initial commit / add all books

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diff --git a/1055/CH24/EX24.4/ch24_4.sce b/1055/CH24/EX24.4/ch24_4.sce
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+// 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");

+

+

+