z =(900-450)*(x( 1)+x(2)+x(3)+x(4))+x(5)+x(6)+x(7)+x(8)+x(9)+x( 10)+x( 1 1))-( 160 * x( 1)+ 130 * x(0)
Constraints include:
x( 1)+x(2)+x(3)+x(4)= 50( 1)
x(5)+x(6)+x(7)+x(8)≤60 (2)
x(9)+x( 10)+x( 1 1)≤50(3)
x( 1)+x(5)+x(9)≤30+50 (4)
x(2)+x(6)+x( 10)≤70+70 (5)
x(3)+x(7)+x( 1 1)≤ 10+20(6)
x(4)+x(8)≤ 10+40 (7)
x( 1)+x(5)+x(9)≥30 (8)
x(2)+x(6)+x( 10)≥70 (9)
x(3)+x(7)+x( 1 1)≥ 10( 10)
x(4)+x(8)≥ 10( 1 1)
x(i)≥0，i= 1，2，3，4，5，6，7，8，9， 10， 1 1 ( 12)
This problem requires the maximum value of the solution z, that is, the minimum value of the solution (-z). This is a problem of linear programming. Solve with MATLAB, the program is as follows:
c =-；
The% plus and minus signs will be maximized and minimized.
part =；
% lower limit
[x，f，exitflag，output，lag]=linprog(c，A 1，b 1，[]，[]，v 1)
The optimal solution is x = [0 0,50,0,0,0,50,0, 10,40,0, 10], the optimal value is f=-47600 (the maximum value is z=-f=47600), and the exit flag is = The list is as follows:
Water supply /kt methyl ethyl propyl butyl
A 0 50 0 0
B 0 50 0 10
C 40 0 10 /
Only by distributing water supply in this way can the company get the maximum profit, which is 47600 yuan.
If the daily maximum water supply of the three reservoirs is doubled, the objective function remains unchanged, the constraint (4) ~ (12) remains unchanged, and (1) ~ (3) changes, as follows:
x( 1)+x(2)+x(3)+x(4)≤ 100( 1)
x(5)+x(6)+x(7)+x(8)≤ 120(2)
x(9)+x( 10)+x( 1 1)≤ 100(3)
Solve with MATLAB, the program is as follows:
c =-；
The% plus and minus signs will be maximized and minimized.
part =；
% lower limit
, optimal value f=-88700 (maximum value z=-f=88700), exitflag= 1 (convergence). The list is as follows:
Water supply /kt methyl ethyl propyl butyl
A 0 100 0
B 30 40 0 50
C 50 0 30
After the maximum daily water supply of the three reservoirs is doubled, the company can make the most profit by distributing the water supply in this way, which is 88,700 yuan, an increase of 4 1 100 yuan.