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?
Related
For example, the analogous two dimensional question, x + y = 2n is easy to solve: one can simply consider pairs (i,2n-i) for i=1,2,...,n and thus index every solution, exactly once. We note that we have n such pairs solving x + y = 2n, for every fixed value of positive integer n, and so the cardinality of such a set is equal to n as expected.
However, trying to repeat the same problem for x + y + z = 2n, it is not clear to me how (or if it is possible) to write down a minimal set {(2n-i-j,i,j)} such that varying i and j over particular intervals precisely produces every such triplet, exactly once. It can be shown that the number of elements in such a minimal set would be equal to the nearest integer to n^2/3.
It is not hard to see how one can obtain such an indexing with repetitions, or how one can algorithmically remove repetitions, but what I would like to know is whether there is a clean, general construction, as for the x + y = 2n case. Is this possible, or will one always have to artificially restrict certain values of the parameters on the intervals for which they are defined?
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.
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
How would you possibly show that 2 is O(1)?
More over, how would you show that a constant is theta(1) hence omega(1) and O(1)?
For O, I am under the impression that you are able to do a simplification for f(n), whereby it can be reduced down to 1, but then how can this prove that 2 is O(1) for some n0? What would be the n0 value in this case?
By definition, a function f is in O(1) if there exist constants n0 and M such that f(n) ≤ M · 1 = M for all n ≥ n0.
If f(n) is defined as 2, then just set M = 2 (or any greater value; it doesn't matter) and n0 = 1 (or any greater value; it doesn't matter), and the condition is met.
[…] that 2 is O(1) for some n0? What would be the n0 value in this case?
n0 is not a parameter here; it's not meaningful to say "O(1) for some n0". You can arbitrarily choose any value of n0 that makes f satisfy the condition; if one exists, then f is O(1), period.
Big Oh and Theta so not indicate the time taken by an algorithm. They indicate the rate of increase in time as the input increases for the algorithm. When you understand this, things become very easy and less mathematical. f(x) = 2 {for all and any x} is always O(1) since the output value (2) does not depend on the input value (x) at all! O(1) represents this independence. So does theta(1) and omega(1).
I'm new to cryptography and modular arithmetic. So, I'm sure it's a silly question, but I can't help it.
How do I calculate a from
pow(a,q) = 1 (mod p),
where p and q are known? I don't get the "1 (mod p)" part, it equals to 1, doesn't it? If so, than what is "mod p" about?
Is this the same as
pow(a,-q) (mod p) = 1?
The (mod p) part refers not to the right hand side, but to the equality sign: it says that modulo p, pow(a,q) and 1 are equal. For instance, "modulo 10, 246126 and 7868726 are equal" (and they are also both equal to 6 modulo 10): two numbers x and y are equal modulo p if they have the same remainder on dividing by p, or equivalently, if p divides x-y.
Since you seem to be coming from a programming perspective, another way of saying it is that pow(a,q)%p=1, where "%" is the "remainder" operator as implemented in several languages (assuming that p>1).
You should read the Wikipedia article on Modular arithmetic, or any elementary number theory book (or even a cryptography book, since it is likely to introduce modular arithmetic).
To answer your other question: there is no general formula for finding such an a (to the best of my knowledge) in general. Assuming that p is prime, and using Fermat's little theorem to reduce q modulo p-1, and assuming that q divides p-1 (or else no such a exists), you can produce such an a by taking a primitive root of p and raising it to the power (p-1)/q. [And more generally, when p is not prime, you can reduce q modulo φ(p), then assuming it divides φ(p) and you know a primitive root (say r) mod p, you can take r to the power of φ(p)/q, where φ is the totient function -- this comes from Euler's theorem.]
Not silly at all, as this is the basis for public-key encryption. You can find an excellent discussion on this at http://home.scarlet.be/~ping1339/congr.htm#The-equation-a%3Csup%3Ex.
PKI works by choosing p and q that are large and relatively prime. One (say p) becomes your private key and the other (q) is your public key. The encryption is "broken" if an attacker guesses p, given aq (the encrypted message) and q (your public key).
So, to answer your question:
aq = 1 mod p
This means aq is a number that leaves a remainder of 1 when divided by p. We don't care about the integer portion of the quotient, so we can write:
aq / p = n + 1/p
for any integer value of n. If we multiply both sides of the equation by p, we have:
aq = np + 1
Solving for a we have:
a = (np+1)1/q
The final step is to find a value of n that generates the original value of a. I don't know of any way to do this other than trial and error -- which equates to a "brute force" attempt to break the encryption.