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问Gekko离散优化标引问题
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Stack Overflow用户

提问于 2022-10-18 08:00:08

回答 1查看 42关注 0票数 2

关于这个问题，我有个问题。我想让模型搜索装载机、loader_size、卡车、truck_size的最优值，以解决总成本和总工期的两个约束。然而，我面临着一个关于成本的问题。由于装载机的大小与装载机的成本有着密切的关系，所以装载机的大小与装载机的成本必须有一定的相关性。例如，在此代码中，如果卡车大小= 20，卡车成本必须等于= 62500，因为它们相互依赖，装载机大小和装载机成本是相同的。请帮我解决这个问题。谢谢。

import numpy as np
from gekko import GEKKO
import random
m = GEKKO()
m.options.SOLVER = 3


Total_Cost = m.Const(15000000)
Total_Duration = m.Const(200)
Q = 10000
k = 1
f = 0.87
E = 0.7
s_load = 7
s_empty = 8
dist = 0.3
dens = 1200
cm = 0.98

#VARIABLE
Loader = m.Var(value = 1, lb=1, ub=2,integer = True)
Truck = m.Var(value = 1, lb=1, ub=5,integer = True)
Loader_Size = m.sos1([2.5, 1.7,1])      
Loader_Cost = m.sos1([21000, 19000,17000])
Truck_Size = m.sos1([20,25.5,15])
Truck_Cost = m.sos1([62500,80000,57000]) 
                             
#DEPENDENT VARIABLE

Travel_T = m.Intermediate((dist/s_load+dist/s_empty)*60)
Unit_Load_Q = m.Intermediate(Truck_Size/dens*1000)
Unit_Load_T = m.Intermediate(Unit_Load_Q/(Loader_Size*k*f*E/cm))
Cycle_Num = m.Intermediate(Q/Unit_Load_Q)
oper_prod = m.min3(Loader/Unit_Load_T,Truck/(Unit_Load_T+Travel_T))
#CONSTRAINTS

a = m.Intermediate(Cycle_Num/oper_prod/60)
b = m.Intermediate((Loader*Loader_Cost)*a)
c = m.Intermediate((Truck*Truck_Cost)*a)
d = m.Intermediate(b+c)


m.Equation(a < Total_Duration)
m.Equation(a > 0)
m.Equation(d < Total_Cost)
m.Equation(d > 0)

#SIMULATION
m.Minimize(d)
m.Minimize(a)
m.solve(disp = False)

print (Loader.value, Loader_Size.value, Loader_Cost.value, Truck.value, Truck_Size.value, Truck_Cost.value)

python

optimization

gekko

discrete

回答 1

Stack Overflow用户

发布于 2022-10-20 05:17:34

使用三次样条(c样条)查找大小和成本的值。它们是点之间的插值，但与整数值的大小和代价完全相等。下面是一个成功解决问题的脚本：

import numpy as np
from gekko import GEKKO
import random
m = GEKKO()

Total_Cost = m.Const(15000000)
Total_Duration = m.Const(200)
a = m.Var(lb=0,ub=Total_Duration)
d = m.Var(lb=0,ub=Total_Cost)
Q = 10000
k = 1
f = 0.87
E = 0.7
s_load = 7
s_empty = 8
dist = 0.3
dens = 1200
cm = 0.98

#VARIABLE
Loader = m.Var(value=1, lb=1, ub=5,integer = True)
Loader_Size=m.Var(1.7)
Loader_Cost=m.Var(19000)
Ldr = np.array([1,2,3,4,5])
Ldrs= np.array([2.5,1.7,1,0.5,0.25])
Ldrc= np.array([21000,19000,17000,13000,9000])
m.cspline(Loader,Loader_Size,Ldr,Ldrs,True)
m.cspline(Loader,Loader_Cost,Ldr,Ldrc,True)
#Loader_Size = m.sos1([2.5, 1.7,1])      
#Loader_Cost = m.sos1([21000, 19000,17000])

Truck = m.Var(value=1, lb=1, ub=4,integer = True)
Truck_Size=m.Var(15)
Truck_Cost=m.Var(57000)
Tr = np.array([1,2,3,4])
Trs= np.array([25.5,20,15,10])
Trc= np.array([80000,62500,57000,32000])
m.cspline(Truck,Truck_Size,Tr,Trs,True)
m.cspline(Truck,Truck_Cost,Tr,Trc,True)
#Truck_Size = m.sos1([20,25.5,15,10])
#Truck_Cost = m.sos1([62500,80000,57000,32000]) 
                             
#DEPENDENT VARIABLE

Travel_T = m.Intermediate((dist/s_load+dist/s_empty)*60)
Unit_Load_Q = m.Intermediate(Truck_Size/dens*1000)
Unit_Load_T = m.Intermediate(Unit_Load_Q/(Loader_Size*k*f*E/cm))
Cycle_Num = m.Intermediate(Q/Unit_Load_Q)
oper_prod = m.min3(Loader/Unit_Load_T,Truck/(Unit_Load_T+Travel_T))
#CONSTRAINTS

b = m.Intermediate((Loader*Loader_Cost)*a)
c = m.Intermediate((Truck*Truck_Cost)*a)

# Declare as variables and equations
#m.Equation(a < Total_Duration)
#m.Equation(a > 0)
#m.Equation(d < Total_Cost)
#m.Equation(d > 0)
m.Equation(oper_prod*a==Cycle_Num/60)
m.Equation(d==b+c)

#SIMULATION
m.Minimize(d)
m.Minimize(a)

# initialize with IPOPT
m.options.SOLVER = 3
m.solve(disp = True)

# solve MINLP with APOPT
m.options.SOLVER = 1
m.solve(disp = True)

print(Loader.value, Loader_Size.value, Loader_Cost.value,
      Truck.value, Truck_Size.value, Truck_Cost.value)

解决办法是：

  70  1.2917685e+07 6.98e-05 7.47e+07  -0.6 1.30e+00    -  1.00e+00 1.99e-01f  2
  71  1.2917684e+07 2.01e-07 7.07e+05  -1.5 4.44e-01    -  1.00e+00 1.00e+00f  1
  72  1.2917684e+07 3.15e-07 6.33e+06  -2.2 3.66e-02    -  1.00e+00 4.63e-01f  2
  73  1.2917684e+07 4.75e-07 5.95e+04  -3.6 2.71e-02    -  1.00e+00 9.91e-01f  1
  74  1.2917684e+07 1.48e-12 6.89e+01  -5.4 2.94e-03    -  1.00e+00 9.99e-01f  1
  75  1.2917684e+07 1.86e-09 1.38e-03  -7.4 2.10e-05    -  1.00e+00 1.00e+00f  1

Number of Iterations....: 75

                                   (scaled)                 (unscaled)
Objective...............:   1.2917684089800559e+07    1.2917684089800559e+07
Dual infeasibility......:   1.3803296452324758e-03    1.3803296452324758e-03
Constraint violation....:   2.4508488805670487e-12    1.8626451492309570e-09
Complementarity.........:   3.9921377093282655e-08    3.9921377093282655e-08
Overall NLP error.......:   1.1157030658549702e-08    1.3803296452324758e-03


Number of objective function evaluations             = 193
Number of objective gradient evaluations             = 72
Number of equality constraint evaluations            = 193
Number of inequality constraint evaluations          = 0
Number of equality constraint Jacobian evaluations   = 77
Number of inequality constraint Jacobian evaluations = 0
Number of Lagrangian Hessian evaluations             = 75
Total CPU secs in IPOPT (w/o function evaluations)   =      0.038
Total CPU secs in NLP function evaluations           =      0.012

EXIT: Optimal Solution Found.
 
 The solution was found.
 
 The final value of the objective function is    12917684.0898006     
 
 ---------------------------------------------------
 Solver         :  IPOPT (v3.12)
 Solution time  :   7.190000000264263E-002 sec
 Objective      :    12917684.0898006     
 Successful solution
 ---------------------------------------------------
 
apm 136.36.4.38_gk_model0 <br><pre> ----------------------------------------------------------------
 APMonitor, Version 1.0.1
 APMonitor Optimization Suite
 ----------------------------------------------------------------
 
 
 --------- APM Model Size ------------
 Each time step contains
   Objects      :            4
   Constants    :            2
   Variables    :           12
   Intermediates:            6
   Connections  :            8
   Equations    :           13
   Residuals    :            7
 
 Number of state variables:             12
 Number of total equations: -            9
 Number of slack variables: -            2
 ---------------------------------------
 Degrees of freedom       :              1
 
 ----------------------------------------------
 Steady State Optimization with APOPT Solver
 ----------------------------------------------
Iter:     1 I:  0 Tm:      0.00 NLPi:   11 Dpth:    0 Lvs:    3 Obj:  1.29E+07 Gap:       NaN
--Integer Solution:   1.47E+07 Lowest Leaf:   1.29E+07 Gap:   1.26E-01
Iter:     2 I:  0 Tm:      0.00 NLPi:    4 Dpth:    1 Lvs:    2 Obj:  1.47E+07 Gap:  1.26E-01
Iter:     3 I:  0 Tm:      0.00 NLPi:   12 Dpth:    1 Lvs:    4 Obj:  1.34E+07 Gap:  1.26E-01
--Integer Solution:   1.47E+07 Lowest Leaf:   1.34E+07 Gap:   9.18E-02
Iter:     4 I:  0 Tm:      0.00 NLPi:    6 Dpth:    1 Lvs:    3 Obj:  1.47E+07 Gap:  9.18E-02
Iter:     5 I:  0 Tm:      0.00 NLPi:    5 Dpth:    2 Lvs:    4 Obj:  1.29E+07 Gap:  9.18E-02
Iter:     6 I: -1 Tm:      0.00 NLPi:    3 Dpth:    3 Lvs:    3 Obj:  1.29E+07 Gap:  9.18E-02
Iter:     7 I: -1 Tm:      0.00 NLPi:    2 Dpth:    3 Lvs:    2 Obj:  1.29E+07 Gap:  9.18E-02
Iter:     8 I: -1 Tm:      0.00 NLPi:    3 Dpth:    2 Lvs:    1 Obj:  1.34E+07 Gap:  9.18E-02
Iter:     9 I: -1 Tm:      0.00 NLPi:    8 Dpth:    2 Lvs:    0 Obj:  1.34E+07 Gap:  9.18E-02
 No additional trial points, returning the best integer solution
 Successful solution
 
 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :   2.300000000104774E-002 sec
 Objective      :    14654721.8680576     
 Successful solution
 ---------------------------------------------------
 
[1.0] [2.5] [21000.0] [1.0] [25.5] [80000.0]

IPOPT查找初始非整数解决方案，然后APOPT查找整数解决方案。

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/74107619

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