1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
|
// illustrate the dynamic programming for preparing an optimal unit commitment.
clear
clc;
function[F1]=F1(P1)
F1=7.1*P1+.00141*(P1^2)
mprintf("F1(%.0f)=%.1f\n",P1,F1);
endfunction
function[f2]=f2(P2)
f2=7.8*P2+.00195*(P2^2)
mprintf("f2(%.0f)=%.0f\n",P2,f2);
endfunction
function[F]=F(P1,P2)
F1=7.1*P1+.00141*(P1^2)
F2=7.8*P2+.00195*(P2^2)
F=F1+F2
mprintf("F1(%.0f)+f2(%.0f)=%.0f\n",P1,P2,F);
endfunction
P1max=600;
P2max=450;
mprintf("Unit Commitment using Load 500MW\n")
F1(500);
mprintf("Since min. Power of second unit is 100MW , we find\n");
F(400,100);
F(380,120);
F(360,140);
mprintf("Therefore for load 500 MW , the load commitment on unit 1 is 400 MW and that on 2 is 100 MW which gives min. cost\n");
mprintf("Next we increase the load by 50 MW and loading unit 1 we get, \n");
F1(550);
mprintf("Also if we distribute a part of load to unit 2 we get ,\n")
F(450,100);
F(400,150);
F(350,200);
mprintf("Therefore for load 550 MW , the load commitment on unit 1 is 400 MW and that on 2 is 150 MW which gives min. cost\n");
|
