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Gekko 優化

[英]Gekko optimization

 原文  2020-06-17 22:21:22  7   1    python/ optimization/ linear-programming/ gekko

問題描述

我正在解決這個優化問題，我需要弄清楚我需要開設多少個配送中心才能滿足 12 家公司設施的需求，同時最大限度地降低運輸成本。 運輸成本只是配送中心之間的距離乘以每英里的成本，但是在這個問題中，每英里的成本是一美元。 我有 5 個選擇，分別是波士頓、納舒厄、普羅維登斯、斯普林菲爾德和伍斯特，這 5 個是公司 12 個設施的一部分。

我解決了這個問題並得到了正確的答案，但后來我嘗試將兩個約束添加到相同的代碼中，但我得到的答案是不正確的。 另外兩個限制條件是配送中心 (DC) 到其他設施 (客戶) 的平均距離必須小於 60 英里； 第二個約束是 50 英里內的客戶百分比必須大於 80% (0.8)。 我知道這個問題的答案，成本必須是66,781 美元，平均客戶距離為15 英里，50 英里內的客戶百分比為90% 。 我的代碼的 output是成本為66289美元，平均客戶距離為15.36英里，50 英里內的客戶百分比為179% ，這沒有意義。

你能幫我弄清楚為什么我得到一個奇怪的 output 嗎？ 提前致謝。

from gekko import GEKKO
import numpy as np
import pandas as pd
import math

m = GEKKO(remote=False) #So that it solves the problem locally
m.options.SOLVER = 1 #MILP

varx = [[0 for col in range(12)] for row in range(5)] #Creates an empty list
for i in range (5):
    for j in range (12):
        varx[i][j] = m.Var(lb = 0, integer = True) 

varx = np.array(varx)    
varxt = np.transpose(varx)



vary = np.empty([]) #Creates an empty array

for i in range(5):
    vary = np.append(vary, m.Var(lb = 0, ub = 1, integer = True)) #Yes/No variables

vary = vary[1:13] 



dists = np.array([[0 , 93, 69, 98, 55, 37, 128, 95, 62, 42, 82, 34], #Boston
                 [37, 65, 33, 103, 20, 0, 137, 113, 48, 72, 79, 41], #Nashua
                 [42, 106, 105, 73, 92, 72, 94, 57, 104, 0, 68, 38], #Providence
                 [82, 59, 101, 27, 93, 79, 63, 57, 127, 68, 0,  47], #Springfield
                 [34, 68, 72, 66, 60, 41, 98, 71, 85, 38, 47,   0]]) #Worcester



max_dist = 60 #Max average distance (in miles)
min_pct = 0.8 #Min percent of demand within 50 miles



aij = np.zeros((5, 12)) #Creates an empty array

for i in range (5):
    for j in range (12):
        if dists[i][j] <= 50:
            aij[i][j] = 1
        else:
            aij[i][j] = 0 #Creates a 0s and 1s array. If the distance to a costumer 
                          #is less than 50, then the matrix element is 1, it is zero
                          #otherwise


dem_consts = np.array([425, 12, 43, 125, 110, 86, 129, 28, 66, 320, 220, 182])

fixd_cost = 10000


sum1 = np.sum(np.multiply(varx, dists))
sum2 = np.sum(vary*fixd_cost)
z = sum1 + sum2 

tot_dem = np.sum(dem_consts)

M = tot_dem



m.Minimize(z)



for i in range(12):
    m.Equation(np.sum(varxt[i, :]) >= dem_consts[i]) #Demand constraints

for i in range(5):
    m.Equation(np.sum(varx[i, :]) <= 2000) #Capacity constraints
    m.Equation(np.sum(varx[i, :]) <= M*vary[i]) #Enforces 0 or 1 value

m.Equation(np.sum(vary[:]) >= 1)


di_sum = np.sum(np.multiply(varx, dists))
di_sumw = di_sum/ tot_dem
m.Equation(di_sumw <= max_dist) #Average (demand) weighted distance from DC to customer

a_sum = np.sum(np.multiply(varx, aij)) 
a_sumw = a_sum/tot_dem
m.Equation(a_sumw >= min_pct) #Percent of demand that is within 50 miles

m.solve(disp = False)


p1 = np.zeros((5, 12))

for i in range (5):
    for j in range (12):
        p1[i][j] = varx[i][j].value[0]
p1t = np.transpose(p1)

p2 = np.zeros((5, )) 

for i in range(5):
    p2[i] = vary[i].value[0] 

mad1 = np.sum(np.multiply(p1, dists)) 
mad2 = mad1/tot_dem
mpi1 = np.sum(np.multiply(p1, aij)) 
mpi2 = mpi1/tot_dem

tot1 = np.sum(np.multiply(p1, dists))
tot2 = np.sum(p2)*fixd_cost
tot = tot1 + tot2 


print('The minimum cost is:' +str(tot))
print('Average customer distance:' +str(mad2))
print('Percent of customers <= 50 miles:' +str(mpi2))


dc = np.array(['Boston', 'Nashua', 'Providence', 'Springfield', 'Worcester'])
cities = ['Boston', 'Brattleboro', 'Concord', 'Hartford', 'Manchester', 'Nashua',
          'New Haven', 'New London', 'Portsmouth', 'Providence', 'Springfield', 'Worcester']
data = {cities[0]: p1t[0], cities[1]: p1t[1], cities[2]: p1t[2], cities[3]: p1t[3],
       cities[4]: p1t[4], cities[5]: p1t[5], cities[6]: p1t[6], cities[7]: p1t[7],
       cities[8]: p1t[8], cities[9]: p1t[9], cities[10]: p1t[10], cities[11]: p1t[11]}

df = pd.DataFrame(data, index = dc)
df

1 個解決方案

解決方案1
 1 已采納  2020-06-20 02:34:55

求解器發出一條消息，當您設置m.solve(disp=True)時，它會在 500 次迭代時提前終止。 它返回一個可行的 integer 解決方案，但它可能不是最好的解決方案。

 Warning: best integer solution returned after maximum MINLP iterations
 Adjust minlp_max_iter_with_int_sol  500  in apopt.opt to change limit
 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  1.3654 sec
 Objective      :  66829.
 Successful solution
 ---------------------------------------------------


The minimum cost is:66829.0
Average customer distance:15.3659793814433
Percent of customers <= 50 miles:1.7943871706758305

如果添加求解器選項：

m.solver_options = ['minlp_gap_tol 1.0e-2',\
                    'minlp_maximum_iterations 10000',\
                    'minlp_max_iter_with_int_sol 5000']

目標 function 提高到 66285：

 Successful solution

 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  1.7178 sec
 Objective      :  66285.
 Successful solution
 ---------------------------------------------------


The minimum cost is:66285.0
Average customer distance:20.781786941580755
Percent of customers <= 50 miles:1.9873997709049256

客戶的百分比是否應改為 <= 50 英里？： mpi3 = mpi1/np.sum(p1)和平均距離是？： mad3 = mad1/np.sum(p1) 。 這使得 <= 50 英里的客戶比例等於 89.94%：

Percent of customers <= 50 miles (mpi3):0.8994297563504406

新的平均距離為：

Average customer distance (mad3):9.405132192846034

這是一個修改后的腳本，它使用了 gekko arrays 和 gekko sum 函數，以便更高效。

from gekko import GEKKO
import numpy as np
import pandas as pd
import math

m = GEKKO(remote=False) #So that it solves the problem locally
m.options.SOLVER = 1 #MILP

varx = m.Array(m.Var,(5,12),lb=0,integer=True)
vary = m.Array(m.Var,5,lb=0,ub=1,integer=True)

dists = np.array([[0 , 93, 69, 98, 55, 37, 128, 95, 62, 42, 82, 34], #Boston
                 [37, 65, 33, 103, 20, 0, 137, 113, 48, 72, 79, 41], #Nashua
                 [42, 106, 105, 73, 92, 72, 94, 57, 104, 0, 68, 38], #Providence
                 [82, 59, 101, 27, 93, 79, 63, 57, 127, 68, 0,  47], #Springfield
                 [34, 68, 72, 66, 60, 41, 98, 71, 85, 38, 47,   0]]) #Worcester

max_dist = 60 #Max average distance (in miles)
min_pct = 0.8 #Min percent of demand within 50 miles

#Creates a 0s and 1s array. If the distance to a costumer 
#is less than 50, then the matrix element is 1, it is zero otherwise
aij = [[1 if dists[i,j]<=50 else 0 for j in range(12)] for i in range(5)]

dem_consts = np.array([425, 12, 43, 125, 110, 86, 129, 28, 66, 320, 220, 182])
fixd_cost = 10000
sum1 = np.sum(np.multiply(varx, dists))
sum2 = np.sum(vary*fixd_cost)
z = sum1 + sum2 
tot_dem = np.sum(dem_consts)
M = tot_dem
m.Minimize(z)

for j in range(12):
    m.Equation(m.sum(varx[:,j]) >= dem_consts[j]) #Demand constraints

for i in range(5):
    m.Equation(m.sum(varx[i,:]) <= 2000) #Capacity constraints
    m.Equation(m.sum(varx[i,:]) <= M*vary[i]) #Enforces 0 or 1 value

m.Equation(m.sum(vary) >= 1)


di_sum = np.sum(np.multiply(varx, dists))
di_sumw = di_sum/ tot_dem
m.Equation(di_sumw <= max_dist) #Average (demand) weighted distance from DC to customer

a_sum = np.sum(np.multiply(varx, aij)) 
a_sumw = m.Intermediate(a_sum/tot_dem)
m.Equation(a_sumw >= min_pct) #Percent of demand that is within 50 miles


m.solver_options = ['minlp_gap_tol 1.0e-2',\
                    'minlp_maximum_iterations 10000',\
                    'minlp_max_iter_with_int_sol 5000']
m.solve(disp = True)


p1 = np.zeros((5, 12))

for i in range (5):
    for j in range (12):
        p1[i][j] = varx[i][j].value[0]
p1t = np.transpose(p1)

p2 = np.zeros(5) 
for i in range(5):
    p2[i] = vary[i].value[0] 

mad1 = np.sum(np.multiply(p1, dists)) 
mad2 = mad1/tot_dem
mad3 = mad1/np.sum(p1)
mpi1 = np.sum(np.multiply(p1, aij)) 
mpi2 = mpi1/tot_dem
mpi3 = mpi1/np.sum(p1)

tot1 = np.sum(np.multiply(p1, dists))
tot2 = np.sum(p2)*fixd_cost
tot = tot1 + tot2 

print(p1)
print(p2)
print('The minimum cost is:' +str(tot))
print('Average customer distance (mad2):' +str(mad2))
print('Average customer distance (mad3):' +str(mad3))
print('Percent of customers <= 50 miles (mpi2):' +str(mpi2))
print('Percent of customers <= 50 miles (mpi3):' +str(mpi3))

dc = np.array(['Boston', 'Nashua', 'Providence', 'Springfield', 'Worcester'])
cities = ['Boston', 'Brattleboro', 'Concord', 'Hartford', 'Manchester', 'Nashua',
          'New Haven', 'New London', 'Portsmouth', 'Providence', 'Springfield', 'Worcester']
data = {cities[0]: p1t[0], cities[1]: p1t[1], cities[2]: p1t[2], cities[3]: p1t[3],
       cities[4]: p1t[4], cities[5]: p1t[5], cities[6]: p1t[6], cities[7]: p1t[7],
       cities[8]: p1t[8], cities[9]: p1t[9], cities[10]: p1t[10], cities[11]: p1t[11]}

df = pd.DataFrame(data, index = dc)
df

這是解決方案：

[[1102.    0.   43.    0.  110.   86.    0.    0.   66.    0.    0.  182.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.   28.    0.  495.    0.    0.]
 [   0.   12.    0.  125.    0.    0.  129.    0.    0.    0. 1480.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]]
[1. 0. 1. 1. 0.]
The minimum cost is:66285.0
Average customer distance (mad2):20.781786941580755
Average customer distance (mad3):9.405132192846034
Percent of customers <= 50 miles (mpi2):1.9873997709049256
Percent of customers <= 50 miles (mpi3):0.8994297563504406

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