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why is my code wrong ?

조회 수: 1 (최근 30일)
Bhavz
Bhavz 2014년 10월 19일
마감: MATLAB Answer Bot 2021년 8월 20일
Please see the files attached and let me know what is wrong with this
l_dp =
0.7101 0 0 0.0447 0.7701 0 0.0379 0.8407
0 0 0 1.5729 0.0635 0 0 0.5264
1.1636 0.4907 0 0 0.2497 0 1.0603 0
0 0 1.1273 0 0.1478 0 0 1.4100
and this not right since demand on a path should total to one ie the sum of each row should be 1
whats wrong with my code is it ub limit or what ?
Please do help
  댓글 수: 2
Geoff Hayes
Geoff Hayes 2014년 10월 19일
Bhavz - where in your code do you force that condition (that each row should sum to one) to be true? Or where do you expect that to be enforced (and why)?
Bhavz
Bhavz 2014년 10월 19일
편집: Bhavz 2015년 4월 28일
^ its not about enforcing the condition but shudnt the demand on a path total to one technically ... thats what the correct functioning wud be ? I am attaching the equation i want to implement constraint wise ,u can see if i have made a mistake there What I have done is taken the network optimization from paper 2 in the network from 1 and I am trying to measure the effect on link utilization. How would you change the given Algorithm to route unsplittable flows on the specified single primary path only
And as always thankyou so much for yr help

답변 (1개)

Matt J
Matt J 2014년 10월 19일
편집: Matt J 2014년 10월 19일
ie the sum of each row should be 1
If the unknown vector x is supposed to satisfy this, then presumably you need to have linear equality constraints,
Aeq=kron(ones(1,8),eye(4));
beq=[1;1;1;1]
[x,fval,exitflag] = ga(@objectiveFun,...
32,[],[],Aeq,beq,lb,[],nonlcon,options);
  댓글 수: 6
Bhavz
Bhavz 2014년 10월 19일
편집: Bhavz 2014년 10월 19일
Oh How would I do that as in pass values from constraints.m to
[x,fval,exitflag] = ga(@objectiveFun,...
32,[],[],Aeq,beq,lb,[],nonlcon,options);
Matt J
Matt J 2014년 10월 19일
편집: Matt J 2014년 10월 19일
You would not use the nonlinear constraint function to handle linear constraints. You would generate appropriate A, b matrices, similar to what I did for you with Aeq and beq.

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