MatLab中求和的線性規劃

我想複製我在Matlab中的研究論文中找到的線性優化。我需要解決以下線性優化問題：MatLab中求和的線性規劃

objective function

其中C1，C2，C3，C4，C5，W1和W2是優先權重。

j取自1至12（一年的12個月）。

以下限制適用：

constraint functions

I（j）的和L（j）是按月記錄。

我使用MatLab來編寫解決方案。這裏是我的代碼（我很新的這一點，所以請原諒任何不好的編碼！）：

%set up the data for the year: 
I = [72.6 26.0 23.2 20.4 15.2 22.0 40.9 45.2 38.7 41.4 142.2 116.8] 
L = [1.6 1.3 0.8 0.6 0.6 0.6 1 1.5 1.8 1.8 1.8 2.0]; 
%set up the initial level: 
S0 = 683 
%set up the priority weightings 
w2= 1; 
w1 = 1.5; 
C1 = 3; 
C2 = 2; 
C3 = 5; 
C4 = 4; 
C5 = -5; 
%set up the constraint equation, lower bond and upper bound 
A = [(C1*w1) C2 (C3*w2) (C4*w2) C5]; 
Aeq = [1 1 1 1 1]; 
lb = [70 0 0 0 0]; 
ub = [815 54.14 13.4 41.8 17345]; 
%set up a blank matrix to store the results 
x=zeros(12,5); 

%for each month calculate the optimum and store in the matrix 
for j = 1:12 
    Beq = [(I(j)+S0-L(j))]; 
    x(j,:) = linprog(-A,[],[],Aeq,Beq,lb,ub); 
    S0 = x(j,1); 
end 

%output the result 
opt = x

問題是，當我比較我的結果的研究論文中，我發現，我得到不同的結果。最終我意識到我正在爲每個月找到最佳狀態，而不是全年的全球最佳狀態。我一直在網上搜索如何找到全年的最佳值（即優化求和函數），但我找不到任何東西。任何人都可以將我指向正確的方向嗎？

來源

2017-04-17 Louis Hall

有一個在Matlab的linprog沒有總和。它完全基於矩陣。對於某些問題，這是沒有問題的，對於其他問題，這是噩夢（參見[這裏]（http://yetanothermathprogrammingconsultant.blogspot.com/2016/10/matlab-vs-gams-integer-programming.html））。在你的情況下，你需要建立一個大矩陣，每一列對應一個變量，每一行對應一個約束。 –

@ErwinKalvelagen非常感謝您的回覆。我用你的例子來解決我的問題。 –

感謝@ErwinKalvelagen我能解決我的問題。這裏是未來的我爲別人解決方案：

%set up the data for the year: 
I = [72.6; 26.0; 23.2; 20.4; 15.2; 22.0; 40.9; 45.2; 38.7; 41.4; 142.2; 116.8;]; 
I = I*1; %allows the inflow to be scaled to test the model for a dry or wet year 
L = [1.6; 1.3; 0.8; 0.6; 0.6; 0.6; 1; 1.5; 1.8; 1.8; 1.8; 2.0;]; 

%set up the priority weightings 
w1 = 2; 
w2= 50; 
C1 = 3; 
C2 = 2; 
C3 = 5; 
C4 = 4; 
C5 = -5; 

%set up the constraint equation, lower bond and upper bound 
A = [(C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5 (C1*w1) C2 (C3*w2) (C4*w2) C5]; 

%set up spare matrix for Aeq 
Aeq = zeros(12,60); 

% Populate Aeq 
% first the positive portions of the monthly data 
row = 1; 
coloumn = 1; 
for counter = 1:12 
    for counter = 1:5 
     Aeq(coloumn,row)=1; 
     row = row + 1; 
    end 
    coloumn = coloumn+1; 
end 
% then the -S0 for each month 
Aeq(1, 56)=-1; 
coloumn = 1; 
for row = 2:12 
    Aeq(row,coloumn)=-1; 
    coloumn = coloumn+5; 
end 

%populate Beq 
Beq = I-L; 

%populate the lower and upper bounds 
Smin = 70; 
Smax_1_9 = 731.2; 
Smax_10_12 = 673.2 
QDmin = 0 
QDmax = 54.14 
Rmin = 0; 
Rmax = 13.4; 
RRmin = 0; 
RRmax = 41.8; 
SPILLmin = 0; 
SPILLmax = 17345; 
%first the lower bound 
lbmonthly = [Smin; QDmin; Rmin; RRmin; SPILLmin;]; 
lb = vertcat(lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly,lbmonthly); 
%second the upper bound 
ubmonthly_1_9 = [Smax_1_9; QDmax; Rmax; RRmax; SPILLmax;]; 
ubmonthly_10_12 = [Smax_10_12; QDmax; Rmax; RRmax; SPILLmax;]; 
ub = vertcat(ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_1_9, ubmonthly_10_12, ubmonthly_10_12, ubmonthly_10_12); 

%find the optimal 
opt = linprog(-A,[],[],Aeq,Beq,lb,ub); 

%output the result as a matrix 
opt = reshape(opt,5,12)'

來源

2017-04-23 07:25:36

MatLab中求和的線性規劃

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