Group Search Optimization Technique for Multi-area Economic Dispatch

Abstract

Group search optimization to solve multi-area economic dispatch (MAED) problem is presented in this paper with transmission losses, constraints in the tie-line with different fuels, the valve point loading effect, and the prohibited operating zones. The method proposed has been examined on two different test systems, large and small, considering a changing degree of complexity. Then, the comparison has been made with evolutionary programming, differential evolution, and real-coded genetic algorithm where the solution quality is considered. The method which is proposed here gives an alternative approach which is very promising solution for solving one of the power system problems like MAED.

Appendices

Appendix 1

See Table 3.

Table 3 Data for two-area system

The formula coefficients of transmission loss of two-area system are

$$\begin{aligned} {\text{B}}_{1} & = \left[ {\begin{array}{*{20}c} {17} & {12} & 7 \\ {12} & {14} & 9 \\ 7 & 9 & {31} \\ \end{array} } \right] \times 10^{ - 6} \\ {\text{B}}_{01} & = \left[ {\begin{array}{*{20}c} { - 0.3908} & { - 0.1297} & {0.7047} \\ \end{array} } \right] \times 10^{ - 3} \\ {\text{B}}_{001} & = 0.045 \\ {\text{B}}_{2} & = \left[ {\begin{array}{*{20}c} {24} & { - 6} & { - 8} \\ { - 5} & {129} & { - 2} \\ { - 8} & { - 2} & {150} \\ \end{array} } \right] \times 10^{ - 6} \\ {\text{B}}_{02} & = \left[ {\begin{array}{*{20}c} {0.0591} & {0.2161} & { - 0.6635} \\ \end{array} } \right] \times 10^{ - 3} \\ {\text{B}}_{002} & = 0.056 \\ \end{aligned}$$

Appendix 2

Formula coefficients of transmission loss of three-area system:

$$\begin{aligned} {\text{B}}_{1} & = \left[ {\begin{array}{*{20}c} {8.70} & {0.43} & { - 4.61} & {0.36} \\ {0.43} & {8.30} & { - 0.97} & {0.22} \\ { - 4.61} & { - 0.97} & {9.00} & { - 2.00} \\ {0.36} & {0.22} & { - 2.00} & {5.30} \\ \end{array} } \right] \times 10^{ - 5} \\ {\text{B}}_{01} & = \left[ {\begin{array}{*{20}c} { - 0.3908} & { - 0.1297} & {0.7047} & {0.0591} \\ \end{array} } \right] \times 10^{ - 3} \\ {\text{B}}_{001} & = 0.056 \\ {\text{B}}_{2} & = \left[ {\begin{array}{*{20}c} {8.60} & { - 0.80} & {0.37} \\ { - 0.80} & {9.08} & { - 4.90} \\ {0.37} & { - 4.90} & {8.24} \\ \end{array} } \right] \times 10^{ - 5} \\ {\text{B}}_{02} & = \left[ {\begin{array}{*{20}c} {0.2161} & { - 0.6635} & {0.5034} \\ \end{array} } \right] \times 10^{ - 3} \\ {\text{B}}_{002} & = 0.045 \\ \end{aligned}$$

$$\begin{aligned} {\text{B}}_{3} & = \left[ {\begin{array}{*{20}c} {1.20} & { - 0.96} & {0.56} \\ { - 0.96} & {4.93} & { - 0.30} \\ {0.56} & { - 0.30} & {5.99} \\ \end{array} } \right] \times 10^{ - 5} \\ {\text{B}}_{03} & = \left[ {\begin{array}{*{20}c} { - 0.3216} & {0.4635} & {0.3503} \\ \end{array} } \right] \times 10^{ - 3} \\ {\text{B}}_{003} & = 0.055 \\ \end{aligned}$$