Script in AMPL

[Alessandro Oscar Gilardino Arias]

Dear Bob and Victor,

I have a question.

In my model in AMPL I have a parameter called generation and I have to run my model in a .run file for 5 times, this means using a for {1..5} but in every loop I have to change the values of all the parameter generation in different values on every iteration, also I have to let the value of a binary variable calles z[j,m] this means that if in any iteration any variable z[j,m] has a value 1, in the following iteration must take a value of 1.

I write the code I think I should write.

set Location := 1..10;

set Arrange:= 1..5;

variable z[Location, Arrange];

the variable z has to accomplish the following constraint:

subject to Containers_in_one_point {j in Location}:

                sum {m in Arrange} z[j,m] <=1;

I think that the .run file should have the following code:

param T := 1..5;

for {T}

if T = 1 then

solve;

let generation := generation * 0.08;

if sum {m in Arrange} z[j,m] = 1 then sum {m in Arrange} z[j,m] = 1 else sum {m in Arrange} z[j,m] <=1;

elseif T = 2 then

solve;

let generation := generation * 0.10;

if sum {m in Arrange} z[j,m] = 1 then sum {m in Arrange} z[j,m] = 1 else sum {m in Arrange} z[j,m] <=1;

elseif T = 3 then

solve;

let generation := generation * 0.12;

if sum {m in Arrange} z[j,m] = 1 then sum {m in Arrange} z[j,m] = 1 else sum {m in Arrange} z[j,m] <=1;

And so on, I am not very sure if that would work in a script of AMPL or if I have to connect Matlab with AMPL.

Also I would like to ask you something about the AMPL.api, when I use the following command: 

ampl = AMPL;

I have the following error:

Undefined variable "com" or class "com.ampl.AMPL".

Error in AMPL (line 21)

                a = com.ampl.AMPL();.

Thanks a lot in advance, and sorry for give you such a long post.

Thanks a lot again!!

 

[Robert Fourer]

If you are going to write out each case individually, then there is no need to a loop ("for") statement or a long series of "if" statements. Also to say that "if in any iteration any variable z[j,m] has a value 1, in the following iteration must take a value of 1" you should use a statement like

for {j in Location, m in Arrange}
if z[j,m] = 1 then fix z[j,m];

which can be written in an AMPL script.

Bob Fourer
am...@googlegroups.com

=======

[Alessandro Oscar Gilardino Arias]

Thanks a lot for your answer Bob

[Alessandro Oscar Gilardino Arias]

Thanks again Bob, but I still have a question, for the same script in previous post.

You told me to use 

 for {j in Location, m in Arrange} 
      if z[j,m] = 1 then fix z[j,m]; 

but here is my question, the index m can take value 1,2 or 3. This means that I can have a variable z[1,1], z[1,2] or z[1,3] what I am trying to do is that if z[1,2] take value 1, for the next iteration I have two options, that z[1,2] or z[1,3] can take value 1, for no reason the variable z[1,1] can take value 1 .

I am trying with this code

 for {j in Location, m in Arrange} 

      if z[j,m] = 1 then fix sum{l in Arrange:l>=m} z[j,m]=1;

but when I try to run in AMPL returns me a syntax error.

The other code I was trying is the following

 for {j in Location, m in Arrange, l in Arrange:l>=m} 

      if z[j,m] = 1 then fix z[j,l];  

but in this case, I feel that is happening the following: if z[1,2] = 1, for the next iteration z[1,2]=1 and z[1,3]=1 

or this last is the correct code I should use???

Thanks a lot in advance for all your help

[Victor Zverovich]

You can't use sum in a fix statement (thus the error), but you can use it in a constraint:

  s.t. c{j in Location, m in Arrange}:

    z[j,m] = 1 ==> sum{l in Arrange:l>=m} z[j,m]=1;

This uses indicator constraints and assumes that z is binary, but a more general formulation is possible.

HTH,

Victor

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[Alessandro Oscar Gilardino Arias]

Dear Victor, thank you for the answer.

I tried to use what you told me

s.t. c{j in P_POTENCIAL, m in C_GENERAL}:

    z[j,m] = 1 ==> sum{l in C_GENERAL:l>=m} z[j,m]=1; 

but I have the followng message

localizacion.run, line 16 (offset 437):

Abandoning compound command to accept declaration of Fijacion_Z.

context:  if z[j,m]=1 then sum{l in C_GENERAL:l>=m}  >>> z[j,l]=1; <<< 

Also I tried with the following constraint:

subject to Fijacion_Z {j in P_POTENCIAL, m in C_GENERAL}:

if z[j,m]=1 then sum{l in C_GENERAL:l>=m} z[j,l]=1

But i have the following message:

localizacion.run, line 16 (offset 437):

Abandoning compound command to accept declaration of Fijacion_Z.

context:  if z[j,m]=1 then sum{l in C_GENERAL:l>=m}  >>> z[j,l]=1; <<< 

Please, could you give me a hand?

Thanks in advance

[Victor Zverovich]

The error "Abandoning compound command to accept declaration of Fijacion_Z" means that you declare Fijacion_Z in a for loop or other compound statement. To fix this move the declaration outside of the loop.

HTH,

Victor

[Alessandro Oscar Gilardino Arias]

Dear all, thanks for your answers.

In our private emailing you told me to use the following:

for {j in P_POTENTIAL, l in 1..min {m in C_GENERAL: z[j,l]=1} - 1} fix z[j,l];

I tried either this and also I tried the following modifications:

I being trying the following:

for {j in P_POTENTIAL, l in 1..(min {m in C_GENERAL: z[j,m]=1} - 1)-1} fix z[j,l]

(min {m in C_GENERAL: z[j,m]=1} - 1) - 1 this will allow me to, if for example, z[1,2] = 1 then for the variable z[1,1] = 0 because any arrange with a smaller capacity than the arrange number 2 shouldn't be activated. Then I am using the statement lines above.

Also I think I can do that with the followin statement:

for {j in P_POTENTIAL, l in ((min {m in C_GENERAL: z[j,m]=1}-1) -1 )..card(C_GENERAL) } fix z[j,l];

but in both cases I get the following message:

Error at _cmdno 226 executing "for" command

(file localizacion.run, line 15, offset 338):

range with infinite lower bound

Also, I being trying to do the following:

for {j in P_POTENTIAL, m in C_GENERAL}

      if z[j,m] = 1 then if (min {m in C_GENERAL: z[j,l]=1}) = 1 then fix (z[j.1] + z[j,2] + z[j,3]) else if (min {m in C_GENERAL: z[j,l]=1}) = 2 then fix (z[j,2] + z[j,3]) else if (min {m in C_GENERAL: z[j,l]=1}) = 3 then fix z[j,3];

    

but I have the following message:

ocalizacion.run, line 16 (offset 414):

syntax error

context:  if z[j,m] = 1 then if (min {m  >>> in  <<< C_GENERAL: z[j,l]=1}) = 1 then fix (z[j.1] + z[j,2] + z[j,3]) else if (min {m in C_GENERAL: z[j,l]=1}) = 2 then fix (z[j,2] + z[j,3]) else if (min {m in C_GENERAL: z[j,l]=1}) = 3 then fix z[j,3];

Thanks in advance for all your help and I am very sorry for being so annoying 

[Robert Fourer]

Actually it should be

for {j in P_POTENTIAL, l in 1..min {m in C_GENERAL: z[j,m]=1} m - 1} fix z[j,l];

But this still has a problem if it is possible to have a case where z[j,m] = 0 for all m. In that case the set {m in C_GENERAL: z[j,m]=1} is empty and the minimum is Infinity -- hence the error message about a range with an infinite bound. If you need to handle this case then I think you can write

for {j in P_POTENTIAL, l in 1..min ({m in C_GENERAL: z[j,m]=1} m, card(C_GENERAL)+1) - 1} fix z[j,l];

though I haven't been able to try this out on a real case. Similarly in your other example you should write

if (min {l in C_GENERAL: z[j,l]=1} l) = 1

since the AMPL "min" operator requires an operand.

[Alessandro Oscar Gilardino Arias]

Dear Bob, thanks a lot for your answer, and I am very sorry for my late response but I've been very busy.

I would like to ask you if you could please explain me what means the following statement:

for {j in P_POTENTIAL, l in 1..min ({m in C_GENERAL: z[j,m]=1} m, card(C_GENERAL)+1) - 1} fix z[j,l];

Thanks a lot again for your answer

[Victor Zverovich]

The expression

  min ({m in C_GENERAL: z[j,m]=1} m, card(C_GENERAL)+1)

returns the minimum index m such that z[j,m] = 1 or, if there is no such index, the number of elements in C_GENERAL + 1. So the statement

  for {j in P_POTENTIAL, l in 1..min ({m in C_GENERAL: z[j,m]=1} m, card(C_GENERAL)+1) - 1} fix z[j,l];

fixes variables z[j,l] for all j in P_POTENTIAL and l ranging from 1 to the minimum index m such that z[j,m] = 1 (see above) to their current values.

HTH,

Victor

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[Alessandro Oscar Gilardino Arias]

Thanks a lot for your help Victor and Bob.

Best regards