I have been trying to solve seriation problem by using GNU. But I couldn't write a summation like the following.
param n, integer, >= 3;
set O := 1..n;
param d{i in O,j in O};
var x{i in O,j in O}, binary, i < j;
var v{i in O,j in O,k in O}, binary, i < j < k;
maximize total: sum{i in O,j in O, i<j}(d[i,j] - d[j,i])* x[i,j] + sum{i in O,j in O, i<j}d[j,i];
s.t. tran{i in O,j in O,k in O, i<j<k}: x[i,j] + x[j,i] - x[i,k] + v[i,j,k] = 1;
Thanks
You should use : instead of , in the "such that" clause i < j:
sum{i in O,j in O: i < j} ...
# ^ note ':' here
Related
I am modeling a production problem in AMPL with start and end inventories to be zero.
Simple 3 products being produced over 4 day span, on a single machine.
however my model keeps having problems with the bigM constraint and no inventory is allocated.Model is infeasible per AMPL presolve.
I would really appreciate if someone could point whats wrong or what i should try. Thanks in advance!
HERE IS THE CODE -
param n;
param m;
param M;
param T;
set I := {1..n};
set J := {1..m};
param r {I} integer; #productionrate
param stp {I} integer; #setuptime
param rev {I} integer; #sellingprice
param h {I} integer; #inventorycost
param d {I,J} integer; #demand
var t {I,J} integer >= 0, <=480 default 50; #production time allocated for product i on day j
var s {I,J} integer >= 0 default 50; #inventory of product i on day j
var z {I,J} binary default 1; #setup for product i done on day j or not
#sum of revenue - sum of inventory cost
maximize K: sum {i in I,j in J}(rev[i] * t[i,j] * r[i] * z[i,j]) - sum {i in I,j in J}(h[i] * s[i,j]);
#inventory equation
s.t. c2 {i in I,j in 2..4}: s[i,j] = s[i,j-1] + t[i,j]*r[i] - d[i,j];
s.t. c14 {i in I}: s[i,1] = t[i,1]*r[i] - d[i,1];
#initial and end inventories
s.t. c3 {i in I}: s[i,1] = 0;
s.t. c4 {i in I}: s[i,4] = 0;
#ensuring time allocation only when setup
s.t. c5 {i in I,j in J}: t[i,j] <= M*z[i,j];
#ensuring demand is satisfied
s.t. c6 {i in I,j in 2..4}: s[i,j-1] + t[i,j]*r[i] = d[i,j];
s.t. c11 {i in I}: t[i,1]*r[i] = d[i,1];
#production time constraint
s.t. c7 {j in J}: sum{i in I}(z[i,j]*stp[i]) + sum{i in I}(t[i,j]) <= T;
#other non-negativity constraints
s.t. c12 {i in I,j in J}: s[i,j] >= 0;
#s.t. c13 {i in I,j in J}: t[i,j] >= 0;
end;
HERE IS THE DATA -
param n :=3;
param m :=4;
param T :=480;
param M :=480;
#param o :=2;
param d: 1 2 3 4 :=
1 400 600 200 800
2 240 440 100 660
3 80 120 40 100;
param r:=
1 5
2 4
3 2;
param stp:=
1 45
2 60
3 100;
param rev:=
1 50
2 70
3 120;
param h:=
1 2
2 1
3 3;
end;
RUN FILE -
#RESET THE AMPL ENVIROMENT
reset;
#LOAD THE MODEL
model 'EDX 15.053 Production problem.mod';
#LOAD THE DATA
data 'EDX 15.053 Production problem.dat';
#DISPLAY THE PROBLEM FORMULATION
expand K;
expand c2;
#expand c3;
expand c4;
expand c5;
expand c6;
expand c7;
expand c11;
#let s[3,3] := 20;
#CHANGE THE SOLVER (optional)
#option solver CBC;
option solver cplex, cplex_options ’presolve 0’;
option show_stats 1;
option send_statuses 0;
#option presolve 0;
#SOLVE
solve;
print {j in 1.._nvars:_var[j].status = "pre"}: _varname[j];
print {i in 1.._ncons:_con[i].status = "pre"}: _conname[i];
display solve_message;
display solve_result_num, solve_result;
#SHOW RESULTS
display t,s,z,K;
end;
I'm trying to find out the Complexity of the given program. Suppose we have;
int a = θ;
for (i=θ; i<n; i++){
for(j = n; j>i; j--)
{
a = a + i + j;
}
}
Complexity: O(N*N)
Explanation:
The code runs total no of times
`= N + (N – 1) + (N – 2) + … 1 + 0
= N * (N + 1) / 2
= 1/2 * N^2 + 1/2 * N
O(N^2) times`
Trying to prove a simple algorithm on Dafny, but I just get an "assertion violation" on the last assertion with no extra details. Can anybody spot what is wrong and how to fix it? Formal methods is not my specialty.
method BubbleSort(a: array?<int>)
modifies a
requires a != null
ensures sorted(a, 0, a.Length -1)
{
var i := a.Length - 1;
while(i > 0)
decreases i
invariant i < 0 ==> a.Length == 0
invariant -1 <= i < a.Length
invariant sorted(a, i, a.Length -1)
invariant partitioned(a, i)
{
var j := 0;
while(j < i)
decreases i - j
invariant 0 < i < a.Length && 0 <= j <= i
invariant sorted(a, i, a.Length -1)
invariant partitioned(a, i)
invariant forall k :: 0 <= k <= j ==> a[k] <= a[j]
{
if(a[j] > a[j+1])
{
a[j], a[j+1] := a[j+1], a[j];
}
j := j + 1;
}
i := i - 1;
}
}
predicate sorted(a: array?<int>, l: int , u: int)
reads a //Sintaxe do Dafny. PRECISA disso para dizer que vai ler o array
requires a != null
{
forall i, j :: 0 <= l <= i <= j <= u < a.Length ==> a[i] <= a[j]
}
predicate partitioned(a: array?<int>, i: int)
reads a
requires a != null
{
forall k, k' :: 0 <= k <= i < k' < a.Length ==> a[k] <= a[k']
}
method testSort()
{
var b := new int[2];
b[0], b[1] := 2, 1;
assert b[0] == 2 && b[1] == 1;
BubbleSort(b);
assert b[0] == 1 && b[1] == 2;
}
The problem is that the postcondition (ensures clause) of Sort gives no information about the state of A. When Dafny does verification, it verifies each method independently, using only the specifications (not the bodies) of other methods. So when Dafny verifies testSort, it doesn't look at the definition of Sort, but only its postcondition true, which isn't enough to prove your assertions.
For more information, see the FAQ and the section on assertions in the tutorial.
I'm trying write in OPL this sum:
I did this, but it is not exactly what I need.
forall (n in cont, t in tempo, o in portos)
sum(i in colunap, j in linhap)b[i][j][n][t] + v[n][t] == 1;
I should be something like, but opl does not accept it:
forall (n in cont[o], t in tempo[o], o in portos)
sum(i in colunap[o], j in linhap[o])b[i][j][n][t] + v[n][t] == 1;
This should work:
int P=3;
int H[1..P-1] = [1 , 2];
range linhap=1..max(o in 1..P-1) H[o];
I have the following model with a variable that is a value from a vector (index of p in objective function)
But AMPL displays an error: subscript variables are not yet allowed.
How can I do to implement this kind of addressing in objective function?
Thanks in advance and best regards.
Gabriel
param dimension;
set T:={1..dimension};
set O:={0};
set V:= O union T;
param c{i in V, j in V};
param p{i in V};
set ady{i in V} within V := {j in V : i<>j and c[i,j] <> -1} ;
# Variables
var x{i in V, j in V} binary;
var u{i in V} integer;
# Objective
minimize costo: sum{i in V, j in V} p[u[i]-1] * x[i,j] * c[i,j];
# Constraints
s.t. grado_a {j in V} : sum{i in ady[j] : j <> i} x[i,j] = 1;
s.t. grado_b {i in V} : sum{j in ady[i] : i <> j} x[i,j] = 1;
s.t. origen {i in O} : u[i] = 0;
s.t. sigo_1 {i in T} : u[i] >=1;
s.t. sigo_2 {i in T} : u[i] <= card(V) -1;
s.t. precedencia {i in T, j in T : i <> j} : u[i] - u[j] + 1 <= (card(V) - 1)*(1 - x[i,j]) ;
AMPL doesn't allow variables in subscripts yet. However, the ilogcp driver for AMPL supports the element constraint, for example:
include cp.ampl;
minimize costo:
sum{i in V, j in V} element({v in V} p[v], u[i] - 1) * x[i,j] * c[i,j];
where element({v in V} p[v], u[i] - 1) is equivalent to p[u[i] - 1] and is translated into an IloElement constraint.