import sympy as sp
Ps=56
P1=4
P
2=18
P3=6,
c1=[
2,
2,
2,1,8,
2]
c
2=[3,3,3,1,1,3]
c3=[3,3,3,
2,
2,3]
c4=[5,5,5,5,5,4
0]
c5=[6,6,3
0,3
0,1
0,1
0]
p1=[
0.1,0.2,
0.1,0.2,
0.1,0.05]
p
2=[
0.1,0.2,
0.1,0.2,
0.2,
0.05]
p3=[
0.1,0.2,
0.1,0.2,
0.1,0.05]
N1=N
2=Nt=1
#N3=sp
.symbols
('N3'
)
#M3=sp
.symbols
('M3'
)
#NYY_YNY情况
def NYY_YNY
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-p
2)
M3=N3*
(1-
(1-p1
)*
(1-p3
))
income=N3*
(1-p1
)*
(1-p3
)*Ps
cost=Nt*
(c
2+P1+P
2)+N3*
(P3+c3
)
#拆分后的收支
pos=
(Nt-Nt*max
(p1,p
2))-
(N3-M3
)
cost_d=M3*
(c4+c1
)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*
(P3+c3
)+pos*p3*c4
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#NYN_YNN情况
def NYN_YNN
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-p
2)
M3=N3*
(1-
(1-p1
)*
(1-p3
))
income=N3*
(1-p1
)*
(1-p3
)*Ps
cost=Nt*
(c
2+P1+P
2)+N3*
(P3+c5
)
#拆分后的收支
pos=
(Nt-Nt*max
(p1,p
2))-
(N3-M3
)
cost_d=M3*
(c4+c1
)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*P3+pos*p3*
(c4+c5
)
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#YNY_NYY情况
def YNY_NYY
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-p1
)
M3=N3*
(1-
(1-p
2)*
(1-p3
))
income=N3*
(1-p
2)*
(1-p3
)*Ps
cost=Nt*
(c1+P1+P
2)+N3*
(P3+c3
)
#拆分后的收支
pos=
(Nt-Nt*max
(p1,p
2))-
(N3-M3
)
cost_d=M3*
(c4+c
2)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*
(P3+c3
)+pos*p3*c4
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#NNN_YYN情况
def NNN_YYN
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt
M3=Nt*
(1-
(1-p1
)*
(1-p
2)*
(1-p3
))
income=N3*
(1-p1
)*
(1-p
2)*
(1-p3
)*Ps
cost=Nt*
(P3+P1+P
2)+M3*c5
#拆分后的收支
pos=M3-Nt*max
(p1,p
2)
cost_d=M3*
(c4+c
2+c1
)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*P3+pos*p3*
(c4+c5
)
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#YYY_NNY情况
def YYY_NNY
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-max
(p1,p
2))
M3=Nt*p3
income=N3*
(1-p3
)*Ps
cost=Nt*
(c1+P1+P
2+c
2)+N3*
(P3+c3
)
#拆分后的收支
pos=N3*p3
cost_d=pos*c4
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*
(P3+c3
)+pos*p3*c4
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#YNN_NYN情况
def YNN_NYN
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-p1
)
M3=N3*
(1-
(1-p
2)*
(1-p3
))
income=N3*
(1-p
2)*
(1-p3
)*Ps
cost=Nt*
(c1+P1+P
2)+N3*P3+M3*c5
#拆分后的收支
pos=
(Nt-Nt*max
(p1,p
2))-
(N3-M3
)
cost_d=M3*
(c4+c
2)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*P3+pos*p3*
(c4+c5
)
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#YYN_NNN情况
def YYN_NNN
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt*
(1-max
(p1,p
2))
M3=Nt*p3
income=N3*
(1-p3
)*Ps
cost=Nt*
(c1+P1+P
2+c
2)+N3*
(P3+c5
)
#拆分后的收支
pos=N3*p3
cost_d=M3*c4
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*P3+pos*p3*
(c4+c5
)
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#NNY_YYY情况
def NNY_YYY
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt
M3=Nt*
(1-
(1-p1
)*
(1-p
2)*
(1-p3
))
income=N3*
(1-p1
)*
(1-p
2)*
(1-p3
)*Ps
cost=Nt*
(P1+P
2+P3+c3
)
#拆分后的收支
pos=
(Nt-Nt*max
(p1,p
2))-
(N3-M3
)
cost_d=M3*
(c4+c1+c
2)
income_d=pos*Ps
#迭代部分
while pos>
0:
cost_d=cost_d+pos*
(P3+c3
)+pos*p3*c4
pos=pos*p3
cost_total=cost_d+cost
income_total=income+income_d
probit=income_total-cost_total
return cost_total,income_total,probit
#NNN情况
def NNN
(p1,p
2,p3,c1,c
2,c3,c4,c5,P1=4,P
2=18,P3=6,Ps=56,Nt=1
):
N3=Nt
M3=Nt*
(1-
(1-p1
)*
(1-p
2)*
(1-p3
))
income=N3*
(1-p1
)*
(1-p
2)*
(1-p3
)*Ps
cost=Nt*
(P3+P1+P
2)+M3*c5
probit=income-cost
return cost,income,probit
#
循环求解
f=[NYY_YNY,NYN_YNN,YNY_NYY,YNN_NYN,YYN_NNN,NNY_YYY,YYY_NNY,NNN_YYN,NNN]
for i in range
(6
):
max_cost=
0
max_income=
0
max_probit=
0
for j in range
(9
):
cost_total,income_total,probit=f[j]
(p1[i],p
2[i],p3[i],c1[i],c
2[i],c3[i],c4[i],c5[i]
)
if max_probit<probit:
max_cost=cost_total
max_income=income_total
max_probit=probit
fy=f[j]
else:
continue
print
('第{:}个情况的最优方法为{:}，期望成本为{:<
20},期望利润为{:}'
.format
(i,str
(fy
),max_cost,max_probit
))
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