How am I going to go about solving this with only a singular constraint and an unknown?
Refer to the following LP Formulation with unknown number S:
Max x1+x2
s.t
Sx1+x2 <= 1
x1,x2 => 0
How do I identify the unknown S to
(a) Having an optimal solution
(b) Being infeasible
(c) Being Unbound
Since the homework due date has probably passed, I thought I'd answer this because it is a little tricky. (b) almost got me..
(a) S>=0
(b) feasible for all S
(c) S<0
Related
Reduced cost give the dual variable corresponding to a box constraint of a variable as was pointed out in the related answer to this question: LP: postive reduced costs corresponding to positive variables?
How do I know whether the lower or upper bound is the active constraint? Of course, I could check whether the difference between the variable value and its bounds is smaller than some epsilon. However, the choice of such an epsilon is unclear to me, because a model could potentially attempt to fix a variable by setting the lower bound equal to the upper bound. In such a case, no epsilon could unambiguously indicate which of the bounds is the active one.
Does cplex provide the information which bound is active in its C++ api? Does any other LP solver do so? Is there another trick to figure out the active bound?
Thank you.
To a large extent you can look at the sign. The rule for reduced costs is:
Basic Non-basic-at-LB Non-basic-at-UB
minimization 0 >= 0 <= 0
maximization 0 <= 0 >= 0
Some problems with this:
Not all solvers may follow this (especially when maximizing).
Degeneracy may make things difficult
Most solvers will give you access to the basis status. E.g. Cplex has BasisStatus which gives you exactly the basis status of a variable: Basic, AtLower, AtUpper or FreeOrSuperbasic.
Question: How I may linearize the following constraint (joined figure), where
r[x][y], c[y,j], a[j,t] and p[x,t+l] are binary variables.
Could someone help me please ?
Best
Quadratic constraint
You can introduce an auxiliary binary variable ap, which holds the product of a and p.
Linearization can be done with:
ap <= a
ap <= p
ap >= a + p - 1
Then your remaining formula is ready to be formulated within gurobi (no more quadratic terms).
i want binary number that only have 0's at the beginning or end, for instance,
1111111
01111110
001111111
000111000
but no:
01001
0011101
do they have an specific name or property to get them?
I'm looking for something like linear integer optimization conditions, my solutions must have this form, but i can't think of any condition i can add to ensure that
Regards,
This is something which is not nice to formulate within mixed-integer programs. Most problems involving this are more suitable for alternative methods (SAT-solving, SMT-solving, Constraint-programming).
It can be done of course, but the solver will have some work as the formulation is non-trivial and introduces a lot of binary-variables (and the basic approach of MIP-solvers won't work amazingly here; bad integrality-gap).
I won't give you a complete solution, but some basic idea on how to formulate this and i also indicate how hard and cumbersome it is (there are alternative formulations; actually infinite many; but nothing much more simple).
The basic idea here is the following:
your binary number is constructed from N binary-variables
let's call them x
you introduce N auxiliary binary-variables l (left)
l[i] == 1 implicates: every l[j] with i<j is 0
you introduce N auxiliary binary-variables r (right)
r[i] == 1 implicates: every r[j] with i>j is 0
you add the following constraint for each position k:
x[k] == 0 implicates: l[i] == 1 for i < k OR r[i] == 1 for i>k
idea::
if there is a zero somewhere, either all on the left-side or all on the right-side are zeroes (or means: at least one side; but can be both)
To formulate this, you need two more ideas:
A: How to formulate the equality-check?
B: How to formulate the implication?
Remark: a -> b == not a or b (propositional calculus)
(this was wrongly stated earlier and corrected by OP)
These are common in MIP and you will find the solution in many integer-programming books, tutorial and papers. Here is an example (start with indicator-variables).
Another small common formulation:
if a is binary, b is binary:
a OR b is equivalent to: a+b >= 1 (the latter is a linear-expression ready to use for MIP)
Remark: The formulas in my idea-setting above might be wrong in regards to indices (i vs. i-1, vs. i+1) and binary-relations (<vs. <=). You will need to do the actual math yourself and just learn from the idea itself!
Remark 2: This kind of constraint is cumbersome in MIP, but more easily formulated within SAT and CP.
I am looking for a simple way to obtain a lot of "good" solutions in a LP problem (not MIP) with CPLEX, and not only (one of the) optimal basic solution(s). By "good" solutions I mean that the corresponding objective values are not so far from the real optimal value. Such pool of solutions could help the decision-maker...
More precisely, given a certain polyedron Ax<=b with x>=0 and an objective function z=cx I want to maximize, after running the LP, I can obtain the optimal value z*. Then I want to enumerate all the extreme points of the polyhedron given by the set of constraints
Ax <= b
cx >= z* - epsilon
x >= 0
when epsilon is a given tolerance.
I know that CPLEX offers way to generate solution pool (see here), but it will not function because this method is for MIP : it enumerates all the solutions of an IP (or one solution for every given set of fixed integer variables if the problem is a MIP).
An interesting efficient way is to visit the adjacent solutions of the optimal basic solution, i.e. all the adjacent extreme points : if I suppose the polyhedron is not degenerative, for each pair of basic variable x_B and non-basic variable x_N, I compute the basic solution obtained when x_B leaves the basis and x_N enters in the basis. Then I throw the solutions with cx < z*-epsilon, and for the others I repeat the procedure. [I know that I could improve this algorithm, but this is the general idea].
The routine CPPXpivot of the Callable Library could help to do this pivoting operation, but I did not find an equivalent in the C++ API (concert technology). Does someone know if such an equivalent exist, or could propose me an other way to answer my original problem ?
Thanks a lot :) !
Rémi L.
There is one interesting way to make this suitable for use with the Cplex solution pool. Use binary variables to encode the current basis, e.g. basis[k] = 0 meaning nonbasic and basis[k] = 1 indicating variable (or row) k is basic. Of course we have sum(k, basis[k]) = m (number of rows). Finally we have x[k] <= basis[k] * upperbound[k] (i.e. if nonbasic then zero -- assuming positive variables). When we add this to the LP model we end up with a MIP and can enumerate (all or some, optimal or near optimal) bases using the Cplex solution pool. See here and here.
I am trying to solve a minimisation problem and I want to minimise an expression
a/b
Where both a & b are variables. Hence this is not a linear problem...
How can I transform this function into an other one (being a linear one).
There is a detailed section on how to handle ratios in Linear Programming on the lpsolve site. It should be general enough to apply to AMPL and CPLEX as well.
There are several ways to do this, but the simplest to explain requires that you solve a series of linear programs. First, remove the objective and add a constraint
a <= c * b
Where c is a known upper bound on the solution. Then do a binary search on c you can a range where c_l, c_u where the problem is infeasible for
a <= c_l * b
but feasible for
a <= c_u * b
The general form of the obj should be a linear fractional function, something like f_{0}(x)=(c^Tx+d)/(e^Tx+f). For your case, X=(a,b),c=(1,0),(e=0,1),d=f=0.
To solve this kind of opt, something called linear fractional programming can be used. it's like linear constrainted version of linear fractional function and Charnes-Cooper transformation is applied to transform into a LP. You can find the main idea from wiki. Many OR books talk more about this such as pp53, pp165 in the Boyd's "convex optimization" (free to download).