不同极限条件下的三节点LP问题.docx

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2017-07-19 发布于浙江
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不同极限条件下的三节点LP问题.docx

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不同极限条件下的三节点LP问题

I set the PL3 to be 6 in per unit value then get these results.1).There are no particular limits of the PG1 and PG2. f=[35;50;0;0;0]A=[0 0 1 -1 0 0 0 1 0 -1 0 00 1 -1]b=[100,200,300]Aeq=[-1 0 15 -10 -5 0 -1 -10 10+1/0.3 -1/0.3 0 0 -5 -1/0.3 5+1/0.3]beq=[0;0;-6]lb=[0;0;-1e-6;-1000;-1000]ub=[1000;1000;1e-6;1000;1000] [x,fval,exitflag,output,lambda]=linprog(f,A,b,Aeq,beq,lb,ub)Result:x = 6.0000 0.0000 0.0000 -0.2000 -0.8000fval = 210.0000exitflag = 1output = iterations: 6algorithm: large-scale: interior pointcgiterations: 0message: Optimization terminated.constrviolation: 2.1498e-008lambda = ineqlin: [3x1 double]eqlin: [3x1 double]upper: [5x1 double]lower: [5x1 double] lambda.eqlinans = 35.0000 35.0000 35.0000No limits means that all power should be sent all comes from PG1. The shadow price in each bus is 35-------------------------------------------I am the dividing line------------------------------------------------------2).I set the upper limit of PG1 is 5 which is 1 less than the total output PL3. It means there should be some power input at PG2.f=[35;50;0;0;0]A=[0 0 1 -1 0 0 0 1 0 -1 0 00 1 -1]b=[100,200,300]Aeq=[-1 0 15 -10 -5 0 -1 -10 10+1/0.3 -1/0.3 0 0 -5 -1/0.3 5+1/0.3]beq=[0;0;-6]lb=[0;0;-1e-6;-1000;-1000]ub=[5;1000;1e-6;1000;1000] [x,fval,exitflag,output,lambda]=linprog(f,A,b,Aeq,beq,lb,ub)Result:x = 5.0000 1.0000 -0.0000 -0.1167 -0.7667fval = 225.0000exitflag = 1output = iterations: 7algorithm: large-scale: interior pointcgiterations: 0message: Optimization terminated.constrviolation: 3.0971e-008lambda = ineqlin: [3x1 double]eqlin: [3x1 double]upper: [5x1 double]lower: [5x1 double] lambda.eqlinans = 50.0000 50.0000 50.0000We can see from the answer of this problem that as much as power should be transmit from PG1 because the price at PG1 is cheaper. We can also find out that the shadow price in each bus rise up to 50.-------------------------------------------I am the dividing line----