[Cmpl] Linearise nonlinear objective function (Indrajit Sen Gupta)

Mon Mar 9 12:39:01 EDT 2015

Indrajit:

Based on the information provided about the formulation, one simple option
you could try is substituting with another variable to represent the
nonlinear term:

Z[j,k] = x[j] * x[k]

The objective function will become linear in Z[j,k] and x[k]. Writing this
in longer form for greater clarity:

Constant *x[k] - Constant * Z[j,k]

Then add simple constraints to enforce the needed relationship between
Z[j,k] and x[j] and x[k].

I will leave this last part for you to do as an exercise. If you're not sure
about the last part, please ask again.

Best regards,
Peter

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Today's Topics:

   1. Linearizing a nonlinear objective function (Indrajit Sen Gupta)
   2.  Benchmark / Statistics (Thorsten)
   3. [CMPL] Readcsv error (Thorsten)

----------------------------------------------------------------------

Message: 1
Date: Mon, 9 Mar 2015 12:05:32 +0530
From: Indrajit Sen Gupta <indrajitsg at gmail.com>
To: "cmpl at list.coin-or.org" <cmpl at list.coin-or.org>
Subject: [Cmpl] Linearizing a nonlinear objective function
Message-ID:
	<CAB-BrmcAJ_sttf_-SthBwvuorvnbPht6J_j+pu6RWYAwhaEEMA at mail.gmail.com>
Content-Type: text/plain; charset="utf-8"

Hi All,

I had a basic question regarding linearizing non-linear objective functions.
Currently the software that we have is running into issues as it does not
support non-linear objective function along with binary decision variables.

One part of the objective function looks something like this:

constant * (1 - x[j]) * x[k]

where x's are binary variables (0/1). Is there a way to linearize the
objective function in such situations?

Regards,
Indrajit
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Message: 2
Date: Mon, 9 Mar 2015 10:07:01 +0100
From: Thorsten <thorten at hotmail.com>
To: "cmpl at list.coin-or.org" <cmpl at list.coin-or.org>
Subject: [Cmpl]  Benchmark / Statistics
Message-ID: <DUB123-W434C9ABE063C9057693783B11B0 at phx.gbl>
Content-Type: text/plain; charset="iso-8859-1"

Hi Mike,

thank you very much indeed for the quick reply! I scanned the appendix for
CBC-Parameters but couldn't find a suitable one which enables me to get the
required information. Even increasing the log(Level) to maximum doesn't
provide the neccessary information. Do you possibly have a hint or idea in
mind for a CBC-Command to get me the programs lower bound and the gap?
Thanks for your support!

Best regards,
Thorsten

Hi Thorsten,

You can find these information in the Cbc output. Cmpl does not provide
functionalities to get it directly within a Cmpl model.

Cheers,

Mike

> Am 06.03.2015 um 16:57 schrieb Thorsten <thorten at hotmail.com>:
> 
> Dear CMPL-Community,
> 
> I have implemented a capacitated vehicle routing problem with time windows
in CMPL and it runs within coliop. In my thesis I want to compare the
performance of CMPL and the CBC-solver with the performance of Fico Xpress.
For the numerical analysis I will need some more statistics than I am able
to get so far. Besides the objective value and the running time of the model
I will need the Lower bound of the program and the gap. Can you probably
help me how I can have these values printed out?
> Thank you very much in advance,
> Thorsten 		 	   		  
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Message: 3
Date: Mon, 9 Mar 2015 10:12:18 +0100
From: Thorsten <thorten at hotmail.com>
To: "cmpl at list.coin-or.org" <cmpl at list.coin-or.org>
Subject: [Cmpl] [CMPL] Readcsv error
Message-ID: <DUB123-W19BD131338FFFD8FDBF555B11B0 at phx.gbl>
Content-Type: text/plain; charset="iso-8859-1"

Hi folks!

I am trying to read parameters from a csv-file and have therefore
implemented the following code:
serv[]:=        readcsv("Dist5C.csv");

Unfortunately I get the following error message:
error (input/output): file Minnertry5C.cmpl line 18: Input file 'Dist5C.csv'
not found

The csv-file is in the same directory as the cmpl-file.
Does anybody have an idea what the problem could be?

Best regards,
Thorsten

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