纸浆产生了一个简单的约束问题不是最优解

Question

纸浆解算犯规得到最佳的结果，而这是可以的。难道是不明确的问题，或者是有在图书馆或约束的问题？

通常这是给最适合大多数的制约可能性，但它不是为这个特定的情况下。

from pulp import *
prob = LpProblem("minimization", LpMinimize)

#introducing vars
a=LpVariable("a", 0, None)
b=LpVariable("b", 0, None)
c=LpVariable("c", 0, None)
d=LpVariable("d", 0, None)
e=LpVariable("e", 0, None)

#introducing main obj and constraints
#basically addition of all vars should be minimized to 1.
prob+=a+b+c+d+e>=1,"main objective"
prob+=a<=0.3,"a const"
prob+=b<=0.3,"b const"
prob+=c<=0.3,"c const"
prob+=d<=0.3,"d const"   #can change to 0
prob+=e==0.1,"e const"

"""
for this const. 
the res is:
a=0.3
b=0.3
c=0.3
d=0.3
e=0.1
a+b+c+d+e=1.3

if you change prob+=d<=0.3 to prob+=d<=0
then:
a=0.3
b=0.3
c=0.3
d=0.0
e=0.1
a+b+c+d+e=1

"""


#solves the problem
status = prob.solve()
#checks if optimal or infeasible
print("status:",LpStatus[prob.status])

#shows all the vars
for variable in prob.variables():
    print("{} = {}".format(variable.name, variable.varValue))

这个常量。

该水库是：

一个= 0.3

B = 0.3

C = 0.3

d = 0.3

E = 0.1

A + B + C + d + E = 1.3

它应该是1，因为它是LpMinimize。

如果更改的概率+ = d <= 0.3〜概率+ = d <= 0

然后：

一个= 0.3

B = 0.3

C = 0.3

d = 0.0

E = 0.1

A + B + C + d + E = 1

Answer 1

这是一种约束，而不是一个目标：

prob+=a+b+c+d+e>=1,"main objective"

从本质上讲，你没有一个客观的，因此求解器只是寻找一个可行的解决方案。

尝试

prob+=a+b+c+d+e,"main objective"
prob+=a+b+c+d+e>=1,"constraint"

现在，你都尽量减少a+b+c+d+e以及具有在此约束。

结论：我认为纸浆是正确的在这里。