Now I have variable set x in my ampl model, and I want to define a default value for each x.
set N := 1..10;
var x {i in N} default 0;
If I build the model like this, the initial value of all x would be set as 0. How can I set different value for each x, like [0,0,0,0,0,1,1,1,1,1]?
If you just want to change some parts of x to a non-default value, this is easy to do. For example:
var x{i in N} default 0;
let{i in 6..10} x[i] := 1;
I'm not aware of any way to have more than one default value for different elements of a var.
Related
I'm having troubles to define the objective fucntion in a SMT problem with z3py.
Long story, short, I have to optimize the placing of smaller blocks inside a board that has fixed width but variable heigth.
I have an array of coordinates (represented by an array of integers of length 2) and a list of integers (representing the heigth of the block to place).
# [x,y] list of integer variables
P = [[Int("x_%s" % (i + 1)), Int("y_%s" % (i + 1))]
for i in range(blocks)]
y = [int(b) for a, b in data[2:]]
I defined the objective function like this:
obj= Int(max([P[i][1] + y[i] for i in range(blocks)]))
It calculates the max height of the board given the starting coordinate of the blocks and their heights.
I know it could be better, but I think the problem would be the same even with a different definition.
Anyway, if I run my code, the following error occurs on the line of the objective function:
" raise Z3Exception("Symbolic expressions cannot be cast to concrete Boolean values.") "
While debugging I've seen that is P[i][1] that gives an error and I think it's because the program reads "y_i + 3" (for example) and they can't be added togheter.
Point is: it's obvious that the objective function depends on the variables of the problem, so how can I get rid of this error? Is there another place where I should define the objective function so it waits to have the P array instantiated before doing anything?
Full code:
from z3 import *
from math import ceil
width = 8
blocks = 4
x = [3,3,5,5]
y = [3,5,3,5]
height = ceil(sum([x[i] * y[i] for i in range(blocks)]) / width) + 1
# [blocks x 2] list of integer variables
P = [[Int("x_%s" % (i + 1)), Int("y_%s" % (i + 1))]
for i in range(blocks)]
# value/ domain constraint
values = [And(0 <= P[i][0], P[i][0] <= width - 1, 0 <= P[i][1], P[i][1] <= height - 1)
for i in range(blocks)]
obj = Int(max([P[i][1] + y[i] for i in range(blocks)]))
board_problem = values # other constraints I've not included for brevity
o = Optimize()
o.add(board_problem)
o.minimize(obj)
if (o.check == 'unsat'):
print("The problem is unsatisfiable")
else:
print("Solved")
The problem here is that you're calling Python's max on symbolic values, which is not designed to work for symbolic expressions. Instead, define a symbolic version of max and use that:
# Return maximum of a vector; error if empty
def symMax(vs):
m = vs[0]
for v in vs[1:]:
m = If(v > m, v, m)
return m
obj = symMax([P[i][1] + y[i] for i in range(blocks)])
With this change your program will go through and print Solved when run.
set P;
set K;
set I:= {i in K};
set J:={j in P};
param C {P} >=0;
param A {K,P} >=0;
param B {K} >=0;
var X{j in P} >=0;
P consists of 4 set values, namely sweatshirt-f, sweatshirtB/F, tshirtf and tshirtBF, but I would like sweatshirt-f only to return an integer:
maximize f: sum{j in P} C[j]*X[j];
s.t. Constraint {i in K}:
sum{j in P} A[i,j]*X[j]<=B[i];
Option 1: declare all of them as integer, then relax integer constraints for the others:
var X{j in P} >=0 integer;
let {i in P: i <> "sweatshirt-f"} X[i].relax := 1;
Option 2: declare a dummy variable as integer, then constrain relevant value to match it:
var X{j in P} >=0;
var int_dummy integer;
s.t. force_integer: X["sweatshirt-f"] = int_dummy;
I'm not able to test those right now, so you may have to do a little debugging, but that should get you in the neighbourhood of a solution.
I'm doing variable frequency clock on AHDL. Algoritm is: one counter (trigger) counts from 0 to x, and when it reaches x - we have pulse. I have another trigger which is used to store that X. Also I have two inputs plus and minus which are used to change frequency (increase or decrease X value).
And I have following code:
constant maximum = 9;
constant minimum = 1;
constant default = 5;
subdesign generator(
plus, minus, clk: input;
pulse, level[3..0], curr_val[3..0]: output;
)
variable
level[3..0]: dff;
curr_val[3..0]: dff;
begin
defaults
level[].d = default; % load 5 as default X value %
end defaults;
level[].clk = clk;
curr_val[].clk = clk;
pulse = (curr_val[] == level[]); % if main counter reached X - send one pulsation %
% main counter %
if curr_val[] < level[] then
curr_val[] = curr_val[] + 1;
elsif curr_val[] == level[] then
curr_val[] = 0;
end if;
% buttons %
if plus then
if (level[].q > minimum) then % if X == maximum ignore button %
level[].d = level[].q - 1;
end if;
end if;
if minus then
if (level[].q < maximum) then
level[].d = level[].q + 1;
end if;
end if;
end;
The problem is - after one tick when I change X (level[]) value - it goes back to default value. Am I missing something?
Syntax highlighting is wrong since wrong tag. % text % is commentary.
Found the problem.
The defaults block works everytime if value was not set. So, if you want to store same value you should set it at every time.
I really hope someone can help with this...
This is what i have in the .mod file
set I := 1..10;
set J := 1..10;
set K := 1..2;
set W := 1..20;
param v{K, W};
param d{I, J};
var x1, integer;
var y1, integer;
var x2, integer;
var y2, integer;
var assist{W}, binary;
and this is the code generating error:
minimize nome: sum{w in W} (if (assist[w] == 0) then
(if (x1 >= v[1,w]) then
(if (y1 >= v[2,w]) then
(d[x1 - v[1,w],y1 - v[2,w]])....
where the error regards the last line and says:
Variables in subscripts are not yet allowed.
context: (d[x1 - v[1,w],y1 - >>> v[2,w]] <<< )
this is one of constraints (others are just the same):
subject to rangex1:
x1 > 0 && x1 <= 10;
As the error message says, you can't use decision variables within a subscript in AMPL. In this case x1 and y1 are decision variables, so d[x1 - v[1,w],y1 - v[2,w]] is not allowed. You'll need to reformulate the problem in a way that avoids this issue.
I have an AMPL model file that I'm working to convert to GLPK. It begins:
param n; # The number of nodes in the graph
set V := {1 .. n}; # The set of vertices in the graph
set E within V cross V; # The set of edges in the graph
set NE within V cross V := {i in V, j in V: i < j} diff E;
set FIXED within V cross V default {}; # The set of demand pairs with fixed flow
When running this, I get the following error:
_test.mod:5: set expression following default must have dimension 2 rather than 1
Context: : i < j } diff E ; set FIXED within V cross V default { } ;
MathProg model processing error
This must be a syntactic difference between MathProg and its superset, AMPL—running the code in AMPL works perfectly. How does one express a 2D empty set in MathProg?
Alright, the hack of a solution is this:
set FIXED within V cross V default {i in V, j in V: 1 < 0};
Put an obviously false condition. It'll have the dimensionality you want and still be empty.