1. Chapter 8. General LP Problems
max 𝑗=1 𝑛 𝑐 𝑗 𝑥 𝑗 max 𝑐 ′ 𝑥
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 , 𝑖=1,…,𝑚 𝐴𝑥=𝑏 (8.3)
𝑙 𝑗 ≤ 𝑥 𝑗 ≤ 𝑢 𝑗 , 𝑗=1,…,𝑛 𝑙≤𝑥≤𝑢
( 𝑥 𝑗 ≤ 𝑢 𝑗 , 𝑥 𝑗 ≥ 𝑙 𝑗 )
−∞≤ 𝑙 𝑗 ≤ 𝑢 𝑗 ≤+∞
𝑥 𝑗 with 𝑙 𝑗 =−∞, 𝑢 𝑗 =+∞: free (unrestricted) variable
Converting other forms to general LP problem :
min 𝑐 ′ 𝑥  − max −𝑐 ′ 𝑥
≤  = by adding a nonnegative slack variable
≥  = by subtracting a nonnegative surplus variable
Note that we do not use the transformation
𝑥 𝑗 = 𝑥 𝑗 + − 𝑥 𝑗 − , 𝑥 𝑗 + , 𝑥 𝑗 − ≥0, for free 𝑥 𝑗
OR-1 Opt. 2018

2. 2) 𝑥 𝑗 ∗ = 𝑙 𝑗 or 𝑢 𝑗 for each bounded nonbasic variable.
Note that the standard problem, augmented with slack variables, is a special case of the general LP with lower bounds 0 and upper bounds +.
Upper bounds on variables need not be treated as explicit constraints, but they can be handled implicitly in the algorithm.
Def: 𝑥 ∗ is a basic solution of (8.3) if the 𝑛 components of 𝑥 ∗ can be partitioned into 𝑚 “basic” and 𝑛−𝑚 “nonbasic” variables in such a way that
1) 𝑚 columns of 𝐴 corresponding to the basic variables form a nonsingular matrix.
2) 𝑥 𝑗 ∗ = 𝑙 𝑗 or 𝑢 𝑗 for each bounded nonbasic variable.
(For free variable, no restriction. But usually 0 is used in the algorithm)
(Note that the choice of basic variables does not determine a basic solution as the standard LP case. Upper and lower bound for nonbasic bounded variables must be determined too.)
Def: A basic solution 𝑥 ∗ is called feasible if 𝑙≤ 𝑥 ∗ ≤𝑢.
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3. The entering nonbasic variable becomes basic.
Modifications of the simplex method to handle the general LP directly :
Given a b.f.s., choose an entering nonbasic variable. Then change (increase or decrease depending on the status of the entering nonbasic variable) the value of the entering nonbasic variable while other nonbasic variables are fixed at their current values. Then the value of the basic variables must be changed to satisfy the equations 𝐴𝑥=𝑏.
Change the value of the entering variable while values of the entering variable and the basic variables satisfy the bound constraints.
If one of the basic variables attains lower or upper bound value, let it become the nonbasic variable.
The entering nonbasic variable becomes basic.
Update the dictionary (tableau).
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4. Candidates for entering nonbasic variable : 2 cases are possible
Suppose we have a b.f.s. 𝑥 ∗ = 𝑥 𝐵 ∗ , 𝑥 𝑁 ∗ and corresponding dictionary (dictionary form not affected by the bound constraints)
𝑧= 𝑦 ′ 𝑏+ 𝑗∈𝑁 𝑐 𝑗 −𝑦′ 𝐴 𝑗 𝑥 𝑗
𝑥 𝐵 = 𝐵 −1 𝑏− 𝑗∈𝑁 𝐵 −1 𝐴 𝑗 𝑥 𝑗
( 𝑦 ′ = 𝑐 𝐵 ′ 𝐵 −1 )
Note that 𝑥 𝐵 ∗ may not be equal to 𝐵 −1 𝑏 and the objective value of the current solution may not be equal to 𝑦 ′ 𝑏.
Candidates for entering nonbasic variable : 2 cases are possible
(1) 𝑐 𝑗 −𝑦′ 𝐴 𝑗 >0 and 𝑥 𝑗 ∗ < 𝑢 𝑗 , 𝑗∈𝑁 (8.6)
(can increase 𝑥 𝑗 and objective value increases)
(2) 𝑐 𝑗 −𝑦′ 𝐴 𝑗 <0 and 𝑥 𝑗 ∗ > 𝑙 𝑗 , 𝑗∈𝑁 (8.7)
(can decrease 𝑥 𝑗 and objective value increases)
OR-1 Opt. 2018

5. Suppose 𝑥 𝑗 is selected as the entering nonbasic variable.
Let 𝑥 𝑗 𝑡 = 𝑥 𝑗 ∗ +𝑡 or 𝑥 𝑗 ∗ −𝑡, 𝑡≥0, and
𝑥 𝐵 𝑡 = 𝑥 𝐵 ∗ −𝑡𝑑 or 𝑥 𝐵 ∗ −(−𝑡)𝑑, 𝑡≥0 (𝑑= 𝐵 −1 𝐴 𝑗 )
while other nonbasic variables are fixed at the current values.
Then the values of the basic variables must be changed to
𝑥 𝐵 𝑡 = 𝑥 𝐵 ∗ − 𝐵 −1 𝐴 𝑗 +𝑡 or 𝑥 𝐵 ∗ − 𝐵 −1 𝐴 𝑗 −𝑡
and 𝑧← 𝑐 ′ 𝑥 ∗ + 𝑐 𝑗 −𝑦′ 𝐴 𝑗 ±𝑡 so that the new solution satisfies the equations.
Let 𝑡 ∗ be the largest 𝑡 such that
𝑙 𝑗 ≤ 𝑥 𝑗 (𝑡) ≤ 𝑢 𝑗 (8.8)
𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵
New solution obtained :
𝑥 𝑗 ← 𝑥 𝑗 ∗ ± 𝑡 ∗ , 𝑥 𝐵 ← 𝑥 𝐵 ∗ − 𝐵 −1 𝐴 𝑗 ± 𝑡 ∗
𝑧← 𝑐 ′ 𝑥 ∗ + 𝑐 𝑗 −𝑦′ 𝐴 𝑗 (±𝑡 ∗ )
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6. See BOX 8.1 for details of a simplex iteration.
( continued )
𝑙 𝑗 ≤ 𝑥 𝑗 (𝑡) ≤ 𝑢 𝑗
𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵
3 cases can happen:
The upper bound on 𝑡 imposed by 𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵 is stricter than the upper bound (if any) imposed by 𝑙 𝑗 ≤ 𝑥 𝑗 (𝑡) ≤ 𝑢 𝑗 .  Determine the leaving (basic) variable. This may be any basic variable 𝑥 𝑖 such that the upper bound imposed on 𝑡 by 𝑙 𝑖 ≤ 𝑥 𝑖 (𝑡) ≤ 𝑢 𝑖 alone is as strict as the upper bound imposed by all the constraints in 𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵 .
The upper bound on 𝑡 imposed by 𝑙 𝑗 ≤ 𝑥 𝑗 (𝑡) ≤ 𝑢 𝑗 is at least as strict as the upper bound (if any) imposed by 𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵 .  The entering (nonbasic) variable changes its status (lower bound  upper bound). But there is no basis change.
Neither 𝑙 𝑗 ≤ 𝑥 𝑗 (𝑡) ≤ 𝑢 𝑗 nor 𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡) ≤ 𝑢 𝐵 impose any upper bound on 𝑡.  The problem is unbounded.
See BOX 8.1 for details of a simplex iteration.
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7. Example maximize 𝑐 ′ 𝑥, subject to 𝐴𝑥=𝑏, 𝑙≤𝑥≤𝑢.
l = [ -5, -, -4, -2, 2, 0, 0, 3, -, -, -, - ]T
u = [ +, 3, -2, 3, 5, 1, +, +, 0, 5, +, + ]T
Let 𝑥 1 , 𝑥 2 be basic variables. Then the following 𝑥 ∗ is a basic feasible solution.
𝑥 ∗ = [ ], 𝑐′ 𝑥 ∗ =−10
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8. We first find 𝑦 vector from 𝑦′𝐵= 𝑐 𝐵 ′ 𝑦′ 3 1 1 0 = 2 1  𝑦′= 1 −1
𝑦′ =  𝑦′= 1 −1
Next, check whether any nonbasic variable satisfies the following conditions.
𝑐 𝑗 −𝑦 𝐴 𝑗 >0 and 𝑥 𝑗 ∗ < 𝑢 𝑗 , 𝑗∈𝑁 (8.6)
𝑐 𝑗 −𝑦 𝐴 𝑗 <0 and 𝑥 𝑗 ∗ > 𝑙 𝑗 , 𝑗∈𝑁 (8.7)
Satisfying (8.6)  𝑥 5 , 𝑥 7 , 𝑥 11
Satisfying (8.7)  𝑥 4 , 𝑥 10 , 𝑥 12
Suppose we choose 𝑥 5 (satisfying (8.6)) as the entering (nonbasic) variable.
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9. Find 𝑑 vector from 𝐵𝑑= 𝐴 𝑗 (𝑑= 𝐵 −1 𝐴 𝑗 ). 3 1 1 0 𝑑= 9 8  𝑑= 8 −15
𝑑=  𝑑= 8 −15
Find 𝑡 in 𝑥 𝑗 𝑡 = 𝑥 𝑗 ∗ +𝑡
For entering variable 𝑥 5 , 2≤2+𝑡≤5
For basic variable 𝑥 1 , −5≤1−8𝑡
For basic variable 𝑥 2 , 𝑡≤3
 𝑡≤0.2 and 𝑥 2 reaches its upper bound, hence leaves basis.
New basic solution is
𝑥 ∗ = [ ]
𝑥 1 and 𝑥 5 are basic variables.
( compare with the previous solution,
𝑥 ∗ = [ ] )
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10. Current dictionary when 𝑥 1 , 𝑥 2 basic:
𝑧= 𝑦 ′ 𝑏+ 𝑗∈𝑁 𝑐 𝑗 −𝑦′ 𝐴 𝑗 𝑥 𝑗
𝑥 𝐵 = 𝐵 −1 𝑏− 𝑗∈𝑁 𝐵 −1 𝐴 𝑗 𝑥 𝑗
Current dictionary when 𝑥 1 , 𝑥 2 basic:
z = x3 - 3x x5 - x6 + 2x x x x x x12
x1 = x3 - 5x x5 - x6 - 2x x x x x x12
x2 = x3 +9x4 + 15x5 - x6 + 3x7 + 17x8 + 17x9 + 15x x11 - 2x12
l = [ -5, -, -4, -2, 2, 0, 0, , , , -,  ]T
u = [ +, 3, -2, 3, 5, , +, +, , , +, + ]T
𝑥 1 ∗ =1, 𝑥 2 ∗ =0 ( 𝑥 5 entering nonbasic, increasing)
𝑥 1 ∗ ←1−8𝑡, 𝑥 2 ∗ ←0+15𝑡, 𝑥 5 ∗ ←2+𝑡,
−5≤1−8𝑡≤+∞, −∞≤0+15𝑡≤3, ≤2+𝑡≤5
 𝑡≤0.2
OR-1 Opt. 2018

11. When 𝑥 10 (satisfying (8.7)) is chosen as the entering variable,
𝑑=  𝑑= 7 −15 .
Now the constraints (8.8) assume the form
5−𝑡≤5 and −5≤1+7𝑡, −15𝑡≤3
imposing no upper bound on 𝑡. Hence the problem is unbounded.
In our case, every
𝑥 ∗ = 1+7𝑡, −15𝑡, −2, 3, 2, 0, 0, 3, 0, 5−𝑡, −1, 1 ′ with 𝑡≥0 constitutes a feasible solution with 𝑐 ′ 𝑥 ∗ =−10+𝑡.
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12. Degeneracy and Termination
Def: A basic feasible solution 𝑥 ∗ is called a degenerate solution if 𝑥 𝑖 ∗ = 𝑙 𝑖 or 𝑥 𝑖 ∗ = 𝑢 𝑖 for one or more basic variables 𝑥 𝑖 .
(When a simplex iteration begins with a degenerate b.f.s. 𝑥 ∗ , the constraints 𝑙 𝐵 ≤ 𝑥 𝐵 (𝑡)≤ 𝑢 𝐵 may force 𝑡 equal to zero. In that case, the entering variable enters the basis and the leaving variable leaves, but the solution 𝑥 ∗ remains unchanged. Such iterations are called degenerate.)
Cycling: appearance of the same basic feasible solution with the same set of basic variables in two different iterations.
Thm 8.1: If the simplex method applied to a problem
maximize 𝑐 ′ 𝑥 subject to 𝐴𝑥=𝑏, 𝑙≤𝑥≤𝑢
fails to terminate, then it must cycle.
Pf) see text.
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13. Thm 8.2: The simplex method applied to a problem
Cycling is very rare in practice, but degenerate solutions and degenerate iterations frequently show up in practice, especially in well structured problems. Need some special pivoting rules to guarantee finite convergence.
Thm 8.2: The simplex method applied to a problem
maximize 𝑐 ′ 𝑥 subject to 𝐴𝑥=𝑏, 𝑙≤𝑥≤𝑢
terminates as long as the entering and the leaving variables are selected by the smallest-subscript rule in each iteration.
Pf) not given here.
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14. Remarks
Note that a free nonbasic variable can always be a candidate for entering variable as long as 𝑐 𝑗 −𝑦′ 𝐴 𝑗 ≠0.
(Although 𝑐 𝑗 −𝑦′ 𝐴 𝑗 =0, we may let it enter the basis without changing the objective value)
Moreover, a free basic variable never become nonbasic in subsequent iterations. (There are no bounds on free variables, see (8.8) )
Hence we may let all free variables become basic in early iterations and let them remain basic in subsequent iterations.
In commercial softwares, simplex algorithm for general LP is used. (also called the bounded variable simplex method) (0≤ 𝑥 𝑗 ≤ 𝑢 𝑗 )
Convenient to use when we want to solve the problem again with a little bit of data change (e.g. fix the values of the variables, change the bounds, etc. Useful when we solve the integer program using branch-and-bound )
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15. Two-phase simplex method
Need initial b.f.s. to apply the simplex method for general LP
Use idea similar to the one we used earlier for standard LP
max 𝑗=1 𝑛 𝑐 𝑗 𝑥 𝑗
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 , 𝑖=1,…,𝑚 (8.9)
𝑙 𝑗 ≤ 𝑥 𝑗 ≤ 𝑢 𝑗 , 𝑗=1,…,𝑛
Introduce artificial variables 𝑥 𝑛+1 , 𝑥 𝑛+2 ,…, 𝑥 𝑛+𝑚 .
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 + 𝑥 𝑛+𝑖 = 𝑏 𝑖 , 𝑖=1,…,𝑚 (8.10)
𝑙 𝑗 ≤ 𝑥 𝑗 ≤ 𝑢 𝑗 , 𝑗=1,…,𝑛+𝑚
( 𝐴 𝐼 𝑥 𝑥 𝑎𝑟𝑡 =𝑏 )
𝑙 𝑗 and 𝑢 𝑗 for artificial variables need to be determined. However, for the above problem (8.10), we can find an initial b.f.s. easily and (8.9) has a feasible solution if and only if (8.10) has a solution with all artificial variables 0.
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16. Let 𝑥 𝑗 = 𝑙 𝑗 or 𝑥 𝑗 = 𝑢 𝑗 for every bounded variable 𝑥 𝑗 , 𝑗=1,…,𝑛
𝑥 𝑗 =0 for every free variable 𝑥 𝑗 , 𝑗=1,…,𝑛.
Set 𝑥 𝑛+𝑖 = 𝑏 𝑖 − 𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 for all 𝑖=1,2,…,𝑚
Now set the bounds for artificial variables
If 𝑥 𝑛+𝑖 ≥0, set 𝑙 𝑛+𝑖 =0, 𝑢 𝑛+𝑖 =+∞
If 𝑥 𝑛+𝑖 <0, set 𝑙 𝑛+𝑖 =−∞, 𝑢 𝑛+𝑖 =0
Above solution is a b.f.s. to (8.10).
Starting with this initial b.f.s. to (8.10), we want to find a solution to (8.10) with all artificial variables having 0 value.
Therefore, we let 𝑤 𝑛+𝑖 =1 if 𝑙 𝑛+𝑖 =0 and 𝑤 𝑛+𝑖 =−1 if 𝑢 𝑛+𝑖 =0
and consider minimize 𝑖=1 𝑚 𝑤 𝑛+𝑖 𝑥 𝑛+𝑖 , subject to (8.10). (Phase 1 problem)
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17. For all feasible solutions to (8.10), we have 𝑖=1 𝑚 𝑤 𝑛+𝑖 𝑥 𝑛+𝑖 ≥0.
If the optimal value of the auxiliary problem is >0, the problem (8.9) is infeasible. (contraposition of the statement (8.9) feasible  auxiliary problem has optimal value 0)
If the optimal value is equal to 0, all 𝑥 𝑛+𝑖 =0, 𝑖=1,…,𝑚, hence the solution 𝑥 𝑗 , 𝑗=1,…,𝑛 is feasible to (8.9).
We restore the original objective function and solve the following problem after resetting the bounds on the artificial variables to 0. (Phase 2 problem)
Note that the solution we have at the end of phase one is a b.f.s.
max 𝑗=1 𝑛 𝑐 𝑗 𝑥 𝑗
s.t. 𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 + 𝑥 𝑛+𝑖 = 𝑏 𝑖 (𝑖=1,2,…,𝑚)
𝑙 𝑗 ≤ 𝑥 𝑗 ≤ 𝑢 𝑗 (𝑗=1,2,…,𝑛)
0≤ 𝑥 𝑛+𝑖 ≤0 (𝑖=1,2,…,𝑚)
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18. We may omit some artificial variables in the Phase 1 problem
We may omit some artificial variables in the Phase 1 problem. Instead, we may use suitably chosen original variables as part of the initial basic variables.
Ex) max 3 𝑥 1 + 𝑥 𝑥 3 +2 𝑥 4
s.t. 3 𝑥 1 +2 𝑥 𝑥 3 +2 𝑥 =2
4 𝑥 𝑥 2 −3 𝑥 3 + 𝑥 4 + 𝑥 =−1
𝑥 1 −3 𝑥 𝑥 3 +3 𝑥 𝑥 =3
2 𝑥 𝑥 𝑥 3 +3 𝑥 𝑥 7 =5
𝑥 1 , 𝑥 2 ,…, 𝑥 7 ≥0
May use Phase 1 problem
max − 𝑥 8 + 𝑥 9
s.t. 3 𝑥 1 +2 𝑥 𝑥 3 +2 𝑥 𝑥 =2
4 𝑥 𝑥 2 −3 𝑥 3 + 𝑥 4 + 𝑥 𝑥 9 =−1
𝑥 1 −3 𝑥 𝑥 3 +3 𝑥 𝑥 =3
2 𝑥 𝑥 𝑥 3 +3 𝑥 𝑥 =5
𝑥 1 , 𝑥 2 ,…, 𝑥 7 ≥0, 𝑥 8 ≥0, 𝑥 9 ≤0
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19. Getting Rid of Artificial Basic Variables
At the end of phase one, the artificial variables which are nonbasic (hence have value 0) can be dropped before phase two is applied.
Artificial variables which are basic cannot be dropped since we need 𝑚 basic variables to denote a basic solution. However, the procedure given in BOX 8.2 replaces each artificial basic variable in the basis by an original variable which is nonbasic.
If the procedure fails to replace an artificial basic variable, it indicates that the constraint to which the artificial variable is attached is a redundant equation, hence the equation can be removed from the formulation and the feasible solution set is not changed. (Also the artificial variable can be dropped from the basis and eliminated from the formulation)
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20. BOX 8.2 Driving Artificial Variables Out of the Basis
Assume each artificial basic variable 𝑥 𝑛+𝑘 appears in the 𝑘−𝑡ℎ position of the basis heading. Note that each column of 𝐵 is either a column of 𝐴 or a column of 𝐼 (for artificial variables).
BOX 8.2 Driving Artificial Variables Out of the Basis
Step 0. Let 𝑆 be the set of all the subscripts 𝑖 such that 𝑥 𝑛+𝑖 is a basic variable.
Step 1. If 𝑆 is empty then stop. Otherwise delete one subscript 𝑘 from 𝑆.
Step 2. Solve the system 𝑟 ′ 𝐵=𝑒′ with 𝑒′ standing for the 𝑘−𝑡ℎ row of the identity matrix. (Hence 𝑟 ′ =𝑒′ 𝐵 −1 , 𝑟′ is 𝑘−𝑡ℎ row of 𝐵 −1 . Also 𝑟 ′ 𝐴 is 𝑘−𝑡ℎ row of updated coefficient matrix 𝐵 −1 𝐴 with current basis 𝐵.)
If 𝑟 ′ 𝐴 is the zero vector, then return to step 1. Otherwise, there is a nonbasic variable 𝑥 𝑗 such that 𝑟′ 𝐴 𝑗 ≠0 for the corresponding column of 𝐴 𝑗 of 𝐴. Replace the 𝑘−𝑡ℎ column of 𝐵 by 𝐴 𝑗 ; replace 𝑥 𝑛+𝑘 by 𝑥 𝑗 in the basis heading and return to step 1.
OR-1 Opt. 2018

21. Driving artificial variables out of the basis (in tableau form)
Note that the updated tableaus are same for the standard LP and general LP for the same basis matrix B.
( −𝑧+ 0 ′ 𝑥 𝐵 + 𝑐 𝑁 ′− 𝑐 𝐵 ′ 𝐵 −1 𝑁 𝑥 𝑁 =− 𝑐 𝐵 ′ 𝐵 −1 𝑏
𝐼 𝑥 𝐵 𝐵 −1 𝑁 𝑥 𝑁 = 𝐵 −1 𝑏 )
𝑥 1 ⋯ ⋯ ⋯ ⋯ 𝑥 𝑗 ⋯ ⋯ ⋯ ⋯ 𝑥 𝑛
𝑥 𝑛+1 ⋯ 𝑥 𝑛+𝑘 ⋯ 𝑥 𝑛+𝑚 −𝑧
𝑐 1 ⋯ ⋯ ⋯ ⋯ 𝑐 𝑗 ⋯ ⋯ ⋯ ⋯ 𝑐 𝑛
𝑐 𝑛+1 ⋯ 0⋯ 𝑐 𝑛+𝑚
− 𝑐 𝐵 ′ 𝐵 −1 𝑏
𝑢 1 ⋮ 𝑢 𝑘
⋮ 𝑢 𝑚
0 ⋮ 1
⋮ 0
𝐵 −1 𝑏
Pivot element ( 𝑢 𝑘 =𝑟′ 𝐴 𝑗 ≠0). Pivot with 𝑥 𝑛+𝑘 leaves basis, 𝑥 𝑗 enters basis.
Solution not changed, only basis change.
OR-1 Opt. 2018

22. (Recall ‘Simultaneous Linear Equations’ in ornote-02.)
Justification: The updated matrix 𝐵 remains nonsingular. Let 𝐸 be the identity matrix whose 𝑘−𝑡ℎ column is replaced by the vector 𝐵 −1 𝐴 𝑗 . Then, 𝐵 =𝐵𝐸. The matrix 𝐸 is nonsingular if and only if the 𝑘−𝑡ℎ entry of the vector 𝐵 −1 𝐴 𝑗 is nonzero. Now 𝑒 ′ 𝐵 −1 𝐴 𝑗 = 𝑒 ′ 𝐵 −1 𝐴 𝑗 =𝑟′ 𝐴 𝑗 ≠0. Hence 𝐵 is nonsingular. Also, the b.f.s. with basis 𝐵 is the same as the b.f.s. with basis 𝐵.
What if the procedure fails to eliminate artificial basic variable 𝑥 𝑛+𝑘 from the basis, i.e. 𝑟 ′ 𝐴=0′ (rows of 𝐴 are linearly dependent) in step 2?  It implies that the 𝑘−𝑡ℎ equation in (8.9) to which artificial variable 𝑥 𝑛+𝑘 is attached is unnecessary (or redundant) to describe the feasible solution set and may be deleted altogether.
(Recall ‘Simultaneous Linear Equations’ in ornote-02.)
OR-1 Opt. 2018

23. We claim that every solution of 𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 (𝑖∈𝐼) (8.12)
Let 𝐽 denote the set of subscripts 𝑘 such that 𝑥 𝑛+𝑘 persists in the basis, and let 𝐼 denote the subscripts 1, 2, … , 𝑚 that do not belong to 𝐽.
We claim that every solution of
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 (𝑖∈𝐼) (8.12)
satisfies all the equations
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 (𝑖=1,2,…,𝑚). (8.13)
Pf) Consider arbitrary 𝑘∈𝐽, along with the vector 𝑟 computed immediately after the deletion of 𝑘 from 𝑆. Since 𝑟′𝐴=0′, we have
𝑖=1 𝑚 𝑟 𝑖 𝑎 𝑖𝑗 =0 for all 𝑗=1,2,…,𝑛.
Since 𝑟 ′ 𝐵=𝑒′ (note that 𝑘th column of 𝐵 is 𝑒), we have 𝑟 𝑘 =1 and 𝑟 𝑗 =0 for all the remaining subscripts 𝑗∈𝐽. Hence (8.13) may be recorded as 𝑎 𝑘𝑗 =− 𝑖∈𝐼 𝑟 𝑖 𝑎 𝑖𝑗 and every solution of (8.12) must satisfy
𝑗=1 𝑛 𝑎 𝑘𝑗 𝑥 𝑗 = 𝑗=1 𝑛 − 𝑖∈𝐼 𝑟 𝑖 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑖∈𝐼 − 𝑟 𝑖 𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑖∈𝐼 − 𝑟 𝑖 𝑏 𝑖 (8.14)
OR-1 Opt. 2018

24. (continued)
In particular, (8.14) must be satisfied by the feasible solution 𝑥 1 ∗ , 𝑥 2 ∗ ,…, 𝑥 𝑛 ∗ of (8.9), and so 𝑖∈𝐼 − 𝑟 𝑖 𝑏 𝑖 = 𝑗=1 𝑛 𝑎 𝑘𝑗 𝑥 𝑗 ∗ = 𝑏 𝑘 .
We conclude that every solution of (8.12) must satisfy 𝑗=1 𝑛 𝑎 𝑘𝑗 𝑥 𝑗 = 𝑏 𝑘 , which is the desired result.
Thus (8.9) and the problem
max 𝑗=1 𝑛 𝑐 𝑗 𝑥 𝑗
𝑗=1 𝑛 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 , (𝑖∈𝐼) (8.15)
𝑙 𝑗 ≤ 𝑥 𝑗 ≤ 𝑢 𝑗 , (𝑗=1,2,…,𝑛)
are equivalent in the sense that they have precisely the same set of feasible solutions. (clearly, every solution of (8.9) also satisfies (8.15).)
OR-1 Opt. 2018

25. 𝑗∈𝐵 𝑎 𝑖𝑗 𝑥 𝑗 + 𝑥 𝑛+𝑖 = 𝑏 𝑖 (𝑖∈𝐽) has a unique solution.
The feasible solution of (8.9) delivered by the first phase of the two-phase simplex method is a feasible solution of (8.15). We claim that this solution is basic in (8.15), with the appropriate set of basic variables having been delivered by the procedure described earlier.
To justify this claim, denote by 𝐵 the subscripts 𝑗 with 1≤𝑗≤𝑛 for which 𝑥 𝑗 appears in the basis heading upon termination of the procedure. We have to verify only that the system
𝑗∈𝐵 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 (𝑖∈𝐼)
has a unique solution. But this claim follows at once from the fact that the system
𝑗∈𝐵 𝑎 𝑖𝑗 𝑥 𝑗 = 𝑏 𝑖 (𝑖∈𝐼)
𝑗∈𝐵 𝑎 𝑖𝑗 𝑥 𝑗 + 𝑥 𝑛+𝑖 = 𝑏 𝑖 (𝑖∈𝐽)
has a unique solution.
OR-1 Opt. 2018

26. Thm 8.3: If (8.9) has a feasible solution, then some set 𝐼 of subscripts 1,2,…,𝑚 has the following two properties: (i) problems (8.9) and (8.15) have precisely the same set of feasible solutions and (ii) problem (8.15) has a basic feasible solution.
OR-1 Opt. 2018

27. Fundamental Theorem of LP Reviewed
maximize 𝑐 ′ 𝑥 subject to 𝐴𝑥=𝑏, 𝑙≤𝑥≤𝑢 (8.16)
Def: a basic solution in which each nonbasic free variable is set at zero is called a normal basic solution. ((8.16) has only a finitely many normal basic solutions.)
Thm 8.4: If (8.16) has no optimal solution, then it is either infeasible or unbounded. Furthermore, if 𝐴 has full row rank, then (8.16) has the following two properties:
If it has a feasible solution, then it has a normal basic feasible solution.
If it has an optimal solution, then it has a normal basic optimal solution.
Pf) See text. Note that if 𝐴 does not have full row rank, there does not exist a basic solution for (8.16). However, the assumption is satisfied whenever (8.16) is obtained from standard form. The 𝑚 columns corresponding to the slack variables constitutes the identity matrix, which makes the 𝐴 matrix full row rank.
OR-1 Opt. 2018