如何解决Python中线性程序的多种解决方案？

问题描述

我有一些 Pyomo 代码（见下文），可以生成线性方程的一种可能解：

3a + 2b + 2c = 15

地点：

-20

a,b,c 是实数

如何重写程序以在每次运行时生成随机解决方案？我也愿意使用 Pulp。

我的代码：

# Coefficients
coeffs = [3,2,2]
y_target = 15

# Define model and variables
model = ConcreteModel()
model.a = Var(bounds=(-20,0))
model.b = Var(bounds=(0,20))
model.c = Var(bounds=(20,50))
model_x = [model.a,model.b,model.c]
    
# Model Constraint
def weight_rule(model):
    return sum(coeffs[i] * model_x[i] for i in range(len(coeffs))) == y_target 
model.weight = Constraint( rule=weight_rule )

# Model Objective
obj = sum(coeffs[i] * model_x[i] for i in range(len(coeffs))) - y_target 
model.objective = Objective(expr=obj,sense=minimize)

# Solve the model
solver = SolverFactory('glpk')
solver.solve(model)

# Print results
print([value(model_x[i]) for i in range(len(coeffs))])

# Current Output:
[-8.33333,20]

编辑 - 新方法（仍然无效）： 下面我使用@Erwin Kalvelagen 描述的方法改编了@kabdulla 提供的一些有用的代码 here，但不能完全让它适用于我的情况。任何建议将不胜感激。

import pulp as pulp

# Accounting:
# n structural varuables (n = 3)
# m constraints (m = 2)
# => No. of basics = 2 (no. of constraints)
# => No. of non-basics = 2 (no. of variables)

A = []    # Empty list for solutions to exclude
nb = 3
M = 100   # large M value - upper bound for x1,x2,x3 * the slacks
numSolutions = 5 # How many solutions we seek by iterating

# Set lower and upper bound ranges for respective variables
lowers = [-20,20]
uppers = [0,20,50]

for _ in range(numSolutions):

    model = pulp.LpProblem('get all basis',pulp.LpMaximize)

    # Variables
    x = pulp.LpVariable.dicts('x',range(3),cat='Continuous')
    for i in x.keys():
         x[i].lowBound = lowers[i]
         x[i].upBound = uppers[i]

    # Non-negative Slack Variables - one for each constraint
    s = pulp.LpVariable.dicts('s',range(1),lowBound=0,upBound=None,cat='Continuous')

    # Basis variables (binary)
    # one for each variable & one for each constraint (& so slack)
    B_x = pulp.LpVariable.dicts('b_x',range(len(x)),cat='Binary')
    B_s = pulp.LpVariable.dicts('b_s',range(len(s)),cat='Binary')

    # Objective
    model += 3*x[0] + 2*x[1] + 2*x[2]

    # Constraints - with explicit slacks
    model += 3*x[0] + 2*x[1] + 2*x[2] + s[0] == 15

    # No. of basics is correct:
    model += pulp.lpSum(B_x) + pulp.lpSum(B_s) == nb

    # Enforce basic and non-basic behavIoUr
    for i in range(len(x)):
        model += x[i] <= M*B_x[i]

    for i in range(len(s)):
        model += s[i] <= M*B_s[i]
 
    for a in A:
        model += (B_x[0]*a[0] + B_x[1]*a[1] + B_x[2]*a[2] + B_s[0]*a[3]) <= nb - 1 

    model.solve()
    
    # Cuts - already discovered solutions
    exc = [v.varValue for v in model.variables()][:4]
    exc = exc[1:4] + exc[0:1]
    
    A.append(exc)

    print('Status:',pulp.LpStatus[model.status])
    print('Objective:',pulp.value(model.objective))

    for v in model.variables():
        print (v.name,"=",v.varValue)

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

如果你已经找到好的解决方法，欢迎将解决方案带上本链接一起发送给小编。

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linear-programming pulp pyomo python