I am working on a supply chain optimisation problem using Pyomo and I need to set up a constraint on specific variables in the model. The constraint is that the variable should be within the set (0,1) or (200, to infinity). However, when I try to set up that constraint I am getting a TypeError, here is my code:
def rail_rule(model):
for route in routes:
if "rail" in route[0].lower():
dest = route[0]
dg = route[1]
sg = route[2]
site = route[3]
return model.x[dest, dg, sg, site]>=200 or model.x[dest, dg, sg, site]<=1
model.railconst = Constraint(rule = rail_rule)
I get this error when I run it :
TypeError: Relational expression used in an unexpected Boolean context.
The inequality expression:
200.0 <= x[RAIL - KENSINGTON,8,8,BROCKLESBY]
contains non-constant terms (variables) that were evaluated in an
unexpected Boolean context at
File '<ipython-input-168-901363ebc86f>', line 8:
return model.x[dest, dg, sg, site]>=200 or model.x[dest, dg, sg, site]<=1
Evaluating Pyomo variables in a Boolean context, e.g.
if expression <= 5:
is generally invalid. If you want to obtain the Boolean value of the
expression based on the current variable values, explicitly evaluate the
expression using the value() function:
if value(expression) <= 5:
or
if value(expression <= 5):
So my understanding is that I cant give Pyomo a boolean expression as a constraint, but I am quite new to Pyomo and not too sure if that's what my issue is or if I am doing it correctly.
This constraint could also be implemented in the variable intialisation as boundaries, but I cant find a way to set up two boundaries on a single variable in Pyomo.
Thanks !
There are different ways to handle this:
(1) Use binary variables. Assume you have a good upper bound on x, i.e., x ∈ [0, U]. Then formulate the constraints
x ≤ 1 + (U-1) δ
x ≥ 200 δ
δ ∈ {0,1} (binary variable)
This is the easiest way.
(2) If you don't have a good upper bound on x, you can use a SOS1 set. (SOS1 means Special Ordered Set of type 1). Assume x,s1,s2 ≥ 0.
x ≤ 1 + s1
x ≥ 200 - s2
s1,s2 ∈ SOS1 (s1,s2 form a SOS1 set)
(3) Use disjunctive programming.
Related
Let's say I have a set:
S: Id X Counter
Id: \nat
Counter: \nat
Need help to define an operation filter which takes in two parameters, x:\nat and y:\nat which I can apply to S. This function will match on the first parameter with \exists a \in S # first(a) = x and then reduce second(a) with y. If the value is less than or equal to zero, a should be removed from the set S, otherwise (first(a), second(a) - y) should replace a.
If the above is indecipherable, just please give me any example definition that operates over a set and modifies it.
Thanks.
I am new in optimisation problem. In my optimisation problem, I have one binary variable y and one integer variable n.
I have L: L= b*c*y
Where b, c are parameters.
and the constraint is: L/n<1 (which means L over n). I can also consider L<n. Another constraint in my problem is: I need that if L==0 then n=0 (n should be equal to 0, too). Moreover, L and N are always non-negative.
How can I linearise these constraints? Is it possible?
I need to rely on the fact that two Z3 variables
can not have the same name.
To be sure of that,
I've used tuple_example1() from test_capi.c in z3/examples/c and changed the original code from:
// some code before that ...
x = mk_real_var(ctx, "x");
y = mk_real_var(ctx, "y"); // originally y is called "y"
// some code after that ...
to:
// some code before that ...
x = mk_real_var(ctx, "x");
y = mk_real_var(ctx, "x"); // but now both x and y are called "x"
// some code after that ...
And (as expected) the output changed from:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
invalid
counterexample:
y -> 0.0
x -> 1.0
to:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
valid
BUG: unexpected result.
However, when I looked closer, I found out that Z3 did not really fail or anything, it is just a naive (driver) print out to console.
So I went ahead and wrote the exact same test with y being an int sort called "x".
To my surprise, Z3 could handle two variables with the same name when they have different sorts:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
invalid
counterexample:
x -> 1.0
x -> 0
Is that really what's going on? or is it just a coincidence??
Any help is very much appreciated, thanks!
In general, SMT-Lib does allow repeated variable names, so long as they have different sorts. See page 27 of the standard. In particular, it says:
Concretely, a variable can be any symbol, while a function symbol
can be any identifier (i.e., a symbol or an indexed symbol). As a
consequence, contextual information is needed during parsing to know
whether an identifier is to be treated as a variable or a function
symbol. For variables, this information is provided by the three
binders which are the only mechanism to introduce variables. Function
symbols, in contrast, are predefined, as explained later. Recall that
every function symbol f is separately associated with one or more
ranks, each specifying the sorts of f’s arguments and result. To
simplify sort checking, a function symbol in a term can be annotated
with one of its result sorts σ. Such an annotated function symbol is a
qualified identifier of the form (as f σ).
Also on page 31 of the same document, it further clarifies "ambiguity" thusly:
Except for patterns in match expressions, every occurrence of an
ambiguous function symbol f in a term must occur as a qualified
identifier of the form (as f σ) where σ is the intended output sort of
that occurrence
So, in SMT-Lib lingo, you'd write like this:
(declare-fun x () Int)
(declare-fun x () Real)
(assert (= (as x Real) 2.5))
(assert (= (as x Int) 2))
(check-sat)
(get-model)
This produces:
sat
(model
(define-fun x () Int
2)
(define-fun x () Real
(/ 5.0 2.0))
)
What you are observing in the C-interface is essentially a rendering of the same. Of course, how much "checking" is enforced by the interface is totally solver specific as SMT-Lib says nothing about C API's or API's for other languages. That actually explains the BUG line you see in the output there. At this point, the behavior is entirely solver specific.
But long story short, SMT-Lib does indeed allow two variables have the same name used so long as they have different sorts.
I'm trying to solve a model using Julia-JuMP. The following is the outline of the model that I created. Here, z[i,j] is a binary variable and d[i,j] is the cost for which z[i,j]=1.
My constraint creates an infinite number of constraint and hence I need to use a separation algorithm to solve it.
First, I solve the model without any constraint, so the answer to all variables z[i,j] and d[i,j] are zero.
Then, I'm including the separation algorithm (which is given inside the if condition). Even though I'm including if z_value == 0, z_values are not passing to it.
Am I missing something in the format of this model?
m = Model(solver=GurobiSolver())
#variable(m, z[N,N], Bin)
#variable(m, d[N,N]>=0)
#objective(m, Min, sum{ d[i,j]*z[i,j], i in N, j in N} )
z_value = getvalue(z)
d_value = getvalue(d)
if z_value == 0
statement
elseif z_value == 1
statement
end
#constraint(m, sum{z[i,j], i in N, j in N}>=2)
solve(m)
println("Final solution: [ $(getvalue(z)), $(getvalue(d)) ]")
You're multiplying z by d which both are variables, hence your model is non-linear,
Are the costs d[i,j] constant or really a variable of the problem ?
If so you need to use a non-linear solver
I'm trying to figure out why I get this error in one of my constraints.
The line is as follows:
subject to resource1{t in time: operation[1,1,t]= 1}:
sum {p in plane, m in t..(t+process_time[p,1])}
(operation[p,1,m] + operation[p,8,m]) <= 1;
This is pretty much saying that for all time where the binary variable "operation" = 1, which means that operation is using a particular machine/resource at given time t, no other operation must start until that operation is done processing.
You can't use variables in indexing expressions in AMPL. If you are using CPLEX, you can formulate this as an indicator constraint instead:
subject to resource1{t in time}:
operation[1,1,t] = 1 ==> sum {p in plane, m in t..(t+process_time[p,1])}
(operation[p,1,m] + operation[p,8,m]) <= 1;
Alternatively, you can use some kind of a Big M formulation.