A new non-monotonic infeasible simplex-type algorithm for Linear Programming

Step-by-step description of iEPSA and pseudocode

The algorithm consists of two phases. In the first phase, the algorithm generates a sequence of points $\left(x_{interior}^i,x_{exterior}^i\right)$, i = 0, 1, 2, ..., T, where $x_{interior}^i$ is a point in the relative interior of the feasible region for i = 0, ..., T, and $x_{exterior}^i$ is a basic solution to LP, that is infeasible to both the primal and the dual problem for i = 0, ..., T − 1. The first phase ends with a pair $\left(x_{interior}^T,x_{exterior}^T\right)$, where the exterior point is either feasible to the primal or the dual problem. If the first phase ands with a basic feasible solution to the primal, then the second phase runs an algorithm called Exterior Point Simplex Algorithm (EPSA) from previous literature (Paparrizos, 1993), to obtain the optimal basic feasible solution. If the first phase ends with a dual feasible solution, the second phase runs an algorithm called Primal-Dual Interior Point Simplex Algorithm (PDIPSA), also from previous literature (Samaras, 2001). We show that the first phase method always ends with a basic solution that is feasible to either the primal or the dual problem. Thus, using the prior results on EPSA and PDIPSA, the overall algorithm is shown to correctly solve LP.

The main idea behind the first phase is the following: the algorithm is initialized with any basic (infeasible) solution $x_{exterior}^0$ and an interior point $x_{interior}^0$, found by a standard Interior Point Solver (in this case MOSEK IPM). At every iteration, i = 0, 1, 2, ... one computes the intersection of the line passing

 
 
 
 
 
1  Data: Eq. (1), Infeasible Basic Partition [B N], Interior Point xinterior 
2  Result: Primal or Dual feasible basic partition [__B __N] 
   _______________________________________________________________________________________________________________________________________________________________ 
  3  (Initialization) Compute: 
 4 (AB)−1,xB,wT, (sN)T 
 5 xcurrent = [ xB  xN] 
 6 d = xinterior − xcurrent 
 7 P = {j ∈ N : sj < 0}, Q = {j ∈ N : sj ≥ 0} 
 8 (sP )T = (cP )T − wTAP , (sQ)T = (cQ)T − wTAQ 
 9 (General loop) 
10  while xB(⁄≥ 0) do 
11       αα = xcurrent + αd : α =  xB[ra] _−dB[ra]  = min{ xB[i] _−dB[i]  : dB[i] < 0} ,∀i = 1,...,m 
12       ββ = xcurrent + βd : β =  xB[rb] _−dB[rb]  = max{ xB[i] _−dB[i]  : xB[i] < 0} ,∀i = 1,...,m 
13       if α = +∞ then 
14                   STOP-Eq. 1 is unbounded. 
         else 
15                   Find xmiddle = αα+ββ    2 
16                   if cTxmiddle < cTxinterior then 
17                      __xinterior = xmiddle 
                     else 
18                       if cTxmiddle = cTxinterior then 
19                            α = min{(xinterior)[i] c[i]         : −c[i] < 0} 
20                            __xinterior = xinterior + α 2 (−cT ) 
                         else 
21                            d = xinterior − xmiddle 
22                            α = min{(xinterior)[i] −(d)[i]      : (d)[i] < 0} 
23                            __xinterior = xinterior + α 2 d 
                         end if 
                     end if 
24                   Compute: 
25                   xB[rb] = xk 
26                   HrP = ((AB)−1)rb.AP 
27                   HrQ = ((AB)−1)rb.AQ, 
28                   θ1 = −sp Hrp = min{−sj Hrj : Hrj < 0,j ∈ P} 
29                   θ2 = −sq Hrq = min{−sj Hrj : Hrj < 0,j ∈ Q} 
30                   (Pivot-Update) 
31                   Find t1,t2 : P(t1) = p , Q(t2) = q. 
32                   if θ1 ≤ θ2 then 
33                                l = p. 
                     else 
34                                l = q 
                     end if  
35                   Find t : N(t) = l. Set N(t) = k , B(rb) = l . Update: 
36                   (AB)−1, xB,wT, (sN)T, (sP )T, (sQ)T , P,Q, __xcurrent = [ xB  xN] , __d = __xinterior −_xcurrent 
         end if 
    end while 
                                                 Algorithm 1: iEPSA    

through $x_{exterior}^i$ and $x_{interior}^i$ with the feasible region. This gives a line segment l (assuming the problem is bounded) with midpoint $x_{middle}^i$. Otherwise, one takes a half-step from the current $x_{interior}^i$ in the direction of $x_{middle}^i$ and sets this as the new $x_{interior}^{i+1}$. One also computes the endpoint of the line segment l closest to $x_{interior}^i$. This endpoint lies on some facet of the feasible region. This facet dictates which nonbasic variable will enter the basis and an appropriate exiting variable is selected. This then gives the new basic solution $x_{exterior}^{i+1}$.

A flow diagram of iEPSA combined with EPSA and PDIPSA to provide an integrated solver for LP is shown in Fig. 1. A formal description of the iEPSA method is given in Algorithm 1.

An example

We will briefly demonstrate the iEPSA method in a simple example. Points {αα, ββ} represent the exiting and entering boundary points correspondingly for each feasible direction spanning the polyhedron. Assume we are given the following linear programming problem: (4)${\displaystyle \begin{array}{cccccc}\hfill max\hfill & \hfill Z=x_1& \hfill +\hfill & \hfill x_2& \hfill \hfill & \hfill \\ \hfill \mathrm{subject to}\hfill & \hfill x_1& \hfill -\hfill & \hfill x_2& \hfill \le \hfill & 2\hfill \\ \hfill \hfill & \hfill -x_1& \hfill +\hfill & \hfill x_2& \hfill \le \hfill & 4\hfill \\ \hfill \hfill & \hfill 3x_1& \hfill +\hfill & \hfill 5x_2& \hfill \le \hfill & 30\hfill \\ \hfill \hfill & \hfill -4x_1& \hfill -\hfill & \hfill 13x_2& \hfill \le \hfill & -23\hfill \\ \hfill \hfill & \hfill x_1& \hfill -\hfill & \hfill 8x_2& \hfill \le \hfill & -12\hfill \\ \hfill \hfill & \hfill 8x_1& \hfill -\hfill & \hfill 5x_2& \hfill \le \hfill & 3\hfill \\ \hfill \hfill & \hfill & \hfill \hfill & \hfill x_i& \hfill \ge \hfill & 0,\forall i\in \left\{1,2\right\}\hfill \end{array}}$

The corresponding feasible region is depicted in Fig. 2. There exist in total eight different variables after the addition of the slack ones. The axis system represents variables x₁ (y=0) and x₂ (x=0). The numbers with P in brackets on the right at each basic solution stand for the number of elements in vector P. The optimal point is [3,4.2] and the optimal objective value is Z = 7.2. After the addition of the slack variables we have: ${\displaystyle A=\left[\begin{array}{cccccccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -4\hfill & \hfill -13\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill -8\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 8\hfill & \hfill -5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right],c=\left[\begin{array}{c}\hfill -1,-1,0,0,0,0,0,0\hfill \\ \hfill \hfill \end{array}\right],b=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 4\hfill \\ \hfill 30\hfill \\ \hfill -23\hfill \\ \hfill -12\hfill \\ \hfill 3\hfill \\ \hfill \hfill \end{array}\right]}$

Initialization

Assume we start with the infeasible partition B = [1, 3, 4, 5, 7, 8], N = [2, 6] and the following interior point x_interior (calculated from MOSEK’s IPM). All the appropriate computations are: ${\displaystyle {\left(A_B\right)}^{-1}=\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.25\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -0.25\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0.75\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.25\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.25\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill \hfill \end{array}\right],{\left(s_N\right)}^T=\left[\begin{array}{c}\hfill 2.25,-0.25\hfill \end{array}\right],w^T=\left[\begin{array}{c}\hfill 0,0,0,0.25,0,0\hfill \end{array}\right]}$ ${\displaystyle x_B=\left[\begin{array}{c}\hfill -3.75\hfill \\ \hfill 9.75\hfill \\ \hfill 12.75\hfill \\ \hfill -17.75\hfill \\ \hfill 5.75\hfill \\ \hfill -43\hfill \end{array}\right],x_{current}=\left[\begin{array}{c}\hfill 5.75\hfill \\ \hfill 0\hfill \\ \hfill -3.75\hfill \\ \hfill 9.75\hfill \\ \hfill 12.75\hfill \\ \hfill 0\hfill \\ \hfill -17.75\hfill \\ \hfill -43\hfill \end{array}\right],x_{interior}=\left[\begin{array}{c}\hfill 0.3189\hfill \\ \hfill 3.0877\hfill \\ \hfill 4.7688\hfill \\ \hfill 1.2312\hfill \\ \hfill 13.6047\hfill \\ \hfill 18.4160\hfill \\ \hfill 12.3829\hfill \\ \hfill 15.8874\hfill \end{array}\right]}$

Here w are the dual variables. The N set of indexes actually represents the current non-basic solution. In our case [2,6] is the 2-D point [5.75,0]. A feasible direction d is then constructed by connecting the interior point with the infeasible basic solution: ${\displaystyle d=x_{interior}-x_{current}=\left[\begin{array}{c}\hfill 0.3189\hfill \\ \hfill 3.0877\hfill \\ \hfill 4.7688\hfill \\ \hfill 1.2312\hfill \\ \hfill 13.6047\hfill \\ \hfill 18.4160\hfill \\ \hfill 12.3829\hfill \\ \hfill 15.8874\hfill \end{array}\right]-\left[\begin{array}{c}\hfill 5.75\hfill \\ \hfill 0\hfill \\ \hfill -3.75\hfill \\ \hfill 9.75\hfill \\ \hfill 12.75\hfill \\ \hfill 0\hfill \\ \hfill -17.75\hfill \\ \hfill -43\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -5.4311\hfill \\ \hfill 3.0877\hfill \\ \hfill 8.5188\hfill \\ \hfill -8.5188\hfill \\ \hfill 0.8547\hfill \\ \hfill 18.4160\hfill \\ \hfill 30.1329\hfill \\ \hfill 58.8874\hfill \end{array}\right]}$

Mapping now the direction d on the basic variables we get d_B: ${\displaystyle d_B=\left[\begin{array}{c}\hfill 8.5188\hfill \\ \hfill -8.5188\hfill \\ \hfill 0.8547\hfill \\ \hfill 30.1329\hfill \\ \hfill -5.4311\hfill \\ \hfill 58.8874\hfill \end{array}\right]}$

Also we have P = [6] and Q = [2]. The direction from the initial infeasible basic solution (5.75,0) to the interior point is shown in Fig. 3. Since x_current is our initial infeasible basic solution and d is our current feasible direction, this direction intersects the feasible region at an exiting point AA1 (as shown in Fig. 3) which can be calculated using the relations below:

General loop - Iteration 1

${\displaystyle \alpha =\frac{x_{B\left[r\right]}}{-d_{B\left[r\right]}}=min\left\{\frac{x_{B\left[i\right]}}{-d_{B\left[i\right]}}:d_{B\left[i\right]}<0\right\}=min\left\{\frac{x_B\left[2,5\right]}{-d_B\left[2,5\right]}\right\}=min\left\{\frac{9.75}{8.5188},\frac{5.75}{5.4311}\right\}=min\left\{1.1445,1.0587\right\}=1.0587}$ ${\displaystyle \alpha \alpha =x_{current}+\alpha d=\left[\begin{array}{c}\hfill 5.75\hfill \\ \hfill 0\hfill \\ \hfill -3.75\hfill \\ \hfill 9.75\hfill \\ \hfill 12.75\hfill \\ \hfill 0\hfill \\ \hfill -17.75\hfill \\ \hfill -43\hfill \end{array}\right]+1.0587\left[\begin{array}{c}\hfill -5.4311\hfill \\ \hfill 3.0877\hfill \\ \hfill 8.5188\hfill \\ \hfill -8.5188\hfill \\ \hfill 0.8547\hfill \\ \hfill 18.4160\hfill \\ \hfill 30.1329\hfill \\ \hfill 58.8874\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 3.2690\hfill \\ \hfill 5.2690\hfill \\ \hfill 0.7310\hfill \\ \hfill 13.6549\hfill \\ \hfill 19.4973\hfill \\ \hfill 14.1522\hfill \\ \hfill 19.3451\hfill \end{array}\right]}$

In a similar manner, the entering point BB1 (as shown in Fig. 3) can be calculated using a maximum ratio test: ${\displaystyle \beta =\frac{-x_{B\left[r\right]}}{d_{B\left[r\right]}}=max\left\{\frac{-x_{B\left[i\right]}}{d_{B\left[i\right]}}:x_{B\left[i\right]}<0\right\}=max\left\{\frac{-x_B\left[1,4,6\right]}{d_B\left[1,4,6\right]}\right\}=max\left\{\frac{-3.75}{-8.5188},\frac{-17.75}{-30.1329},\frac{-43}{-58.8874}\right\}=max\left\{0.4402,0.5891,0.7302\right\}=0.7302}$ Hence, r_b = 3, then the 3^rd element of [1, 4, 6] is r = 6.

The entering point then is: ${\displaystyle \beta \beta =x_{current}+\beta d=\left[\begin{array}{c}\hfill 5.75\hfill \\ \hfill 0\hfill \\ \hfill -3.75\hfill \\ \hfill 9.75\hfill \\ \hfill 12.75\hfill \\ \hfill 0\hfill \\ \hfill -17.75\hfill \\ \hfill -43\hfill \end{array}\right]+0.7302\left[\begin{array}{c}\hfill -5.4311\hfill \\ \hfill 3.0877\hfill \\ \hfill 8.5188\hfill \\ \hfill -8.5188\hfill \\ \hfill 0.8547\hfill \\ \hfill 18.4160\hfill \\ \hfill 30.1329\hfill \\ \hfill 58.8874\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1.7842\hfill \\ \hfill 2.2547\hfill \\ \hfill 2.4705\hfill \\ \hfill 3.5295\hfill \\ \hfill 13.3741\hfill \\ \hfill 13.4475\hfill \\ \hfill 4.2533\hfill \\ \hfill 0\hfill \end{array}\right]}$

It is easy now to compute the middle point MIDDLE1 (as shown in Fig. 3) between αα − ββ: ${\displaystyle x_{middle}=\frac{\alpha \alpha +\beta \beta }{2}=\left[\begin{array}{c}\hfill 0.8921\hfill \\ \hfill 2.7618\hfill \\ \hfill 3.8697\hfill \\ \hfill 2.1303\hfill \\ \hfill 13.5145\hfill \\ \hfill 16.4724\hfill \\ \hfill 9.2027\hfill \\ \hfill 9.6726\hfill \\ \hfill \hfill \end{array}\right]}$

We observe the following:

• Both exiting and entering points have only one zero element which was expected since both points are boundary in a 2-D problem

• Maximum step size β is less than the minimum step α so as the entering point is closer to the infeasible basic solution and the exiting furthest as expected, since the direction will intersect the feasible region

• These two boundary points define a unique feasible ray segment from ββ to αα (ββ → αα)

• The objective function value at ββ is better than in αα (direction is non-improving) and the objective function value at the middle point is better than the initial interior point.

Since the middle point has better objective value than the initial interior point (it is: Z_middle = (c^Tmiddle) = [ − 0.8921,  − 2.7618] =  − 3.6539 and Z_xinterior = (c^Tx_interior) = [ − 0.3189,  − 3.0877] =  − 3.4066) we keep this middle point as an improved interior point for the next iteration. We now choose the leaving variable according to ββ boundary point: k = 8, r = 6. The variable x_B(r) = x_k = x₈ is leaving the basis. We can now move on, to select the entering variable x_l:

${\displaystyle H_{rP}=B^{-1}{A}_{.P}=\left[2\right]>0}$ ${\displaystyle H_{rQ}=B^{-1}{A}_{.Q}=\left[-31\right]<0}$ ${\displaystyle l=2,t2=1,P=\left[6\right],Q=\left[8\right]}$ ${\displaystyle {\theta }_1=\left[\right],{\theta }_2=\left[0.0726\right]}$

Since only θ₂ could be computed, the selection is done using the set Q. Variable x_l = x₂ is entering the basis. The pivot operation now updates the basic and non-basic index lists as shown below: ${\displaystyle \overline{B}=\left[\begin{array}{c}\hfill 3,4,5,7,1,2\hfill \end{array}\right],\overline{N}=\left[\begin{array}{c}\hfill 8,6\hfill \end{array}\right],{\left(A_B\right)}^{-1}=\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.0242\hfill & \hfill 0\hfill & \hfill -0.1371\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0.0242\hfill & \hfill 0\hfill & \hfill 0.1371\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0.4435\hfill & \hfill 0\hfill & \hfill -0.1532\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.4758\hfill & \hfill 1\hfill & \hfill -0.3629\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.0403\hfill & \hfill 0\hfill & \hfill 0.1048\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.0645\hfill & \hfill 0\hfill & \hfill -0.323\hfill \end{array}\right],{\left(s_N\right)}^T=\left[\begin{array}{c}\hfill 0.0726,-0.1048\hfill \end{array}\right],w^T=\left[\begin{array}{c}\hfill 0,0,0,0.1048,0,-0.0726\hfill \\ \hfill \hfill \end{array}\right],x_B=\left[\begin{array}{c}\hfill 2.1452\hfill \\ \hfill 3.8548\hfill \\ \hfill 19.3387\hfill \\ \hfill -2.1452\hfill \\ \hfill 1.2419\hfill \\ \hfill 1.3871\hfill \end{array}\right],{\overline{x}}_{current}=\left[\begin{array}{c}\hfill 1.2419\hfill \\ \hfill 1.3871\hfill \\ \hfill 2.1452\hfill \\ \hfill 3.8548\hfill \\ \hfill 19.3387\hfill \\ \hfill 0\hfill \\ \hfill -2.1452\hfill \\ \hfill 0\hfill \end{array}\right]}$

Note that x_B ≤ 0 in this pivot. The new direction $\overline{d}={\overline{x}}_{middle}-{\overline{x}}_{current}$ now is: ${\displaystyle \overline{d}=\left[\begin{array}{c}\hfill 0.8921\hfill \\ \hfill 2.7618\hfill \\ \hfill 3.8697\hfill \\ \hfill 2.1303\hfill \\ \hfill 13.5145\hfill \\ \hfill 16.4724\hfill \\ \hfill 9.2027\hfill \\ \hfill 9.6726\hfill \end{array}\right]-\left[\begin{array}{c}\hfill 1.2419\hfill \\ \hfill 1.3871\hfill \\ \hfill 2.1452\hfill \\ \hfill 3.8548\hfill \\ \hfill 19.3387\hfill \\ \hfill 0\hfill \\ \hfill -2.1452\hfill \\ \hfill 0\hfill \end{array}\right]=\left[\begin{array}{c}\hfill -0.3498\hfill \\ \hfill 1.3747\hfill \\ \hfill 1.7246\hfill \\ \hfill -1.7246\hfill \\ \hfill -5.8242\hfill \\ \hfill 16.4724\hfill \\ \hfill 11.3479\hfill \\ \hfill 9.6726\hfill \end{array}\right]⟹\overline{d_B}=\left[\begin{array}{c}\hfill 1.7246\hfill \\ \hfill -1.7246\hfill \\ \hfill -5.8242\hfill \\ \hfill 11.3479\hfill \\ \hfill -0.3498\hfill \\ \hfill 1.3747\hfill \\ \hfill \hfill \end{array}\right]}$

General loop: Iteration 2

The new exiting boundary point AA2 (as shown in Fig. 4) is:

${\displaystyle \alpha =\frac{x_{B\left[r\right]}}{-d_{B\left[r\right]}}=min\left\{\frac{x_{B\left[i\right]}}{-d_{B\left[i\right]}}:d_{B\left[i\right]}<0\right\}=min\left\{\frac{x_B\left[2,3,5\right]}{-d_B\left[2,3,5\right]}\right\}=min\left\{\frac{3.8548}{1.7246},\frac{19.3387}{5.8242},\frac{1.2419}{0.3498}\right\}=min\left\{2.2352,3.3204,3.5499\right\}=2.2352}$ ${\displaystyle \alpha \alpha =x_{current}+\alpha d=\left[\begin{array}{c}\hfill 1.2419\hfill \\ \hfill 1.3871\hfill \\ \hfill 2.1452\hfill \\ \hfill 3.8548\hfill \\ \hfill 19.3387\hfill \\ \hfill 0\hfill \\ \hfill -2.1452\hfill \\ \hfill 0\hfill \end{array}\right]+2.2352\left[\begin{array}{c}\hfill 0.3498\hfill \\ \hfill 1.3747\hfill \\ \hfill 1.7246\hfill \\ \hfill -1.7246\hfill \\ \hfill -5.8242\hfill \\ \hfill 16.4724\hfill \\ \hfill 11.3479\hfill \\ \hfill 9.6726\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 0.4599\hfill \\ \hfill 4.4599\hfill \\ \hfill 6\hfill \\ \hfill 0\hfill \\ \hfill 6.3203\hfill \\ \hfill 36.8196\hfill \\ \hfill 23.2200\hfill \\ \hfill 21.6204\hfill \end{array}\right]}$

Correspondingly, the entering point BB2 (as shown in Fig. 4) can be calculated using the maximum ratio test: ${\displaystyle \beta =\frac{-x_{B\left[r\right]}}{d_{B\left[r\right]}}=max\left\{\frac{-x_{B\left[i\right]}}{d_{B\left[i\right]}}:x_{B\left[i\right]}<0\right\}=max\left\{\frac{-x_B\left[4\right]}{d_B\left[4\right]}\right\}=max\left\{\frac{-2.1452}{-11.3479}\right\}=0.1890}$ Hence, r_b = 4  so  r = 4. The entering point then is:

${\displaystyle \beta \beta =x_{current}+\beta d=\left[\begin{array}{c}\hfill 1.2419\hfill \\ \hfill 1.3871\hfill \\ \hfill 2.1452\hfill \\ \hfill 3.8548\hfill \\ \hfill 19.3387\hfill \\ \hfill 0\hfill \\ \hfill -2.1452\hfill \\ \hfill 0\hfill \end{array}\right]+0.1890\left[\begin{array}{c}\hfill -0.3498\hfill \\ \hfill 1.3747\hfill \\ \hfill 1.7246\hfill \\ \hfill -1.7246\hfill \\ \hfill -5.8242\hfill \\ \hfill 16.4724\hfill \\ \hfill 11.3479\hfill \\ \hfill 9.6726\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1.1758\hfill \\ \hfill 1.6470\hfill \\ \hfill 2.4712\hfill \\ \hfill 3.5288\hfill \\ \hfill 18.2377\hfill \\ \hfill 3.1139\hfill \\ \hfill 0\hfill \\ \hfill 1.8285\hfill \end{array}\right]}$

The new middle point MIDDLE2 (as shown in Fig. 4) between αα − ββ is now: ${\displaystyle {\overline{x}}_{middle}=\frac{\alpha \alpha +\beta \beta }{2}=\left[\begin{array}{c}\hfill 0.8179\hfill \\ \hfill 3.0535\hfill \\ \hfill 4.2356\hfill \\ \hfill 1.7644\hfill \\ \hfill 12.2790\hfill \\ \hfill 19.9667\hfill \\ \hfill 11.6100\hfill \\ \hfill 11.7244\hfill \end{array}\right]}$

The variable x_B(r) = x_k = x₇, r = 4 is leaving the basis. We can now move on to select the entering variable x_l:

${\displaystyle P=\left[6\right],Q=\left[8\right]}$ ${\displaystyle H_{rP}={\left(A_B\right)}^{-1}{A}_{.P}=\left[-0.4758\right]<0}$ ${\displaystyle H_{rQ}={\left(A_B\right)}^{-1}{A}_{.Q}=\left[-0.3629\right]<0}$ ${\displaystyle l=2,t2=1,P=\left[7\right],Q=\left[8\right]}$ ${\displaystyle {\theta }_1=\left[-0.2203\right],{\theta }_2=\left[0.2\right]}$

Since θ₁ ≤ θ₂ the selection is done using the set P. Variable x_l = x₆ is entering the basis. The pivot operation now updates the basic and non-basic index lists as shown below:

${\displaystyle \overline{B}=\left[\begin{array}{c}\hfill 3,4,5,6,1,2\hfill \end{array}\right],\overline{N}=\left[\begin{array}{c}\hfill 8,7\hfill \end{array}\right]}$ ${\displaystyle {\left(A_B\right)}^{-1}=\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.0508\hfill & \hfill -0.1186\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0.0508\hfill & \hfill 0.1186\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0.9322\hfill & \hfill -0.4915\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -2.1017\hfill & \hfill 0.7627\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.0847\hfill & \hfill 0.1356\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -0.1356\hfill & \hfill 0.0169\hfill \end{array}\right],{\left(s_N\right)}^T=\left[\begin{array}{c}\hfill 0.1525,-0.2203\hfill \end{array}\right],}$ ${\displaystyle w^T=\left[\begin{array}{c}\hfill 0,0,0,0,0.2203,-0.1525\hfill \end{array}\right],x_B=\left[\begin{array}{c}\hfill 2.2542\hfill \\ \hfill 3.7458\hfill \\ \hfill 17.3390\hfill \\ \hfill 4.5085\hfill \\ \hfill 1.4237\hfill \\ \hfill 1.6780\hfill \end{array}\right]}$

Note that x_B ≥ 0 in this pivot. Method iEPSA stops here. The basic solutions constructed as shown in Fig. 4 are C₁ → G → A₁. In practice, EPSA would take over and finish the optimization moving with one extra iteration to the optimal vertex from the feasible vertex A₁. Note that in the second pivot, the middle point (MIDDLE2) constructed could have worse objective function than MIDDLE1 (although it seems better in this case). We did not calculate it on purpose here, since the second basic solution constructed is feasible. The sequence of the objective function value at each pair of basic solutions with the corresponding interior points is shown below:

${\displaystyle \text{basic solutions}:\left[\begin{array}{ccc}\hfill 5.75,\hfill & \hfill 2.629,\hfill & \hfill 3.1017\hfill \end{array}\right]}$ ${\displaystyle \text{interior points}:\left[\begin{array}{c}\hfill 3.4066\to improved\to 3.6539\hfill \\ \hfill 3.6539\to improved\to 3.8714\hfill \end{array}\right].}$