Sensitivity analysis LI Xiao-lei

A graphical introduction to sensitivity analysis Sensitivity analysis is concerned with how changes in an LP’s parameters affect the LP’s optimal solution.

A graphical introduction to sensitivity analysis The Giapetto problem max z=3x 1 +2x 2 s.t. 2x 1 + x 2 ≤100 (finishing constraint) x 1 + x 2 ≤80 (carpentry constraint) x 1 ≤40 (demand constraint) x 1, x 2 ≥0 Where x 1 =number of soldiers produced per week x 2 =number of trains produced per week

A graphical introduction to sensitivity analysis The optimal solution is z=180, x 1 =20, x 2 =60

Graphical analysis of the effect of a change in an objective function coefficient Let c 1 be the contribution to profit by each soldier. For what values of c 1 does the current basis remain optimal? At present, c 1 =3 and the profit line has the form 3x 1 +2x 2 =constant, or x 2 =-3x 1 /2 + constant/2.

Figure 1 B C D A E

Graphical analysis of the effect of a change in an objective function coefficient If a change in c 1 cause the profit lines to be flatter than the carpentry constraint, the optimal solution will change from point B to a new optimal solution (point A). The slope of each profit line is –c 1 /2 The profit line will be flatter than the carpentry constraint if –c 1 /2>-1, or c 1 <2, and the new optimal solution will be (0,80), point A.

Graphical analysis of the effect of a change in an objective function coefficient If the profit line s are steeper than the finishing constraint, the optimal solution will change from point B to point C. The slope of the finishing constraint is - 2. If –c 1 /2 4, and the new optimal solution will be (40,20).

Graphical analysis of the effect of a change in an objective function coefficient In summary, if all other parameters remain unchanged, the current basis remains optimal for 2≤c 1 ≤4. and Giapetto should still manufacture 20 soldiers and 60 trains. But Giapetto’s profit will change. For instance, if c1=4, Giapetto’s profit will now be 4(20)+2(60)=$200 instead of $180.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Whether a change in the right-hand side of a constraint will make the current basis no longer optimal? Let b1 the number of available finishing hours. Currently, b1=100.

Figure 2 A B C D

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution A change in b1 shifts the finishing constraint parallel to its current position. The current optimal solution is where the carpentry and finishing constraints are binding. Then as long as the point where the finishing and carpentry constraints are binding remains feasible, the optimal solution will still occur where these constraints intersect.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution From figure 2, for 80≤b 1 ≤120, the current basis remains optimal. With the changing of b 1, the values of the decision variables and the objective function value change. For example, if 80≤b 1 ≤100, the optimal solution will change from point B to some point on the line segment AB. Similarly, if 100≤b 1 ≤120, the optimal solution will change from point B to some point on the line segment BD.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution To determine how a change in the right- hand side of a constraint changes the values of the decision variables. Let b 1 =number of available finishing hours. If we change b 1 to 100+Δ, the current basis remains optimal for -20≤Δ≤20. Note: as b 1 changes, the optimal solution to the LP is still the point where the finishing- hour and carpentry-hour constraints are binding.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Thus, we can find the new values of the decision variables by solving 2x 1 +x 2 =100+Δ and x 1 +x 2 =80 This yields x 1 =20+Δ and x 2 =60-Δ Thus, a increase in the number of available finishing hours results in an increase in the number of soldiers produced and a decrease in the number of trains produced.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Let b 2 =the number of available carpentry hours If we change b 2 to 80+Δ, the current basis remains optimal for -20≤Δ≤20. Note: as b 2 changes, the optimal solution to the LP is still the point where the finishing-hour and carpentry-hour constraints are binding.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Thus, if b 2 =80+ Δ, the optimal solution to the LP is the solution to 2x 1 +x 2 =100 and x 1 +x 2 =80+ Δ This yields x 1 =20-Δ and x 2 =60+2Δ. It shows that an increase in the amount of available carpentry hours decrease the number of soldiers produced and increases the number of trains produced.

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Let b 3 =the demand for soldiers. If b 3 is changed to 40+Δ, the current basis remains optimal for Δ≥-20. for Δ in this range, the optimal solution to the LP will still occur where the finishing and carpentry constrains are binding. Thus, the optimal solution to the LP is the solution to 2x 1 +x 2 =100 and x 1 +x 2 =80 This yields x 1 =20 and x 2 =60

Graphical analysis of the effect of a change in a right-hand side on the LP’s optimal solution Note: In a constraint with positive slack( or positive excess) in an LP’s optimal solution, if we change the right- hand side of the constraint to a value in the range where the current basis remains optimal, the optimal solution to the LP is unchanged.

Shadow prices We define the shadow price for the i th constraint of an LP to be the amount by which the optimal z-value is improved if the right-hand side of the i th constraint is increased by 1. This definition applies only if the change in the right-hand side of constraint i leaves the current basis optimal.

Shadow prices For example, for the Δchanging in the finishing hours, the optimal solution is x 1 =20+Δ and x 2 =60-Δ. Then the optimal z- value will equal 3x 1 +2x 2 =3(20 +Δ) +2(60-Δ) =180+ Δ. Thus, as long as the current basis remains optimal, a unit increase in the number of available finishing hours will increase the optimal z-value by $1.so the shadow price of the first constraint is $1.

Shadow prices For the second constraint, if Δchanging in the finishing hours, the optimal solution is x 1 =20-Δ and x 2 =60+2Δ. Then the optimal z- value will equal 3x 1 +2x 2 =3(20 -Δ) +2(60+2Δ) =180+ Δ. Thus, as long as the current basis remains optimal, a unit increase in the number of available finishing hours will increase the optimal z-value by $1.so the shadow price of the second constraint is $1.

Shadow prices For the third constraint, the optimal values of the decision variables remain unchanged, as long as the current basis remains optimal. Then the optimal z-value will also remain unchanged, which shows that the shadow price of the third constraint is $0. It turns out that whenever the slack variable or excess variable for a constraint is positive in an LP’s optimal solution, the constraint will have a zero shadow price.

Shadow prices Each unit by which constraint i ’s right-hand side is increased will increase the optimal z- value( for a max problem) by the shadow price. Suppose we increase the rhs of the i th constraint of an LP by Δb i, thus, the new optimal z-value is given by (new optimal z-value)=(old optimal z-value) +(constraint i ’s shadow price)Δb i For a minimization problem, (new optimal z-value)=(old optimal z-value) -(constraint i ’s shadow price)Δb i

Importance of sensitivity analysis If a parameter changes, sensitivity analysis often makes it unnecessary to solve the problem again. A knowledge of sensitivity analysis often enables the analyst to determine from the original solution how changes in an LP’s parameters change the optimal solution.

Some important formulas Assume a max LP problem with m constraints and n variables. Although some of these variables may be slack, excess, or artificial ones, we choose to label them x 1,x 2,…,x n. max z=c 1 x 1 +c 2 x 2 +…+c n x n s.t. a 11 x 1 +a 12 x 2 +…+a 1n x n =b 1 a 21 x 1 +a 22 x 2 +…+a 2n x n =b 2 (1) : : : : a m1 x 1 +a m2 x 2 +…+a mn x n =b m x i ≥0 (i=1,2,…,n)

Some important formulas For the Dakota problem, max z=60x 1 +30x 2 +20x 3 +0s 1 +0s 2 +0s 3 s.t. 8x 1 + 6x 2 + x 3 + s 1 =48(lumber) 4x 1 + 2x x 3 + s 2 =20 (finishing) (1’) 2x x x 3 + s 3 =8 (carpentry) x 1,x 2,x 3,s 1,s 2,s 3 ≥0

Some important formulas Let BV i be the basic variable for row i of the optimal tableau. Also define BV={BV 1, BV 2, …, BV m } to be the set of basic variables in the optimal tableau, and define the m × 1 vector x Bv =[x Bv1, x Bv2, …, x Bvm ] T We also define NBV=the set of nonbasic variables in the optimal tableau x NBv =(n-m) ×1 vector listing the nonbasic variables.

Some important formulas The optimal tableau for the LP 1’ z + 5x 2 +10s 2 +10s 3 = x 2 + s 1 +2 s 2 - 8s 3 =24 - 2x 2 + x 3 + 2s 2 - 4s 3 =8 x x s s 3 =2 For this optimal tableau, BV 1 =s 1, BV 2 =x 3, and BV 3 =x 1, then x BV =[s 1,x 3,x 1 ]’ We may choose NBV={x 2,s 2,s 3 }, then x NBV =[x 2,s 2,s 3 ]’

Some important formulas DEFINATION c BV is the 1×m row vector [c BV1 c BV2 … c BVm ]. Thus, the elements of c BV are the objective function coefficients for the optimal tableau’s basic variables. For the Dakota problem, BV ={s 1,x 3,x 1 }, then form (1’) we find that c BV =[ ].

Some important formulas DEFINATION c NBV is the 1×(n-m) row vector whose elements are the coefficients of the nonbasic variables( in the order of NBV). If we choose to list the nonbasic variables for the Dakota problem in the order NBV ={x 2,s 2,s 3 }, then form (1’) we find that c NBV =[30 0 0].

Some important formulas DEFINATION The m×m matrix B is the matrix whose j th column is the column for BV j in (1). For the Dakota problem, the first column of B is the s 1 column in (1’); the second, the x 3 column; and the third, the x 1 column. Thus,Dakota problem

Some important formulas DEFINATION a j is the column (in the constraints) for the variable x j in (1). For example, in the Dakota problem,Dakota problem

Some important formulas DEFINATION N is the m×(n-m) matrix whose columns are the columns for the nonbasic variables (in the NBV order) in (1). For the Dakota problem, we write NBV={x 2,s 2,s 3 }, thenDakota problem

Some important formulas DEFINATION The m×1 column vector b is the right-hand side of the constraints in (1). For the Dakota problem,Dakota problem

Matrix notation (1) may be written as (3)

Matrix notation The Dakota problem can be written as

Matrix notation Multiplying the constraints in (3) through by B -1, we obtain or(4) In (4),BV i occurs with a coefficient of 1 in the i th constraint and a zero coefficient in each other constraint. Thus, BV is the set of the basic variables for (4), and (4) yields the constraints for the optimal tableau.

Matrix notation For the Dakota problem, Then, Of course, these are the constraints for the optimal tableau. optimal tableau.

Matrix notation From (4), we see that the column of a nonbasic variable x j in the constraints of the optimal tableau is given by B -1 (column for x j in (1) )=B -1 a j The following two equations summarize the preceding discussion: Column for x j in optimal tableau’s constraints=B -1 a j (5) Right-hand side of optimal tableau’s constraints=B -1 b (6)

Matrix notation How to express row 0 of the optimal tableau in terms of BV? We multiply the constraints through by the vector c BV B -1 : (7) And rewrite the original objective function as, (8) Adding (7) to (8), we can eliminate the optimal tableau’s basic variables, (9)

Matrix notation From (9), the coefficient of x j in row 0 is c BV B -1 (column of N for x j )- (coefficient for x j in c NBV )=c BV B -1 a j -c j the right hand side of row 0 is c BV B -1 b To summarize, we let be the coefficient of x j in the optimal tableau’s row 0. then we have shown that (10) and right-hand side of optimal tableau’s row 0=c BV B -1 b (11)

Matrix notation To illustrate, we determine row 0 of the Dakota problem’s optimal tableau.Dakota problem’s optimal tableau Then c BV B -1 =[ ], from (10) we find that the coefficients of the nonbasic variables in row 0 are

Matrix notation and Of course, the optimal tableau’s basic variables ( x 1,x 3,and s 1 ) will have zero coefficients in row 0.

Matrix notation From (11), the right-hand side of row 0 is Putting it all together, we see that row 0 is z+5x 2 +10s 2 +10s 3 =280

Matrix notation Simplifying formula (10) for slack, excess, and artificial variables If x j is the slack variable s i, the coefficient of s i in the objective function is 0, and the column for s i in the original tableau has 1 in row I and 0 in all other rows. Then (10) yields Coefficient of s i in optimal row 0 = i th element of c BV B = i th element of c BV B -1 (10’)

Matrix notation Similarly, if x j is the excess variable e i, the coefficient of e i in the objective function is 0 and the column for e i in the original tableau has -1 in row i and 0 in all other rows. Then (10) reduces to Coefficient of e i in optimal row 0 =-( i th element of c BV B -1 )-0 =-( i th element of c BV B -1 ) (10’’)

Matrix notation Finally, if x j is an artificial variable a i, the objective function coefficient of a i ( for a max problem) is –M and the original column for a i has 1 in row i and 0 in all other rows. Then (10) reduces to Coefficient of a i in optimal row 0 =( i th element of c BV B -1 )-(-M) =( i th element of c BV B -1 )+(M) (10’’’)

Summary of formulas for computing the optimal tableau from the initial LP Column for x j in optimal tableau’s constraints=B -1 a j (5) Right-hand side of optimal tableau’s constraints=B -1 b (6) Coefficient of s i in optimal row 0=ith element of c BV B -1 (10’) Coefficient of e i in optimal row 0=-(ith element of c BV B -1 ) (10’’) Coefficient of a i in optimal row 0=-(ith element of c BV B -1 )+(M) (10’’’) right-hand side of optimal tableau’s row 0=c BV B -1 b (11) We must first find B -1 because it is necessary in order to compute all parts of the optimal tableau.

Example 1 Solution After adding slack variables s 1 and s 2, we obtain: max z=x 1 +4x 2 s.t. x 1 +2x 2 +s 1 =6 2x 1 + x 2 +s 2 =8

Example 1 For the following LP, the optimal basis is BV={x 2,s 2 }, computer the optimal tableau. max z=x 1 +4x 2 s.t. x 1 +2x 2 ≤6 2x 1 + x 2 ≤8 x 1,x 2 ≥0

Example 1 First we compute B -1,

Example 1 Use (5) and (6) to determine the optimal tableau’s constraints. The column for x 1 in the optimal tableau is The other nonbasic variable is s 1,

Example 1 From (6) Since BV is listed as { x 2,s 2 },

Example 1 Then (10) yields From (10’)

Example 1 (11) yields Of course, the basic variables x 2 and s 2 will have zero coefficients in row 0. Thus, the optimal tableau’s row 0 is z+x 1 +2s 1 =12

Example 1 The complete optimal tableau is z +x 1 +2s 1 =12 ½x 1 +x 2 + ½ s 1 =3 3/2x 1 - ½ s 1 +s 2 =5

Sensitivity analysis The study of how an LP’s optimal solution depends on its parameters is called sensitivity analysis. We let BV be the set of basic variables in the optimal tableau. Given a change in an LP, we want to determine whether BV remains optimal. Important observation: A simplex tableau( for a max problem) for a set of BV is optimal if and only if each constraint has a nonnegative right-hand side and each variable has a nonnegative coefficient in row 0.

Sensitivity analysis Suppose we have solved an LP and have found that BV is an optimal basis. We can use the following procedure to determine if any change in the LP will cause BV to be no longer optimal. Step 1 Using the preceding formulas, determine how changes in the LP’s parameters change the right-hand side and row 0 of the optimal tableau. Step 2 If each variable in row 0 has a nonnegative coefficient and each constraint has a nonnegative right-hand side, BV is still optimal. Otherwise, BV is no longer optimal.

Sensitivity analysis Two reasons why a change in an LP’s parameters causes BV to be no longer optimal. A variable in row 0 may have a negative coefficient. In this case, a better bfs can be obtained by pivoting in a nonbasic variable with a negative coefficient in row 0. we say that the BV is now a suboptimal basis. A constraint may now have a negative right-hand side. In this case, at least one member of BV will now be negative and BV will no longer yield a bfs. We say that BV is now an infeasible basis.

Illustrate in example The Dakota furniture example x 1 =number of desks manufactured x 2 =number of tables manufactured x 3 =number of chairs manufactured The objective function was max z=60x 1 +30x 2 +20x 3

Illustrate in example The initial tableau was z -60x 1 -30x x 3 =0 8x 1 + 6x 2 + x 3 +s 1 =48 (lumber constraint) 4x 1 + 2x x 3 +s 2 =20 (finishing constraint) 2x x x 3 +s 3 =8 (carpentry constraint) The optimal tableau was z +5x 2 +10s 2 +10s 3 =280 -2x 2 + s 1 + 2s 2 -8s 3 =24 -2x 2 +x 3 +2s 2 -4s 3 =8 x x s s 3 =2 BV={s 1,x 3,x 1 } and NBV={x 2,s 2,s 3 }. The optimal bfs is z=280,s 1 =24,x 3 =8,x 1 =2,x 2 =0,s 2 =0,s 3 =0

Illustrate in example Six types of changes in an LP’s parameters: Change 1 changing the objective function coefficient of a nonbasic variable Change 2 changing the objective function coefficient of a basic variable Change 3 changing the right-hand side of a constraint Change 4 changing the column of a nonbasic variable Change 5 adding a new variable or activity Change 6 adding a new constraint

Illustrate in example Changing the objective function coefficient of a nonbasic variable The only nonbasic decision variable is x 2. the objective function coefficient of x 2 is c 2 =30. How would a change in c 2 affect the optimal solution?

Illustrate in example Suppose we change c 2 from 30 to 30+Δ. From (6), since the B -1 and b are unchanged, the right-hand side of BV’s tableau (B -1 b) has not changed, so BV is still feasible.(6), From (10), BV will remain optimal if c 2 ≥0, and BV will be suboptimal if c 2 <0.(10), From preceding, we know that c BV B -1 =[ ].

Illustrate in example Thus, c2≥0 holds if 5-Δ≥0,or Δ≤5. This means that if the price of tables is decreased or increased by $5 or less, BV remains optimal. If BV remains optimal after a change in a nonbasic variable’s objective function coefficient, the values of the decision variables and the optimal z-value remain unchanged.

Illustrate in example For example, if the price of tables increase to $33(c 2 =33), the optimal solution to the Dakota problem remains unchanged. On the other hand, if c 2 >35, BV will no longer be optimal, because c 2 <0. For example, if c 2 =40,

Illustrate in example Now the BV “final” tableau is, Basic variables ratio z -5x 2 +10s 2 +10s 3 =280 z=280 -2x 2 +s 1 + 2s 2 -8s 3 =24 s 1 =24 none -2x 2 +x 3 +2s 2 -4s 3 =8 x 3 =8 none x x s s 3 =2 x 1 =2 1.6* This is not an optimal tableau, and we can increase z by making x 2 a basic variable in row 3.

Illustrate in example Changing the objective function coefficient of a basic variable In the Dakota problem, x 1 and x 3 are basic variables. Since we are not changing B or b, (6) shows that the right-hand side of each constraint will remain unchanged, and BV will remain feasible.(6) Since we are changing c BV. From (10), this may change more than one coefficient in row 0.(10),

Illustrate in example The Dakota problem,Dakota problem Suppose that c 1 is changed from 60 to 60+Δ.

Illustrate in example Noting that, The coefficient of each nonbasic variable in the new row 0 is,

Illustrate in example Thus, row 0 of the optimal tableau is now z+(5+1.25Δ)x 2 +(10-0.5Δ)s 2 +(10+1.5Δ)s 3 =? BV will remain optimal if anf only if the following hold: Δ≥0 Δ≥ Δ≥0 Δ≤ Δ≥0 Δ≥-20/3 the current basis will remain optimal if and only if -4≤Δ≤20.

Illustrate in example Thus, as long as 56=60- 4≤c 1 ≤60+20=80, the current basis remains optimal. Suppose c 1 =70, z=70x 1 +30x 2 +20x 3. revenue should increase by 2(10)=$20, and new revenue=280+20=$300.

Illustrate in example If c1 80, the current basis is no longer optimal. Suppose c1=100(Δ=40), how can we determine the new optimal solution?

Illustrate in example The BV tableau is, Basic variables ratio z +55x 2 -10s 2 +70s 3 =360 z=360 -2x 2 +s 1 + 2s 2 -8s 3 =24 s 1 = x 2 +x 3 +2s 2 -4s 3 =8 x 3 =8 4* x x s s 3 =2 x 1 =2 none

Illustrate in example Changing the right-hand side of a constraint Effect on the current basis Since b does not appear in (10), changing the right-hand side of a constraint will leave row 0 of the optimal tableau unchanged; changing a right-hand side cannot cause the current basis to become suboptimal. Will affect the right-hand side of the constraints in the optimal tableau.

Illustrate in example Note As long as the right-hand side of each constraint in the optimal tableau remains nonnegative, the current basis remains feasible and optimal. If at least one right-hand side in the optimal tableau becomes negative, the current basis is no longer feasible and therefore no longer optimal.

Illustrate in example Suppose we change the amount of finishing hours (b 2 ) in the Dakota problem.Dakota problem Currently,b 2 =20, we change b 2 to 20+Δ, from (6) The current basis will remain optimal if and only if the following hold: 24+2Δ≥0 Δ≥ Δ≥0 Δ≥ Δ≥0 Δ≤4

Illustrate in example For -4≤Δ≤4, 20-4≤b 2 ≤20+4, or 16≤b 2 ≤24, the current basis remains feasible and therefore optimal.

Illustrate in example Effect on decision variables and z Even if the current basis remains optimal, the values of the decision variables and z change. The new values of the basic variables are as follows From (11),

Illustrate in example When the current basis is no longer optimal Suppose we change b 2 to 30. from (6), the right- hand side of the constraints in the tableau for BV={s 1,x 3,x 1 } is From (11), the right-hand side of row 0 is

Illustrate in example The tableau is, Basic variables z +5x 2 +10s 2 +10s 3 =380 z=380 -2x 2 +s 1 + 2s 2 -8s 3 =44 s 1 =44 -2x 2 +x 3 +2s 2 -4s 3 =28 x 3 =28 x x s s 3 =-3 x 1 =-3 Since x1=-3, BV is no longer feasible or optimal.

Illustrate in example Changing the column of a nonbasic variable x 2 is a nonbasic variable in the optimal solution, this means that Dakota should not manufacture any tables at present. Suppose, the price of tables increased to $43 and, because of changes in production technology, a table required 5 board feet of lumber, 2 finishing hours and 2 carpentry hours.

Illustrate in example Here we are changing elements of the column for x 2 in the original problem. The change leaves B and b unchanged. Thus, the right-hand side of the optimal tableau remains unchanged. From (10),the only part of row 0 that is changed is c 2 ; the current basis will remain optimal if and only if c 2 ≥0 holds.(10),

Illustrate in example Use (10) to compute the new coefficient of x 2 in row 0. Since c 2 <0, the current basis is no longer optimal. c 2 =-3 means that each table now increases revenues by $3.

Illustrate in example From (5), The tableau is now, Basic variables z -3x 2 +10s 2 +10s 3 =280 z=280 -7x 2 +s 1 + 2s 2 -8s 3 =24 s 1 =24 -4x 2 +x 3 +2s 2 -4s 3 =8 x 3 =8 x 1 +2x s s 3 =2 x 1 =2*

Illustrate in example Adding a new activity Suppose that Dakota is considering making footstools. A stool sells for $15 and requires 1 board foot of lumber, 1 finishing hour, and 1 carpentry hour. Should the company manufacture any stools?

Illustrate in example Our new initial tableau is z -60x 1 -30x 2 -20x 3 -15x 4 =0 8x 1 +6x 2 +x 3 +x 4 +s 1 =48 4x 1 +2x x 3 +x 4 +s 2 =20 2x x x 3 +x 4 +s 3 =8 We call the addition of the x 4 column to the problem adding a new activity.

Illustrate in example From (6), we see that the right-hand sides of all constraints in the optimal tableau will remain unchanged.(6), From (10), we see that the coefficient of each of the old variables in row 0 will remain unchanged.(10), Compute c 4, the current basis will remain optimal if c 4 ≥0.

Illustrate in example Since The current basis is still optimal. This means that each stool manufactured will decrease revenues by $5. for this reason, we choose not to manufacture any stools.

Summary of sensitivity analysis Change in initial problem Effect on optimal tableau Current basis is still optimal if Changing nonbasic objective function coefficient c j Coefficient of x j in optimal row 0 is changed Coefficient of c j in row 0 for current basis is still nonnegative Changing basic objective function coefficient c j Entire row 0 may change Each variable still has a nonnegative coefficient in row 0 Changing right-hand side of a constraint Right-hand side of constraints and row 0 are changed Right-hand side of each constraint is still nonnegative Changing the column of a nonbasic variable x j or adding a new variable x j Changes the coefficient for x j in row 0 and x j ’s constraint column in optimal tableau The coefficient of x j in row 0 is still nonnegative