6. Illustrative example

Return to the transportation problem of the model and to the descriptions of cells. Consider the graph Γð Þ B<sup>k</sup> . This graph can have more than one connected components. Let τ be number of connected components, and i∈Iν, ð Þ m þ j for j∈J<sup>ν</sup> be the vertices of ν-th component. It is easy

The linear system (1) for the cell Ξ<sup>k</sup> defines coordinates qj on each connected component up to

(i) <sup>r</sup><sup>k</sup> <sup>∉</sup>Ξk. Since <sup>r</sup><sup>k</sup> is a maximum point on Mk for the strictly concave function <sup>ψ</sup>ð Þ<sup>q</sup> , the value of the function will increase for the moving point q tðÞ¼ ð Þ <sup>1</sup> � <sup>t</sup> qk <sup>þ</sup> tr<sup>k</sup><sup>Þ</sup> when <sup>t</sup> increases in [0,1].

inequalities (2) for <sup>p</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ is fulfilled as equality. Choose one of them. Corresponding edge ð Þ l; m þ j will be added to graph. It unites two of connected components. We obtain

It should be noted that the dimension of the cell Ξ reduces. It will certainly be r<sup>k</sup> ∈Ξ<sup>k</sup> when the

an increase of the function'<sup>s</sup> <sup>ψ</sup>ð Þ<sup>q</sup> value. We verify qk <sup>∈</sup> <sup>Ω</sup>k? For this, we obtain from the equations of the transportation problem the variables zij, ið Þ ; j ∈ Bk, as linear functions zijð Þp and check zij q<sup>k</sup> � � ≥ 0. If it is true, the point q<sup>k</sup> is the required fixed point. Otherwise, we have

<sup>j</sup> , j∈ Jν:

λi, ν ¼ 1, …, τ: (6)

, we need to put pj ¼ qj in corresponding Eq. (6), which

<sup>∗</sup> < 1. Some of corresponding

. But it can occur earlier.

. Otherwise, we can simply replace qk by r<sup>k</sup> with

to verify that the linear system (3) for Lk is going to be equivalent to this one:

pj <sup>¼</sup> <sup>X</sup> i ∈I<sup>ν</sup>

qj <sup>¼</sup> <sup>t</sup>νq<sup>k</sup>

X j∈ J<sup>ν</sup>

a positive multiplier:

36 Optimization Algorithms - Examples

gives the multiplier tν.

To obtain the coordinates of the point r<sup>k</sup>

For the obtained point, r<sup>k</sup> can be realized in two cases.

In considered case, this point reaches a face of Ξ<sup>k</sup> at some t ¼ t

<sup>B</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>B</sup>k∪f g ð Þ <sup>l</sup>; <sup>j</sup> , accept <sup>q</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ and pass to the next step.

current cell <sup>Ξ</sup><sup>k</sup> degenerates into a point, and we have <sup>r</sup><sup>k</sup> <sup>¼</sup> <sup>q</sup><sup>k</sup>

zsl <sup>q</sup><sup>k</sup> � � <sup>&</sup>lt; 0. We accept <sup>B</sup><sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>B</sup><sup>k</sup> f g ð Þ <sup>s</sup>; <sup>l</sup> , q<sup>k</sup>þ<sup>1</sup> <sup>¼</sup> <sup>q</sup><sup>k</sup> and pass to the next step.

(ii) <sup>r</sup><sup>k</sup> <sup>∈</sup>Ξk. In this case, we can assume qk <sup>¼</sup> <sup>r</sup><sup>k</sup>

Figure 4. Illustration of one step of the algorithm.

We show how the described method works on the Fisher's model example of Section 3.

For the start, we need a structure B<sup>1</sup> ∈ B and a point q<sup>1</sup> ∈Ξ B<sup>1</sup> � �: We depict the structures as matrices m � n with elements from f g �; � , and � corresponds to an element of B. For example, the structure B<sup>12</sup> ¼ f g ð Þ 1; 2 ;ð Þ 1; 3 ;ð Þ 2; 1 ;ð Þ 2; 2 will be depicted as the matrix

$$\mathcal{B}\_{12} = \begin{pmatrix} \cdot & \times & \times\\ \times & \times & \cdot \end{pmatrix}$$

(this is the structure for the cell Ω12). Let us start with the structure

$$\mathcal{B}^1 = \begin{pmatrix} \times & \cdot & \cdot\\ \times & \cdot & \cdot \end{pmatrix}$$

It means that both consumers prefer only first good. Let us choose as q<sup>1</sup> the price vector <sup>q</sup><sup>1</sup> <sup>¼</sup> ð Þ <sup>0</sup>:05; <sup>0</sup>:35; <sup>0</sup>:<sup>6</sup> . It is easy to verify that <sup>B</sup><sup>1</sup> <sup>∈</sup> <sup>B</sup> and <sup>q</sup><sup>1</sup> <sup>∈</sup><sup>Ξ</sup> <sup>B</sup><sup>1</sup> � �.

Step 1. The graph Γ B<sup>1</sup> � � has three connected components and the system (6) has the form

$$p\_1 = 1, p\_2 = 0, p\_3 = 0.$$

Thus, we have <sup>r</sup><sup>1</sup> <sup>¼</sup> ð Þ <sup>1</sup>; <sup>0</sup>; <sup>0</sup> . The cell <sup>Ξ</sup> <sup>B</sup><sup>1</sup> � � is given by the system

$$\frac{q\_1}{1} \le \frac{q\_2}{2} \,\, :\,\tag{7}$$

$$\frac{q\_1}{1} \le \frac{q\_3}{3},\tag{8}$$

We have q<sup>1</sup> ∈Ξ B<sup>1</sup> � � and r<sup>1</sup> ∉Ξ B<sup>1</sup> � �. It is the case (i) in the description of algorithm. We have to move the point <sup>q</sup><sup>1</sup> to the point <sup>r</sup>1. For the moving point q tð Þ it will be:

$$q\_1(t) = 0.05 + 0.95t, \quad q\_2(t) = (1 - t)0.35, \quad q\_3(t) = (1 - t)0.6.$$

This point reaches a face of <sup>Ξ</sup> <sup>B</sup><sup>1</sup> � � at <sup>t</sup> <sup>¼</sup> <sup>t</sup> <sup>∗</sup> <sup>¼</sup> <sup>0</sup>:1111: the inequality (7) for <sup>q</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ is fulfilled as equality. We obtain <sup>B</sup><sup>2</sup> <sup>¼</sup> <sup>B</sup><sup>1</sup> <sup>∪</sup>f g ð Þ <sup>1</sup>; <sup>2</sup> and <sup>q</sup><sup>2</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ.

$$\begin{aligned} \mathcal{B}^2 &= \begin{pmatrix} \times & \times & \cdot\\ \times & \cdot & \cdot \end{pmatrix} \\ q^2 &= (0.1556, 0.3111, 0.5333). \end{aligned}$$

Hence,

At t

The equilibrium price vector is

Figure 5. Movement to equilibrium in the example model.

r

q2 <sup>2</sup> <sup>¼</sup> <sup>q</sup><sup>3</sup>

q1 3 ≥ q2 2 :

It is easy to see that the description of the cell Ξ B<sup>4</sup> � � has the form:

<sup>4</sup> <sup>¼</sup> ð Þ <sup>0</sup>:5; <sup>0</sup>:2; <sup>0</sup>:<sup>3</sup> :

<sup>3</sup> , <sup>q</sup><sup>1</sup>

For <sup>q</sup> <sup>¼</sup> <sup>r</sup>4, the inequality (9) is violated, so <sup>r</sup><sup>4</sup> <sup>∉</sup><sup>Ξ</sup> <sup>B</sup><sup>4</sup> � �. For the new moving point q tð Þ we have:

q1ðÞ¼ t 0:1667 þ 0:3333t, q2ðÞ¼ t 0:3333 � 0:1333t, q<sup>3</sup> ¼ 0:5 � 0:2t:

<sup>B</sup><sup>5</sup> <sup>¼</sup> � �� ��� !

~p ¼ ð Þ 0:375; 0:25; 0:375

and new point <sup>q</sup><sup>5</sup> <sup>¼</sup> <sup>c</sup>12. It is easy to verify that we obtain the equilibrium of the model.

<sup>c</sup><sup>12</sup> <sup>¼</sup> ð Þ <sup>0</sup>:375; <sup>0</sup>:25; <sup>0</sup>:<sup>375</sup> that is the boundary of <sup>Ξ</sup> <sup>B</sup><sup>4</sup> � �. We obtain the new structure:

<sup>∗</sup> <sup>¼</sup> <sup>0</sup>:625 the inequality (9) becomes equality, the point q tð Þ attains to the point

<sup>1</sup> <sup>≥</sup> <sup>q</sup><sup>3</sup> 3 ,

Polyhedral Complementarity Approach to Equilibrium Problem in Linear Exchange Models

http://dx.doi.org/10.5772/intechopen.77206

(9)

39

Step 2. The graph Γ B<sup>2</sup> � � has two connected components and the system (6) has the form

$$p\_1 + p\_2 = 1,\\ p\_3 = 0.$$

For the point r2, we have to consider this system with the additional equation

$$\frac{p\_1}{1} = \frac{p\_2}{2}\_{\prime\prime}$$

corresponding to (7), this gives <sup>r</sup><sup>2</sup> <sup>¼</sup> ð Þ <sup>0</sup>:3333; <sup>0</sup>:6667; <sup>0</sup> . We have <sup>r</sup><sup>2</sup> <sup>∉</sup><sup>Ξ</sup> <sup>B</sup><sup>2</sup> � � since the inequality (8) is violated. The new moving point q tðÞ¼ ð Þ <sup>1</sup> � <sup>t</sup> <sup>q</sup><sup>2</sup> <sup>þ</sup> tr<sup>2</sup> has the coordinates:

$$q\_1(t) = 0.1556 + 0.1777t, \quad q\_2(t) = 0.3111 + 0.3556t, \quad q\_3(t) = 0.5333 - 0.5333t.$$

At t <sup>∗</sup> <sup>¼</sup> <sup>0</sup>:0625 this point reaches the boundary of the cell <sup>Ξ</sup> <sup>B</sup><sup>2</sup> � �. It is the point <sup>c</sup><sup>1</sup> in the simplex <sup>σ</sup>. The inequality (8) for <sup>q</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ is fulfilled as equality. Thus, we obtain:

$$\mathcal{B}^3 = \begin{pmatrix} \times & \times & \times \\ \times & \cdot & \cdot \end{pmatrix}$$

$$q^3 = (0.1667, 0.3333, 0.5).$$

Step 3. Now, we have the case (ii) in the description of algorithm: the cell Ξ B<sup>3</sup> � � contains unique point <sup>q</sup><sup>3</sup> and thus <sup>r</sup><sup>3</sup> <sup>¼</sup> <sup>q</sup>3. We have to verify <sup>r</sup><sup>3</sup> <sup>∈</sup> <sup>Ω</sup> <sup>B</sup><sup>3</sup> � �? For this, we obtain from the equations of the transportation problem the variables zij, ið Þ ; <sup>j</sup> <sup>∈</sup>B<sup>3</sup> , and check zij q<sup>3</sup> � � ≥ 0. For these variables, we have the system:

$$z\_{12} = q\_{2'}^3 \quad z\_{13} = q\_{3'}^3 \quad z\_{21} = 0.5, \quad z\_{11} = q\_1^3 - 0.5$$

We obtain z<sup>11</sup> = 0.1667–0.5 = �0.3333 < 0. Thus the element 1ð Þ ; 1 should be removed from the structure B<sup>3</sup> :

$$\begin{aligned} \mathcal{B}^4 &= \begin{pmatrix} \cdot & \times & \times\\ \times & \cdot & \cdot \end{pmatrix},\\ q^4 &= q^3. \end{aligned}$$

Step 4. We have to obtain the point r4. The graph Γ B<sup>4</sup> � � has two connected components and the system for this point has the form:

$$\begin{array}{ll}p\_1 = 0.5, & p\_2 + p\_3 = 0.5, \\ p\_2 = \frac{p\_3}{3}. \end{array}$$

Hence,

<sup>B</sup><sup>2</sup> <sup>¼</sup> ���

For the point r2, we have to consider this system with the additional equation

(8) is violated. The new moving point q tðÞ¼ ð Þ <sup>1</sup> � <sup>t</sup> <sup>q</sup><sup>2</sup> <sup>þ</sup> tr<sup>2</sup> has the coordinates:

<sup>σ</sup>. The inequality (8) for <sup>q</sup> <sup>¼</sup> q t<sup>∗</sup> ð Þ is fulfilled as equality. Thus, we obtain:

equations of the transportation problem the variables zij, ið Þ ; <sup>j</sup> <sup>∈</sup>B<sup>3</sup>

<sup>2</sup>, z<sup>13</sup> <sup>¼</sup> <sup>q</sup><sup>3</sup>

p2 <sup>2</sup> <sup>¼</sup> <sup>p</sup><sup>3</sup> 3 :

<sup>z</sup><sup>12</sup> <sup>¼</sup> <sup>q</sup><sup>3</sup>

these variables, we have the system:

system for this point has the form:

At t

structure B<sup>3</sup>

:

38 Optimization Algorithms - Examples

�� �

Step 2. The graph Γ B<sup>2</sup> � � has two connected components and the system (6) has the form

p1 <sup>1</sup> <sup>¼</sup> <sup>p</sup><sup>2</sup> 2 ,

<sup>q</sup><sup>2</sup> <sup>¼</sup> ð Þ <sup>0</sup>:1556; <sup>0</sup>:3111; <sup>0</sup>:<sup>5333</sup> :

p<sup>1</sup> þ p<sup>2</sup> ¼ 1, p<sup>3</sup> ¼ 0:

corresponding to (7), this gives <sup>r</sup><sup>2</sup> <sup>¼</sup> ð Þ <sup>0</sup>:3333; <sup>0</sup>:6667; <sup>0</sup> . We have <sup>r</sup><sup>2</sup> <sup>∉</sup><sup>Ξ</sup> <sup>B</sup><sup>2</sup> � � since the inequality

q1ðÞ¼ t 0:1556 þ 0:1777t, q2ðÞ¼ t 0:3111 þ 0:3556t, q3ðÞ¼ t 0:5333 � 0:5333t:

!

<sup>q</sup><sup>3</sup> <sup>¼</sup> ð Þ <sup>0</sup>:1667; <sup>0</sup>:3333; <sup>0</sup>:<sup>5</sup> :

Step 3. Now, we have the case (ii) in the description of algorithm: the cell Ξ B<sup>3</sup> � � contains unique point <sup>q</sup><sup>3</sup> and thus <sup>r</sup><sup>3</sup> <sup>¼</sup> <sup>q</sup>3. We have to verify <sup>r</sup><sup>3</sup> <sup>∈</sup> <sup>Ω</sup> <sup>B</sup><sup>3</sup> � �? For this, we obtain from the

We obtain z<sup>11</sup> = 0.1667–0.5 = �0.3333 < 0. Thus the element 1ð Þ ; 1 should be removed from the

!

<sup>B</sup><sup>4</sup> <sup>¼</sup> � �� �� �

Step 4. We have to obtain the point r4. The graph Γ B<sup>4</sup> � � has two connected components and the

p<sup>1</sup> ¼ 0:5, p<sup>2</sup> þ p<sup>3</sup> ¼ 0:5,

<sup>q</sup><sup>4</sup> <sup>¼</sup> <sup>q</sup><sup>3</sup>:

<sup>3</sup>, z<sup>21</sup> <sup>¼</sup> <sup>0</sup>:5, z<sup>11</sup> <sup>¼</sup> <sup>q</sup><sup>3</sup>

<sup>1</sup> � 0:5

, and check zij q<sup>3</sup> � � ≥ 0. For

<sup>B</sup><sup>3</sup> <sup>¼</sup> ��� �� �

<sup>∗</sup> <sup>¼</sup> <sup>0</sup>:0625 this point reaches the boundary of the cell <sup>Ξ</sup> <sup>B</sup><sup>2</sup> � �. It is the point <sup>c</sup><sup>1</sup> in the simplex

!

$$r^4 = (0.5, 0.2, 0.3).$$

It is easy to see that the description of the cell Ξ B<sup>4</sup> � � has the form:

$$\begin{array}{ll} \frac{q\_2}{2} = \frac{q\_3}{3}, & \frac{q\_1}{1} \ge \frac{q\_3}{3},\\ \frac{q\_1}{3} \ge \frac{q\_2}{2}. \end{array} \tag{9}$$

For <sup>q</sup> <sup>¼</sup> <sup>r</sup>4, the inequality (9) is violated, so <sup>r</sup><sup>4</sup> <sup>∉</sup><sup>Ξ</sup> <sup>B</sup><sup>4</sup> � �. For the new moving point q tð Þ we have:

$$q\_1(t) = 0.1667 + 0.3333t, \quad q\_2(t) = 0.3333 - 0.1333t, \quad q\_3 = 0.5 - 0.2t.$$

At t <sup>∗</sup> <sup>¼</sup> <sup>0</sup>:625 the inequality (9) becomes equality, the point q tð Þ attains to the point <sup>c</sup><sup>12</sup> <sup>¼</sup> ð Þ <sup>0</sup>:375; <sup>0</sup>:25; <sup>0</sup>:<sup>375</sup> that is the boundary of <sup>Ξ</sup> <sup>B</sup><sup>4</sup> � �. We obtain the new structure:

$$\mathcal{B}^5 = \begin{pmatrix} \cdot & \times & \times\\ \times & \times & \cdot \end{pmatrix}$$

and new point <sup>q</sup><sup>5</sup> <sup>¼</sup> <sup>c</sup>12. It is easy to verify that we obtain the equilibrium of the model. The equilibrium price vector is

$$\vec{p} = (0.375, 0.25, 0.375)$$

Figure 5. Movement to equilibrium in the example model.

The optimal solutions of the consumer's problems are:

$$
\tilde{\mathfrak{x}}^1 = (0, 0.5, 1), \quad \tilde{\mathfrak{x}}^2 = (1, 0.5, 0).
$$

Figure 5 shows the moving of the point q tð Þ to the equilibrium.
