Good suggestions, Shri. I've made some changes accordingly that will be in the 
next version.

-- 
Ray Zimmerman
Senior Research Associate
B30 Warren Hall, Cornell University, Ithaca, NY 14853
phone: (607) 255-9645





On Jun 6, 2013, at 4:41 PM, Shri <[email protected]> wrote:

> Perhaps MATPOWER should (i) avoid calling printpf if the PF/OPF does not 
> converge or (ii) print the 'Converged/Not converged' line after printpf. The 
> problem is that 'converged/Not converged' line is not glaringly visible. So 
> even if the OPF does not converge printpf  produces output which might fool 
> some users to think that the system has actually converged. This is certainly 
> an issue when running large systems because users have to scroll 3-4 screens 
> up to check whether the PF/OPF has converged when the output is directed to 
> stdout.
> 
> Shri
> On Jun 6, 2013, at 1:38 PM, Ray Zimmerman wrote:
> 
>> Are you sure the OPF converged?
>> 
>> -- 
>> Ray Zimmerman
>> Senior Research Associate
>> B30 Warren Hall, Cornell University, Ithaca, NY 14853
>> phone: (607) 255-9645
>> 
>> 
>> 
>> 
>> 
>> On Jun 5, 2013, at 12:35 PM, Alexandra Kapetanaki 
>> <[email protected]> wrote:
>> 
>>> Dear Ray,
>>> 
>>> I run DC opf for the 24 bus IEEE network and I have negative virtual gen 
>>> for the representation of the load  and for an intact system with 2850 MW 
>>> load there is 2850 MW dispatchable load.
>>> However, in the case I add extra virtual gen to all the 24 buses with zero 
>>> output (also zero Pmin and Pmax) the dispatchable load is 1036MW instead of 
>>> 2850MW.
>>> Basically, the network is the same eg generation capacity, load, 
>>> transmission lines's capacity in both cases.  
>>> The only difference is the extra generators with zero output in the second 
>>> case.
>>> Can you please insight me why the load is not satisfied in the second case?
>>> 
>>> Thank you in advance,
>>> Alexandra
>>> 
>>> Alexandra Kapetanaki
>>> PhD Student
>>> Electrical Energy & Power Systems Group, School of Electrical & Electronic 
>>> Engineering
>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United 
>>> Kingdom
>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179
>>> From: [email protected] 
>>> [[email protected]] on behalf of Alexandra 
>>> Kapetanaki [[email protected]]
>>> Sent: 04 June 2013 15:39
>>> To: MATPOWER discussion forum
>>> Subject: RE: transmission losses-DC opf
>>> 
>>> How can I exclude the Q losses in AC opf?
>>> Because the formula is applicable in a DC model without the effect of Q.
>>> 
>>> Thank you in advance!    
>>> 
>>> Alexandra Kapetanaki
>>> PhD Student
>>> Electrical Energy & Power Systems Group, School of Electrical & Electronic 
>>> Engineering
>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United 
>>> Kingdom
>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179
>>> From: [email protected] 
>>> [[email protected]] on behalf of Ray Zimmerman 
>>> [[email protected]]
>>> Sent: 04 June 2013 15:05
>>> To: MATPOWER discussion forum
>>> Subject: Re: transmission losses-DC opf
>>> 
>>> I'm not sure, but you could try your formula to estimate the losses for the 
>>> AC model as well. That would give you an idea of how good your estimate is.
>>> 
>>> -- 
>>> Ray Zimmerman
>>> Senior Research Associate
>>> B30 Warren Hall, Cornell University, Ithaca, NY 14853
>>> phone: (607) 255-9645
>>> 
>>> 
>>> 
>>> 
>>> 
>>> On Jun 4, 2013, at 7:10 AM, Alexandra Kapetanaki 
>>> <[email protected]> wrote:
>>> 
>>>> Dear Ray,
>>>> 
>>>> Thank you for your support!
>>>> 
>>>> I run DC OPF in the 24 bus IEEE network and my aim is to include a 
>>>> transmission losses model in the system. 
>>>> I followed the procedure you insisted to me and for the losses per line I 
>>>> used the already proved equation Ploss = rij(Pk)^2 which is applicable in 
>>>> a p.u. system. 
>>>> So I divided the Pk of each line by baseMVA(=100) whereas the rij is 
>>>> already in p.u..
>>>> However, I compared the losses of the branches for an intact 
>>>> network(24bus) by using  AC opf  to the corresponding losses by using DC 
>>>> opf and they seem to be different as you can see below:
>>>> Losses for each branch_DC(real values)=
>>>> [4,32
>>>> 6,306
>>>> 2,4
>>>> 2,21
>>>> 2,114
>>>> 1,173
>>>> 1,318
>>>> 5,486
>>>> 2,31
>>>> 2,4
>>>> 5,67
>>>> 12,4
>>>> 6,18
>>>> 3,27
>>>> 6,791
>>>> 3,83
>>>> 14,3655
>>>> 4,268
>>>> 5,459
>>>> 3,761
>>>> 5,155
>>>> 5,11
>>>> 3,94
>>>> 4,13]
>>>> 
>>>> Losses for each branch_AC(real values)=
>>>> [0,0022
>>>> 0,13
>>>> 0,864
>>>> 0,644
>>>> 1,25
>>>> 0,156
>>>> 0,933
>>>> 0,245 
>>>> 0,059 
>>>> 0,94 
>>>> 1,63
>>>> 0,756 
>>>> 0,423 
>>>> 0,285 
>>>> 0,35 
>>>> 0,50 
>>>> 0,611 
>>>> 0,537 
>>>> 1,391
>>>> 0,309 
>>>> 5,80
>>>> 4,69
>>>> 6,129
>>>> 0,030 
>>>> 2,187
>>>> 2,187
>>>> 2,692
>>>> 4,1436
>>>> 0,230 
>>>> 0,833 
>>>> 2,697
>>>> 0,191 
>>>> 0,191 
>>>> 0,100
>>>> 0,100
>>>> 0,313
>>>> 0,31 
>>>> 1,829]
>>>> 
>>>> I would like to let you know that Plosses_AC_opf were obtained by the 
>>>> following command: 
>>>> results_2(N).Plossesbranch_AC(br)=results.branch(br,14)+results.branch(br,16);
>>>>  
>>>> 
>>>> Why do you think the losses are considerable different between the AC,DC 
>>>> models?
>>>> In DC the Q is not included so this is one reason of the deviation but is 
>>>> that enough? 
>>>> 
>>>> Thank you in advance,
>>>> 
>>>> Alexandra Kapetanaki
>>>> PhD Student
>>>> Electrical Energy & Power Systems Group, School of Electrical & Electronic 
>>>> Engineering
>>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United 
>>>> Kingdom
>>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179
>>>> From: [email protected] 
>>>> [[email protected]] on behalf of Ray Zimmerman 
>>>> [[email protected]]
>>>> Sent: 28 May 2013 17:58
>>>> To: MATPOWER discussion forum
>>>> Subject: Re: transmission losses-piece wise linear approach
>>>> 
>>>> - Solve a DC OPF with no losses.
>>>> - Compute the loss for each branch, Ploss = rij(Pk)^2.
>>>> - Add half of the loss to the load at the buses connected by the branch.
>>>> - Re-solve the DC OPF.
>>>> - Recompute the loss for each branch based on the new flow.
>>>> - Adjust the loads at each end of the branch to reflect the change in 
>>>> losses.
>>>> - Repeat, until the change from one iteration to the next is smaller than 
>>>> some threshold.
>>>> 
>>>> -- 
>>>> Ray Zimmerman
>>>> Senior Research Associate
>>>> 419A Warren Hall, Cornell University, Ithaca, NY 14853
>>>> phone: (607) 255-9645
>>>> 
>>>> 
>>>> 
>>>> 
>>>> On May 28, 2013, at 12:25 PM, Alexandra Kapetanaki 
>>>> <[email protected]> wrote:
>>>> 
>>>>> Dear Ray,
>>>>> 
>>>>> Thank you for your help!
>>>>> Can you please explain me more what do you mean with the  "adjusting the 
>>>>> set of dummy loads used to represent losses based on flows in the  
>>>>> previous iteration".
>>>>> For example, I want to represent the losses as dummy loads according to 
>>>>> the following figure:
>>>>> 
>>>>> 
>>>>> <losses.png>
>>>>> 
>>>>> 
>>>>> 
>>>>> 
>>>>> and  express the losses by using the quadratic equation : Ploss=rij(Pk)^2 
>>>>> .
>>>>> How can I practically implement that within "few iterations", as you 
>>>>> recommend?
>>>>> 
>>>>> Thank you once again for your support,
>>>>> 
>>>>> 
>>>>> Alexandra Kapetanaki
>>>>> PhD Student
>>>>> Electrical Energy & Power Systems Group, School of Electrical & 
>>>>> Electronic Engineering
>>>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United 
>>>>> Kingdom
>>>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179
>>>>> From: [email protected] 
>>>>> [[email protected]] on behalf of Ray Zimmerman 
>>>>> [[email protected]]
>>>>> Sent: 28 May 2013 16:55
>>>>> To: MATPOWER discussion forum
>>>>> Subject: Re: transmission losses-piece wise linear approach
>>>>> 
>>>>> Dear Alexandra,
>>>>> 
>>>>> Since the network equations in MATPOWER's DC OPF assume no losses, it 
>>>>> seems you would have to introduce the losses as dummy (dispatchable) 
>>>>> loads at the downstream end of each branch. You would have to add an 
>>>>> additional set of constraints for each of these dummy loads constraining 
>>>>> the consumption lie above the piecewise linear constraints you propose. 
>>>>> These constraints could be added using the mechanism for user-defined 
>>>>> constraints described in section 5.3.2 and chapter 6 of the User's Manual.
>>>>> 
>>>>> Another approach to using MATPOWER to solve a "DC OPF with losses" is to 
>>>>> simply run the DC OPF iteratively, each time adjusting the set of dummy 
>>>>> loads used to represent losses based on flows in the previous iteration. 
>>>>> I'm guessing it wouldn't require more than a few iterations to converge 
>>>>> to a pretty good solution. This approach is a bit brute-force, but may be 
>>>>> simpler to implement and allows you to use whatever function you like 
>>>>> (e.g. a quadratic) to compute the losses. Just another idea.
>>>>> 
>>>>> -- 
>>>>> Ray Zimmerman
>>>>> Senior Research Associate
>>>>> 419A Warren Hall, Cornell University, Ithaca, NY 14853
>>>>> phone: (607) 255-9645
>>>>> 
>>>>> 
>>>>> 
>>>>> 
>>>>> On May 27, 2013, at 3:16 PM, Alexandra Kapetanaki 
>>>>> <[email protected]> wrote:
>>>>> 
>>>>>> Dear Dr Ray,
>>>>>> 
>>>>>> As the AC power flow requires high computation time for the losses to be 
>>>>>> calculated, a DC opf can be used in conjuction with a linear model for 
>>>>>> transmission losses.
>>>>>> My aim is to adjust the piece wise linear approach of Matpower in order 
>>>>>> to accommodate the losses of a transmission line.
>>>>>> More particularly, the losses can be expressed through the following 
>>>>>> equation: Ploss=rij(Pk)^2 [where rij the resistance of the line and Pk 
>>>>>> the power flow in the transmission line].
>>>>>> However, the above equation is a quadratic function but can be expressed 
>>>>>> with a piecewise linear model. The figure below shows a linear model 
>>>>>> consists of N line pieces 
>>>>>>  <piecewiselinear.png>
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> the equation of the nth line piece is:
>>>>>> 
>>>>>> <bbbbbbbbbbbbbbb.png>
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> -How can I modify the piece wise linear approach of Matpower with the 
>>>>>> view to include the losses model?
>>>>>> 
>>>>>> 
>>>>>> Thank you in advance,
>>>>>> 
>>>>>> Alexandra Kapetanaki
>>>>>> PhD Student
>>>>>> Electrical Energy & Power Systems Group, School of Electrical & 
>>>>>> Electronic Engineering
>>>>>> Ferranti Building (B18), The University of Manchester, M13 9PL, United 
>>>>>> Kingdom
>>>>>> Tel: +44 (0) 161 306 2263; Mobile: +44 (0) 7857 598179
>>>>> 
>>>>> 
>>>> 
>>>> 
>>> 
>>> 
>> 
> 

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