# 14 Analyzing Infeasible Problems¶

When developing and implementing a new optimization model, the first attempts will often be either infeasible, due to specification of inconsistent constraints, or unbounded, if important constraints have been left out.

In this section we will

• go over an example demonstrating how to locate infeasible constraints using the MOSEK infeasibility report tool,
• discuss in more general terms which properties may cause infeasibilities, and
• present the more formal theory of infeasible and unbounded problems.

## 14.1 Example: Primal Infeasibility¶

A problem is said to be primal infeasible if no solution exists that satisfies all the constraints of the problem.

As an example of a primal infeasible problem consider the problem of minimizing the cost of transportation between a number of production plants and stores: Each plant produces a fixed number of goods, and each store has a fixed demand that must be met. Supply, demand and cost of transportation per unit are given in Fig. 2.

Fig. 2 Supply, demand and cost of transportation.

The problem represented in Fig. 2 is infeasible, since the total demand

$2300 = 1100+200+500+500$

exceeds the total supply

$2200 = 200+1000+1000$

If we denote the number of transported goods from plant $$i$$ to store $$j$$ by $$x_{ij}$$ , the problem can be formulated as the LP:

(1)$\begin{split}\begin{array}{lccccccccccccccl} \mbox{minimize} & x_{11} & + & 2x_{12} & + & 5x_{23} & + & 2x_{24} & + & x_{31} & + & 2x_{33} & + & x_{34} & & \\ \mbox{subject to} & x_{11} & + & x_{12} & & & & & & & & & & & \leq & 200, \\ & & & & & x_{23} & + & x_{24} & & & & & & & \leq & 1000,\\ & & & & & & & & & x_{31} & + & x_{33} & + & x_{34} & \leq & 1000,\\ & x_{11} & & & & & & & + & x_{31} & & & & & = & 1100,\\ & & & x_{12} & & & & & & & & & & & = & 200, \\ & & & & & x_{23} & + & & & & & x_{33} & & & = & 500, \\ & & & & & & & x_{24} & + & & & & & x_{34} & = & 500, \\ & x_{ij} \geq 0. & & & & & & & & & & & & & & \end{array}\end{split}$

Solving problem (1) using MOSEK will result in a solution, a solution status and a problem status. Among the log output from the execution of MOSEK on the above problem are the lines:

Basic solution
Problem status  : PRIMAL_INFEASIBLE
Solution status : PRIMAL_INFEASIBLE_CER


The first line indicates that the problem status is primal infeasible. The second line says that a certificate of the infeasibility was found. The certificate is returned in place of the solution to the problem.

## 14.2 Locating the cause of Primal Infeasibility¶

Usually a primal infeasible problem status is caused by a mistake in formulating the problem and therefore the question arises: What is the cause of the infeasible status? When trying to answer this question, it is often advantageous to follow these steps:

• Remove the objective function. This does not change the infeasibility status but simplifies the problem, eliminating any possibility of issues related to the objective function.
• Consider whether your problem has some necessary conditions for feasibility and examine if these are satisfied, e.g. total supply should be greater than or equal to total demand.
• Verify that coefficients and bounds are reasonably sized in your problem.

If the problem is still primal infeasible, some of the constraints must be relaxed or removed completely. The MOSEK infeasibility report (Sec. 14.4 (The Infeasibility Report)) may assist you in finding the constraints causing the infeasibility.

Possible ways of relaxing your problem nclude:

• Increasing (decreasing) upper (lower) bounds on variables and constraints.
• Removing suspected constraints from the problem.

Returning to the transportation example, we discover that removing the fifth constraint

$x_{12} = 200$

makes the problem feasible.

## 14.3 Locating the Cause of Dual Infeasibility¶

A problem may also be dual infeasible. In this case the primal problem is often unbounded, meaning that feasbile solutions exists such that the objective tends towards infinity. An example of a dual infeasible and primal unbounded problem is:

$\begin{split}\begin{array}{lcl} \mbox{minimize} & x_1 & \\ \mbox{subject to} & x_1 \leq 5. & \end{array}\end{split}$

To resolve a dual infeasibility the primal problem must be made more restricted by

• Adding upper or lower bounds on variables or constraints.
• Removing variables.
• Changing the objective.

### 14.3.1 A cautionary note¶

The problem

$\begin{split}\begin{array}{lcl} \mbox{minimize} & 0 & \\ \mbox{subject to} & 0 \leq x_1, & \\ & x_j \leq x_{j+1}, & j=1,\ldots ,n-1, \\ & x_n \leq -1 & \end{array}\end{split}$

is clearly infeasible. Moreover, if any one of the constraints is dropped, then the problem becomes feasible.

This illustrates the worst case scenario where all, or at least a significant portion of the constraints are involved in causing infeasibility. Hence, it may not always be easy or possible to pinpoint a few constraints responsible for infeasibility.

## 14.4 The Infeasibility Report¶

MOSEK includes functionality for diagnosing the cause of a primal or a dual infeasibility. It can be turned on by setting the MSK_IPAR_INFEAS_REPORT_AUTO to "MSK_ON". This causes MOSEK to print a report on variables and constraints involved in the infeasibility.

The MSK_IPAR_INFEAS_REPORT_LEVEL parameter controls the amount of information presented in the infeasibility report. The default value is $$1$$.

### 14.4.1 Example: Primal Infeasibility¶

We will keep working with the problem (1) written in LP format:

Listing 34 The code for problem (1). Click here to download.
\
\ An example of an infeasible linear problem.
\
minimize
obj: + 1 x11 + 2 x12
+ 5 x23 + 2 x24
+ 1 x31 + 2 x33 + 1 x34
st
s0: + x11 + x12       <= 200
s1: + x23 + x24       <= 1000
s2: + x31 + x33 + x34 <= 1000
d1: + x11 + x31        = 1100
d2: + x12              = 200
d3: + x23 + x33        = 500
d4: + x24 + x34        = 500
bounds
end



### 14.4.2 Example: Dual Infeasibility¶

The following problem is dual to (1) and therefore it is dual infeasible.

Listing 35 The dual of problem (1). Click here to download.
maximize + 200 y1 + 1000 y2 + 1000 y3 + 1100 y4 + 200 y5 + 500 y6 + 500 y7
subject to
x11: y1+y4 < 1
x12: y1+y5 < 2
x23: y2+y6 < 5
x24: y2+y7 < 2
x31: y3+y4 < 1
x33: y3+y6 < 2
x34: y3+y7 < 1
bounds
-inf <= y1 < 0
-inf <= y2 < 0
-inf <= y3 < 0
y4 free
y5 free
y6 free
y7 free
end


This can be verified by proving that

$(y_1,\ldots,y_7)=(-1,0,-1,1,1,0,0)$

is a certificate of dual infeasibility (see Sec. 16.1.2.2 (Dual Infeasible Problems)) as we can see from this report:

MOSEK DUAL INFEASIBILITY REPORT.

Problem status: The problem is dual infeasible

The following constraints are involved in the infeasibility.

Index    Name             Activity         Objective        Lower bound      Upper bound
5        x33              -1.000000e+00                     NONE             2.000000e+00
6        x34              -1.000000e+00                     NONE             1.000000e+00

The following variables are involved in the infeasibility.

Index    Name             Activity         Objective        Lower bound      Upper bound
0        y1               -1.000000e+00    2.000000e+02     NONE             0.000000e+00
2        y3               -1.000000e+00    1.000000e+03     NONE             0.000000e+00
3        y4               1.000000e+00     1.100000e+03     NONE             NONE
4        y5               1.000000e+00     2.000000e+02     NONE             NONE

Interior-point solution summary
Problem status  : DUAL_INFEASIBLE
Solution status : DUAL_INFEASIBLE_CER
Primal.  obj: 1.0000000000e+02    nrm: 1e+00    Viol.  con: 0e+00    var: 0e+00


Let $$y^*$$ denote the reported primal solution. MOSEK states

• that the problem is dual infeasible,
• that the reported solution is a certificate of dual infeasibility, and
• that the infeasibility measure for $$y^*$$ is approximately zero.

Since the original objective was maximization, we have that $$c^T y^* >0$$. See Sec. 16.1.2 (Infeasibility for Linear Optimization) for how to interpret the parameter values in the infeasibility report for a linear program. We see that the variables y1, y3, y4, y5 and the constraints x33 and x34 contribute to infeasibility with non-zero values in the Activity column.

One possible strategy to fix the infeasibility is to modify the problem so that the certificate of infeasibility becomes invalid. In this case we could do one the following things:

• Add a lower bound on y3. This will directly invalidate the certificate of dual infeasibility.
• Increase the object coefficient of y3. Changing the coefficients sufficiently will invalidate the inequality $$c^Ty^*>0$$ and thus the certificate.
• Add lower bounds on x11 or x31. This will directly invalidate the certificate of infeasibility.

Please note that modifying the problem to invalidate the reported certificate does not imply that the problem becomes dual feasible — the reason for infeasibility may simply move, resulting a problem that is still infeasible, but for a different reason.

More often, the reported certificate can be used to give a hint about errors or inconsistencies in the model that produced the problem.

## 14.5 Theory Concerning Infeasible Problems¶

This section discusses the theory of infeasibility certificates and how MOSEK uses a certificate to produce an infeasibility report. In general, MOSEK solves the problem

(2)$\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x+c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x \end{array}\end{split}$

where the corresponding dual problem is

(3)$\begin{split}\begin{array}{lccl} \mbox{maximize} & (l^c)^T s_l^c - (u^c)^T s_u^c & & \\ & + (l^x)^T s_l^x - (u^x)^T s_u^x + c^f & & \\ \mbox{subject to} & A^T y + s_l^x - s_u^x & = & c,\\ & -y + s_l^c - s_u^c & = & 0, \\ & s_l^c,s_u^c,s_l^x,s_u^x \leq 0. & & \end{array}\end{split}$

We use the convension that for any bound that is not finite, the corresponding dual variable is fixed at zero (and thus will have no influence on the dual problem). For example

$l_j^x = -\infty \quad \Rightarrow \quad (s_l^x)_j=0$

## 14.6 The Certificate of Primal Infeasibility¶

A certificate of primal infeasibility is any solution to the homogenized dual problem

$\begin{split}\begin{array}{lccl} \mbox{maximize} & (l^c)^T s_l^c - (u^c)^T s_u^c & & \\ & + (l^x)^T s_l^x - (u^x)^T s_u^x & & \\ \mbox{subject to} & A^T y + s_l^x - s_u^x & = & 0,\\ & -y + s_l^c - s_u^c & = & 0, \\ & s_l^c,s_u^c,s_l^x,s_u^x \leq 0. & & \end{array}\end{split}$

with a positive objective value. That is, $$(s_l^{c*},s_u^{c*},s_l^{x*},s_u^{x*})$$ is a certificate of primal infeasibility if

$(l^c)^T s_l^{c*} - (u^c)^T s_u^{c*} + (l^x)^T s_l^{x*} - (u^x)^T s_u^{x*} > 0$

and

$\begin{split}\begin{array}{lcl} A^T y + s_l^{x*} - s_u^{x*} & = & 0, \\ -y + s_l^{c*} - s_u^{c*} & = & 0, \\ s_l^{c*},s_u^{c*},s_l^{x*},s_u^{x*} \leq 0. & & \end{array}\end{split}$

The well-known Farkas Lemma tells us that (2) is infeasible if and only if a certificate of primal infeasibility exists.

Let $$(s_l^{c*},s_u^{c*},s_l^{x*},s_u^{x*})$$ be a certificate of primal infeasibility then

$(s_l^{c*})_i > 0 ((s_u^{c*})_i > 0)$

implies that the lower (upper) bound on the $$i$$ th constraint is important for the infeasibility. Furthermore,

$(s_l^{x*})_j > 0 ((s_u^{x*})_i > 0)$

implies that the lower (upper) bound on the $$j$$ th variable is important for the infeasibility.

## 14.7 The certificate of dual infeasibility¶

A certificate of dual infeasibility is any solution to the problem

$\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x & & \\ \mbox{subject to} & \bar{l}^c & \leq & A x & \leq & \bar{u}^c, \\ & \bar{l}^ x & \leq & x & \leq & \bar{u}^x \end{array}\end{split}$

with negative objective value, where we use the definitions

$\begin{split}\bar{l}_i^c := \left\lbrace \begin{array}{ll} 0, & l_i^c > -\infty ,\\ -\infty , & \mbox{otherwise}, \end{array} \right\rbrace,~ \bar{u}_i^c := \left\lbrace \begin{array} {ll} 0, & u_i^c < \infty , \\ \infty , & \mbox{otherwise}, \end{array} \right\rbrace\end{split}$

and

$\begin{split}\bar{l}_i^x := \left\lbrace \begin{array}{ll} 0, & l_i^x > -\infty , \\ -\infty , & \mbox{otherwise}, \end{array} \right\rbrace\quad \mbox{and} \quad \bar{u}_i^x := \left\lbrace \begin{array}{ll} 0, & u_i^x < \infty , \\ \infty , & \mbox{otherwise}. \end{array} \right\rbrace\end{split}$

Stated differently, a certificate of dual infeasibility is any $$x^*$$ such that

(4)$\begin{split}\begin{array}{lcccl} & & c^T x^* & < & 0, \\ \bar{l}^c & \leq & A x^* & \leq & \bar{u}^c, \\ \bar{l}^x & \leq & x^* & \leq & \bar{u}^x \end{array}\end{split}$

The well-known Farkas Lemma tells us that (3) is infeasible if and only if a certificate of dual infeasibility exists.

Note that if $$x^*$$ is a certificate of dual infeasibility then for any $$j$$ such that

$x_j^* \leq 0,$

variable $$j$$ is involved in the dual infeasibility.

The code in Listing 36 will form the repaired problem and solve it.

Listing 36 Feasibility repair example. Click here to download.
function feasrepairex1(inputfile)

cmd = sprintf('read(%s)', inputfile);
[r,res]=mosekopt(cmd);

res.prob.primalrepair = [];
res.prob.primalrepair.wux = [1,1];
res.prob.primalrepair.wlx = [1,1];
res.prob.primalrepair.wuc = [1,1,1,1];
res.prob.primalrepair.wlc = [1,1,1,1];

param.MSK_IPAR_LOG_FEAS_REPAIR = 3;
[r,res]=mosekopt('minimize primalrepair',res.prob,param);
fprintf('Return code: %d\n',r);

end


The parameter MSK_IPAR_LOG_FEAS_REPAIR controls the amount of log output from the repair. A value of 2 causes the optimal repair to printed out. If the fields wlx, wux, wlc or wuc are not specified, they are all assumed to be 1-vectors of appropriate dimensions.

The output from running the commands above is:

MOSEK Version 8.0.0.32(BETA) (Build date: 2016-7-17 10:54:55)
Copyright (c) MOSEK ApS, Denmark. WWW: mosek.com
Platform: Linux/64-X86

Open file '../feasrepair.lp'
Reading terminated. Time: 0.00

MOSEK Version 8.0.0.32(BETA) (Build date: 2016-7-17 10:54:55)
Copyright (c) MOSEK ApS, Denmark. WWW: mosek.com
Platform: Linux/64-X86

Problem
Name                   :
Objective sense        : min
Type                   : LO (linear optimization problem)
Constraints            : 4
Cones                  : 0
Scalar variables       : 2
Matrix variables       : 0
Integer variables      : 0

Primal feasibility repair started.
Optimizer started.
Interior-point optimizer started.
Presolve started.
Linear dependency checker started.
Linear dependency checker terminated.
Eliminator started.
Freed constraints in eliminator : 2
Eliminator terminated.
Eliminator - tries                  : 1                 time                   : 0.00
Lin. dep.  - tries                  : 1                 time                   : 0.00
Lin. dep.  - number                 : 0
Presolve terminated. Time: 0.00
Optimizer  - threads                : 20
Optimizer  - solved problem         : the primal
Optimizer  - Constraints            : 2
Optimizer  - Cones                  : 0
Optimizer  - Scalar variables       : 5                 conic                  : 0
Optimizer  - Semi-definite variables: 0                 scalarized             : 0
Factor     - setup time             : 0.00              dense det. time        : 0.00
Factor     - ML order time          : 0.00              GP order time          : 0.00
Factor     - nonzeros before factor : 3                 after factor           : 3
Factor     - dense dim.             : 0                 flops                  : 5.00e+01
ITE PFEAS    DFEAS    GFEAS    PRSTATUS   POBJ              DOBJ              MU       TIME
0   2.7e+01  1.0e+00  4.0e+00  1.00e+00   3.000000000e+00   0.000000000e+00   1.0e+00  0.00
1   2.5e+01  9.1e-01  1.4e+00  0.00e+00   8.711262850e+00   1.115287830e+01   2.4e+00  0.00
2   2.4e+00  8.8e-02  1.4e-01  -7.33e-01  4.062505701e+01   4.422203730e+01   2.3e-01  0.00
3   9.4e-02  3.4e-03  5.5e-03  1.33e+00   4.250700434e+01   4.258548510e+01   9.1e-03  0.00
4   2.0e-05  7.2e-07  1.1e-06  1.02e+00   4.249996599e+01   4.249998669e+01   1.9e-06  0.00
5   2.0e-09  7.2e-11  1.1e-10  1.00e+00   4.250000000e+01   4.250000000e+01   1.9e-10  0.00
Basis identification started.
Primal basis identification phase started.
ITER      TIME
0         0.00
Primal basis identification phase terminated. Time: 0.00
Dual basis identification phase started.
ITER      TIME
0         0.00
Dual basis identification phase terminated. Time: 0.00
Basis identification terminated. Time: 0.00
Interior-point optimizer terminated. Time: 0.01.

Optimizer terminated. Time: 0.03
Basic solution summary
Problem status  : PRIMAL_AND_DUAL_FEASIBLE
Solution status : OPTIMAL
Primal.  obj: 4.2500000000e+01    nrm: 6e+02    Viol.  con: 1e-13    var: 0e+00
Dual.    obj: 4.2499999999e+01    nrm: 2e+00    Viol.  con: 0e+00    var: 9e-11
Optimal objective value of the penalty problem: 4.250000000000e+01

Repairing bounds.
Increasing the upper bound -2.25e+01 on constraint 'c4' (3) with 1.35e+02.
Decreasing the lower bound 6.50e+02 on variable 'x2' (4) with 2.00e+01.
Primal feasibility repair terminated.
Optimizer started.
Presolve started.
Presolve terminated. Time: 0.00
Optimizer terminated. Time: 0.00

Interior-point solution summary
Problem status  : PRIMAL_AND_DUAL_FEASIBLE
Solution status : OPTIMAL
Primal.  obj: -5.6700000000e+03   nrm: 6e+02    Viol.  con: 0e+00    var: 0e+00
Dual.    obj: -5.6700000000e+03   nrm: 1e+01    Viol.  con: 0e+00    var: 0e+00

Basic solution summary
Problem status  : PRIMAL_AND_DUAL_FEASIBLE
Solution status : OPTIMAL
Primal.  obj: -5.6700000000e+03   nrm: 6e+02    Viol.  con: 0e+00    var: 0e+00
Dual.    obj: -5.6700000000e+03   nrm: 1e+01    Viol.  con: 0e+00    var: 0e+00
Optimizer summary
Optimizer                 -                        time: 0.00
Interior-point          - iterations : 0         time: 0.00
Basis identification  -                        time: 0.00
Primal              - iterations : 0         time: 0.00
Dual                - iterations : 0         time: 0.00
Clean primal        - iterations : 0         time: 0.00
Clean dual          - iterations : 0         time: 0.00
Simplex                 -                        time: 0.00
Primal simplex        - iterations : 0         time: 0.00
Dual simplex          - iterations : 0         time: 0.00
Mixed integer           - relaxations: 0         time: 0.00


reports the optimal repair. In this case it is to increase the upper bound on constraint c4 by 1.35e2 and decrease the lower bound on variable x2 by 20.