I have defined the variables w in AMPL in the following way:
var w[F,J,T] binary;
where F, J and T are sets.
After solving the problem, I would like to print the tables w[f,*,*] for each f in F. What is the command that does this?
If I just write
display w;
it returns me the tables w[*,j,*] for each j in J (I guess it's a question of the size of the results).
Thanks to anyone who will answer
By the way, for future reference, the answer is:
display {f in F} : {j in J, t in T} w[f, j, t];
Related
I'm trying to solve a linear relaxation of a problem I've already solved with a Python library in order to see if it behaves in the same way in Xpress Mosel.
One of the index sets I'm using is not the typical c=1..n but a set of sets, meaning I've taken the 1..n set and have created all the combinations of subsets possible (for example the set 1..3 creates the set of sets {{1},{2},{3},{1,2},{2,3},{1,2,3}}).
In one of my constraints, one of the indexes must run inside each one of those subsets.
The respective code in Python is as follows (using the Gurobi library):
cluster=[1,2,3,4,5,6]
cluster1=[]
for L in range(1,len(cluster)+1):
for subset in itertools.combinations(cluster, L):
clusters1.append(list(subset))
ConstraintA=LinExpr()
ConstraintB=LinExpr()
for i in range(len(nodes)):
for j in range(len(nodes)):
if i<j and A[i][j]==1:
for l in range(len(clusters1)):
ConstraintA+=z[i,j]
for h in clusters1[l]:
restricao2B+=(x[i][h]-x[j][h])
model.addConstr(ConstraintA,GRB.GREATER_EQUAL,ConstraintB)
ConstraintA=LinExpr()
ConstraintB=LinExpr()
(In case the code above is confusing, which I suspect it to be)The constraint I'm trying to write is:
z(i,j)>= sum_{h in C1}(x(i,h)-x(j,h)) forall C1 in C
in which the C1 is each of those subsets.
Is there a way to do this in Mosel?
You could use some Mosel code along these lines (however, independently of the language that you are using, please be aware that the calculated 'set of all subsets' very quickly grows in size with an increasing number of elements in the original set C, so this constraint formulation will not scale up well):
declarations
C: set of integer
CS: set of set of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
CS:=union(i in C) {{i}}
forall(j in 1..C.size-1)
forall(s in CS | s.size=j, i in C | i > max(k in s) k ) CS+={s+{i}}
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
Giving this some more thought, the following version working with lists in place of sets is more efficient (that is, faster):
uses "mmsystem"
declarations
C: set of integer
L: list of integer
CS: list of list of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
L:=list(C)
qsort(SYS_UP, L) ! Making sure L is ordered
CS:=union(i in L) [[i]]
forall(j in 1..L.size-1)
forall(s in CS | s.size=j, i in L | i > s.last ) CS+=[s+[i]]
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
I want to know the time complexity of the code attached.
I get O(n^2logn), while my friends get O(nlogn) and O(n^2).
SomeMethod() = log n
Here is the code:
j = i**2;
for (k = 0; k < j; k++) {
for (p = 0; p < j; p++) {
x += p;
}
someMethod();
}
The question is not very clear about the variable N and the statement i**2.
i**2 gives a compilation error in java.
assuming someMethod() takes log N time(as mentioned in question), and completely ignoring value of N,
lets call i**2 as Z
someMethod(); runs Z times. and time complexity of the method is log N so that becomes:
Z * log N ----------------------------------------- A
lets call this expression A.
Now, x+=p runs Z^2 times (i loop * j loop) and takes constant time to run. that makes the following expression:
( Z^2 ) * 1 = ( Z^2 ) ---------------------- B
lets call this expression B.
The total run time is sum of expression A and expression B. which brings us to:
O((Z * log N) + (Z^2))
where Z = i**2
so final expression will be O(((i**2) * log N) + ((i**2)^2))
if we can assume i**2 is i^2, the expression becomes,
O(((i^2) * log N) + (i^4))
Considering only the higher order variables, like we consider n^2 in n^2 + 2n + 5, the complexity can be expressed as follows,
i^4
Based on the picture, the complexity is O(logNI2 + I4).
We cannot give a complexity class with one variable because the picture does not explain the relationship between N and I. They must be treated as separate variables.
And likewise, we cannot eliminate the logNI2 term because the N variable will dominate the I variable in some regions of N x I space.
If we treat N as a constant, then the complexity class reduces to O(I4).
We get the same if we treat N as being the same thing as I; i.e. there is a typo in the question.
(I think there is mistake in the way the question was set / phrased. If not, this is a trick question designed to see if you really understood the mathematical principles behind complexity involving multiple independent variables.)
I'm trying to set up a ampl model which clusters given points in a 2-dimensional space according to the model of Saglam et al(2005). For testing purposes I want to generate randomly some datapoints and then calculate the euclidian distance matrix for them (since I need this one). I'm aware that I could only make the distance matrix without the data points but in a later step the data points will be given and then I need to calculate the distances between each the points.
Below you'll find the code I've written so far. While loading the model I keep getting the error message "i is not defined". Since i is a subscript that should run over x1 and x1 is a parameter which is defined over the set D and have one subscript, I cannot figure out why this code should be invalid. As far as I understand, I don't have to define variables if I use them only as subscripts?
reset;
# parameters to define clustered
param m; # numbers of data points
param n; # numbers of clusters
# sets
set D := 1..m; #points to be clustered
set L := 1..n; #clusters
# randomly generate datapoints
param x1 {D} = Uniform(1,m);
param x2 {D} = Uniform(1,m);
param d {D,D} = sqrt((x1[i]-x1[j])^2 + (x2[i]-x2[j])^2);
# variables
var x {D, L} binary;
var D_l {L} >=0;
var D_max >= 0;
#minimization funcion
minimize max_clus_dis: D_max;
# constraints
subject to C1 {i in D, j in D, l in L}: D_l[l] >= d[i,j] * (x[i,l] + x[j,l] - 1);
subject to C2 {i in D}: sum{l in L} x[i,l] = 1;
subject to C3 {l in L}: D_max >= D_l[l];
So far I tried to change the line form param x1 to
param x1 {i in D, j in D} = ...
as well as
param d {x1, x2} = ...
Alas, nothing of this helped. So, any help someone can offer is deeply appreciated. I searched the web but I found nothing useful for my task.
I found eventually what was missing. The line in which I calculated the parameter d should be
param d {i in D, j in D} = sqrt((x1[i]-x1[j])^2 + (x2[i]-x2[j])^2);
Retrospectively it's clear that the subscripts i and j should have been mentioned on the line, I don't know how I could miss that.
I have a model for an optimization problem using three different sets of ordered pairs. Using this constraint:
subject to Constraint8 {t in T, j in J, a in A}:
sum{i in I: (i,j) in LINKS} l[i,t,a] >= sum{v in V, (ih,jh) in REV:(ih,j) not in LINKS} x[ih,j,v,t,a]+1;
Actually works and gives a correct solution. I want to alter this constraint though, so I comment it out, and then write the modified version:
subject to Constraint8 {t in T, j in J, a in A}:
sum{i in I: (i,j) in LINKS} l[i,t,a] >= sum{g in T: g <= t, v in V, (ih,jh) in REV: (ih,j) not in LINKS} x[ih,j,v,g,a] +1;
(with the difference that x is now summed over the set T as well, however on the element g in T: g <= t. However, AMPL does not seem to like this and this gives a syntax error. I've been trying it on some different "places" within the sum, but it does not work.
Does anyone have a clue regarding this error?
In AMPL, as well as in mathematical notation, you should specify the "such that" condition at the end of the indexing expression:
subject to Constraint8 {t in T, j in J, a in A}:
sum{i in I: (i,j) in LINKS} l[i,t,a] >=
sum{g in T, v in V, (ih,jh) in REV: g <= t and (ih,j) not in LINKS}
x[ih,j,v,g,a] + 1;
I am new in GLPK. This is some of my code:
set I := setof{(i,r,p,d) in T} i;
var Y{I,I}, binary;
s.t. c1{i in I, j in I}: sum{Y[i,j]} = 6;
I want to have only six values in Y that are 1. Can anyone tell me how to do it in proper way? Because s.t. c1{i in I, j in I}: sum{Y[i,j]} = 6;always produces an error.
Thank you.
This is just a syntax problem. The constraint should look like the following:
s.t. c1: sum{i in I, j in I}(Y[i,j]) = 6;
The first brackets after the name of your constraints imply that the constraint is applied to every single [I, I]. What you want is to fix the sum of all Y in your problem, so you need the constraint to only apply once to your problem (so delete these brackets).
In the sum-syntax don't put the variable you want to sum in the brackets, they belong after them. Inside the brackets you can define the range of the sum.