Fast computation of global solutions to the single-period unit commitment problem

Abstract

The single-period unit commitment problem has significant applications in electricity markets. An efficient global algorithm not only provides the optimal schedule that achieves the lowest cost, but also plays an important role for deriving the market-clearing price. As of today, the problem is mainly solved by using a general-purpose mixed-integer quadratic programming solver such as CPLEX or Gurobi. This paper proposes an extremely efficient global optimization algorithm for solving the problem. We propose a conjugate function based convex relaxation and design a special dual algorithm to compute a tight lower bound of the problem in ${\mathcal {O}}(n\log n)$ complexity. Then, a branch-and-bound algorithm is designed for finding a global solution to the problem. Computational experiments show that the proposed algorithm solves test instances with 500 integer variables in less than 0.01 s, whereas current state-of-the-art solvers fail to solve the same test instances in one hour. This superior performance of the proposed algorithm clearly indicates its potential in day-ahead and real-time electricity markets.

Appendix: NP-hardness of problem (SPUC)

In this section, we prove that problem (SPUC) is NP-hard. It is know that the “partition problem” is NP-hard (Garey and Johnson 1979). The partition problem is given as follows: “Given a set of positive integers $\{\alpha _1,\ldots ,\alpha _n\}$, whether there exists a subset $S \subseteq \{1,\ldots ,n\}$ such that $\sum _{i\in S} \alpha _i =\frac{1}{2}\sum _{i=1}^n \alpha _i$.”

We consider a decision problem of (SPUC) given as follows: “Let $a_i=b_i=0$ and $c_i=u_i=\alpha _i$ for $i=1,\ldots ,n$, and $d=\frac{1}{2}\sum _{i=1}^n \alpha _i$. Whether there exist a feasible solution $({\bar{x}},{\bar{y}})$ of problem (SPUC) such that $\sum _{i=1}^n(a_i{\bar{x}}_i^2+b_i {\bar{x}}_i+c_i {\bar{y}}_i)\le \frac{1}{2}\sum _{i=1}^n \alpha _i$.” Then, we show that an instance of partition problem is a yes instance if and only if the corresponding decision problem of (SPUC) is a yes instance.

If the instance of partition problem is a yes instance, i.e., it has a solution S such that $\sum _{i\in S} \alpha _i =\frac{1}{2}\sum _{i=1}^n \alpha _i$, then let ${\bar{x}}_i=\alpha _i,$ ${\bar{y}}_i=1$ for $i\in S$, and ${\bar{x}}_i={\bar{y}}_i=0$ for $i\in \{1,\ldots ,n\}-S$. Obviously, we have $\sum _{i=1}^n {\bar{x}}_i=d$ and $\sum _{i=1}^n(a_i {\bar{x}}_i^2+b_i {\bar{x}}_i+c_i {\bar{y}}_i)=\sum _{i\in S} c_i=\frac{1}{2}\sum _{i=1}^n \alpha _i$. Thus the corresponding decision problem of (SPUC) is a yes instance.

Conversely, if the decision problem of (SPUC) is a yes instance, i.e., there exist a feasible solution $({\bar{x}},{\bar{y}})$ of problem (SPUC) such that $\sum _{i=1}^n(a_i {\bar{x}}_i^2+b_i {\bar{x}}_i+c_i {\bar{y}}_i)\le \frac{1}{2}\sum _{i=1}^n \alpha _i$. Let $S=\{i| {\bar{x}}_i>0,~i=1,\ldots ,n\}$. Then we have ${\bar{y}}_i=1$ for all $i\in S$. Based on the relationship $0\le {\bar{x}}_i\le u_i {\bar{y}}_i$ and $c_i=u_i=\alpha _i$ for $i=1,\ldots ,n$, we have $\frac{1}{2}\sum _{i=1}^n \alpha _i\ge \sum _{i=1}^n(a_i {\bar{x}}_i^2+b_i {\bar{x}}_i+c_i {\bar{y}}_i)=\sum _{i=1}^n c_i {\bar{y}}_i\ge \sum _{i\in S} c_i{\bar{y}}_i\ge \sum _{i\in S} {\bar{x}}_i=d=\frac{1}{2}\sum _{i=1}^n \alpha _i$. Thus, $\sum _{i\in S} c_i{\bar{y}}_i=\sum _{i\in S} \alpha _i=\frac{1}{2}\sum _{i=1}^n \alpha _i$, and the instance of the partition problem is a yes instance.

Based on the above analysis, we conclude that problem (SPUC) is NP-hard.