最小化问题的reformulation

以一个最简单的线性规划问题为例。
$$\begin{array}{rl}\min & 0.02x1+0.07x2+0.04x3+0.03x4+0.05x5\\ \text{s.t.}& 0.3x1+2x2+x3+0.6x4+1.8x5>=70\\ & 0.1x1+0.05x2+0.02x3+0.2x4+0.05x5>=3\\ & 0.05x1+0.1x2+0.02x3+0.2x4+0.08x5>=9.1\end{array}\text{\hspace{1em}(1)}$$
使用Gurobi进行求解，python代码如下：

from gurobipy import *

# Create a new model
m = Model()

# Create variables
x1 = m.addVar(name="x1")
x2 = m.addVar(name="x2")
x3 = m.addVar(name="x3")
x4 = m.addVar(name="x4")
x5 = m.addVar(name="x5")
phi = m.addVar(name="phi")

# Set objective function
m.setObjective(0.02 * x1 + 0.07 * x2 + 0.04 * x3 + 0.03 * x4 + 0.05 * x5)

# Add constraints
m.addConstr(0.3 * x1 + 2 * x2 + x3 + 0.6 * x4 + 1.8 * x5 >= 70, "c2")
m.addConstr(0.1 * x1 + 0.05 * x2 + 0.02 * x3 + 0.2 * x4 + 0.05 * x5 >= 3, "c3")
m.addConstr(0.05 * x1 + 0.1 * x2 + 0.02 * x3 + 0.2 * x4 + 0.08 * x5 >= 9.1, "c4")

# Optimize model
m.optimize()

# Print values for decision variables
for v in m.getVars():
    print(v.varName, v.x)

# Print maximized profit value
print('Maximized profit:',  m.objVal)

问题（1）等价于：
$$\begin{array}{rl}\min & \varphi \\ \text{s.t.}& \varphi \ge 0.02x1+0.07x2+0.04x3+0.03x4+0.05x5\\ & 0.3x1+2x2+x3+0.6x4+1.8x5>=70\\ & 0.1x1+0.05x2+0.02x3+0.2x4+0.05x5>=3\\ & 0.05x1+0.1x2+0.02x3+0.2x4+0.08x5>=9.1\end{array}\text{\hspace{1em}(2)}$$
使用Gurobi进行求解，python代码如下：

from gurobipy import *

# Create a new model
m = Model()

# Create variables
x1 = m.addVar(name="x1")
x2 = m.addVar(name="x2")
x3 = m.addVar(name="x3")
x4 = m.addVar(name="x4")
x5 = m.addVar(name="x5")
phi = m.addVar(name="phi")

# Set objective function
m.setObjective(phi)

# Add constraints
m.addConstr(phi >= 0.02 * x1 + 0.07 * x2 + 0.04 * x3 + 0.03 * x4 + 0.05 * x5, "c1")
m.addConstr(0.3 * x1 + 2 * x2 + x3 + 0.6 * x4 + 1.8 * x5 >= 70, "c2")
m.addConstr(0.1 * x1 + 0.05 * x2 + 0.02 * x3 + 0.2 * x4 + 0.05 * x5 >= 3, "c3")
m.addConstr(0.05 * x1 + 0.1 * x2 + 0.02 * x3 + 0.2 * x4 + 0.08 * x5 >= 9.1, "c4")

# Optimize model
m.optimize()

# Print values for decision variables
for v in m.getVars():
    print(v.varName, v.x)

# Print maximized profit value
print('Maximized profit:',  m.objVal)