_⫽14. By definition, any feasible solution whose value of Z equals the optimal value is also an optimal solution. Hence, we have an alternate optimal solution to this linear program.

This can be obtained by making x₅ a basic variable. By the minimum-ratio rule, x₃leaves the basis, and we have an alternative optimal tableau, as shown in Tableau 4. Thus the alternate optimal solution is given by x_{1 ⫽} –⁵₂, x_{2 ⫽}

x_{3 ⫽} 0, x_{4 ⫽} 0, and x_5⫽

13 15

––₄, ––₄.

In general, an alternate optimal solution is indicated whenever there exists a nonbasic variable whose relative profit(cjcoefficient) is zero in the optimal tableau. Referring to Figure 4.5, the movement from Tableau 2 to Tableau 3 by the simplex method corresponds to moving along the edge BC of the feasible region. Using the objective function Z _⫽ 3x_{1 ⫹} 2 x₂, it is clear that corner points C and D, as well as the line CD, are optimal.

Summary of Computational Steps

In summary, the computational steps of the simplex method in tableau form for a maximization problem are as follows:

In document Engineering Optimization (Page 179-186)