Consider the following problem.Maximize Z = 3x₁ + x₂ + 2x₃,subject to x₁ - x₂ + 2x₃ ≤ 20 2x₁ + x₂ - x₃ ≤ 10 and x₁ ≥ 0,x₂ ≥ 0,x₃ ≥ 0.Let x₄ and x₅ denote the slack variables for the respective functional constraints.After we apply the simplex method,the final simplex tableau is
(a)Perform sensitivity analysis to determine which of the 11 parameters of the model are sensitive parameters in the sense that even small changes in just that parameter's value might change the optimal solution.
(b)Use algebraic analysis to find the allowable range to stay optimal for each c_j..
(c)Use algebraic analysis to find the allowable range for each b_i.

Question

Consider the following problem.Maximize Z = 3x₁ + x₂ + 2x₃,subject to x₁ - x₂ + 2x₃ ≤ 20 2x₁ + x₂ - x₃ ≤ 10 and x₁ ≥ 0,x₂ ≥ 0,x₃ ≥ 0.Let x₄ and x₅ denote the slack variables for the respective functional constraints.After we apply the simplex method,the final simplex tableau is

Consider the following problem.Maximize Z = 3x<sub>1</sub> + x<sub>2</sub> + 2x<sub>3</sub>,subject to x<sub>1</sub> - x<sub>2</sub> + 2x<sub>3</sub> ≤ 20 2x<sub>1</sub> + x<sub>2</sub> - x<sub>3</sub> ≤ 10 and x<sub>1</sub> ≥ 0,x<sub>2</sub> ≥ 0,x<sub>3</sub> ≥ 0.Let x<sub>4</sub> and x<sub>5</sub> denote the slack variables for the respective functional constraints.After we apply the simplex method,the final simplex tableau is (a)Perform sensitivity analysis to determine which of the 11 parameters of the model are sensitive parameters in the sense that even small changes in just that parameter's value might change the optimal solution. (b)Use algebraic analysis to find the allowable range to stay optimal for each c<sub>j</sub>.. (c)Use algebraic analysis to find the allowable range for each b<sub>i</sub>.

(a)Perform sensitivity analysis to determine which of the 11 parameters of the model are sensitive parameters in the sense that even small changes in just that parameter's value might change the optimal solution.
(b)Use algebraic analysis to find the allowable range to stay optimal for each c_j..
(c)Use algebraic analysis to find the allowable range for each b_i.

Quizplus · Accepted Answer

The Answer is (a)If b₁ or b₂ is changed,then $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ .So both b₁ and b₂ are sensitive parameters.If c₁ changes,then the coefficient of x₁ in Eq.(0)of the final tableau becomes $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ ,so the current optimal solution remains optimal as long as

\Delta

c₁ ≤ 8.Since small changes in c₁ will not affect the optimal solution,it is not a sensitive parameter.If either c₂ or c₃ changes,then the coefficient of x₂ and x₃ in Eq.(0)of the final tableau becomes 0 - $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ and 0 - $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ .Therefore,since x₂ and x₃ are basic variables,any change in $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ or $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ or $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ and $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ are not sensitive parameters.If a₁₁ or a₂₁ changes,it will only affect $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ , $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ ,and $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ .The optimal solution will change only if the new value of $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ is negative.However,since the original value of $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ is such a large positive number (8),rather large changes in a₁₁ or a₂₁ would be needed to drive the value of $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ negative.So small changes will not affect the optimal solution.Hence,a₁₁ and a₂₁ are not sensitive parameters.Since x₂ and x₃ are basic variables,changes in a₁₂,a₁₃,a₂₂,and a₂₃ will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c₁ only affect the value of $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ (the coefficient of x₁ in Eq.(0)in the final tableau).If c₁ = 3 changes to $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ = 3 +

\Delta

c₁,then $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ = 8 changes to $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ = 8 -

\Delta

c₁.The current optimal solution remains optimal as long as $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ = 8 -

\Delta

_ ≥ 0,so

\Delta

c₁ ≤ 8,so $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ = 3 +

\Delta

c₁ ≤ 11.Therefore,the allowable range to stay optimal for c₁ is c₁

\le

11.For changes in c₂,we noted in part (a)that the fact that x₂ is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c₂ changes by

\Delta

c₂,then

\Delta

c₂*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 +

\Delta

c₂ * Row 2

\ge

0.This is equivalent to $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ Since c₂ = 1 +

\Delta

c₂, the allowable range to stay optimal for c₂ is c₂

\ge

-3/5.For changes in c₃,since x₃ also is a basic variable,the same reasoning as for changes in c₂ implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 +

\Delta

c₃ * Row 1

\ge

0.This is equivalent to $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ Since c₃ = 2 +

\Delta

c₃,the allowable range to stay optimal for c₃ is c₃

\ge

-2/3.
(c)Changes in the right-hand side only affect the final right-hand side b^*.We require b^*

\ge

0.For b₁,we need b* + S* $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ or,equivalently, $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ Since b₁ = 20 +

\Delta

b₁, the allowable range for b₁ is b₁

\ge

-10.For b₂,we need b* + S* $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ or,equivalently, $(a)If b1 or b2 is changed,then .So both b1 and b2 are sensitive parameters.If c1 changes,then the coefficient of x1 in Eq.(0)of the final tableau becomes ,so the current optimal solution remains optimal as long as \Delta c1 ≤ 8.Since small changes in c1 will not affect the optimal solution,it is not a sensitive parameter.If either c2 or c3 changes,then the coefficient of x2 and x3 in Eq.(0)of the final tableau becomes 0 - and 0 - .Therefore,since x2 and x3 are basic variables,any change in or will require using elementary row operations to restore a value of 0 in the corresponding coefficient in the final tableau.However,since the coefficients of the nonbasic variables in Eq.(0)of the final tableau are rather large,a small change in either or will not cause any of these coefficients of the nonbasic variables to become negative,so the current optimal solution would remain optimal.Consequently, and are not sensitive parameters.If a11 or a21 changes,it will only affect , ,and .The optimal solution will change only if the new value of is negative.However,since the original value of is such a large positive number (8),rather large changes in a11 or a21 would be needed to drive the value of negative.So small changes will not affect the optimal solution.Hence,a11 and a21 are not sensitive parameters.Since x2 and x3 are basic variables,changes in a12,a13,a22,and a23 will require elementary row operations to restore proper form from Gaussian elimination.In the process,the entire row 1 and/or row 2,including the right-hand sides,will change.Therefore,the optimal solution will change.So these parameters are sensitive.(b)We have seen in part (a)that changes in c1 only affect the value of (the coefficient of x1 in Eq.(0)in the final tableau).If c1 = 3 changes to = 3 + \Delta c1,then = 8 changes to = 8 - \Delta c1.The current optimal solution remains optimal as long as = 8 - \Delta  ≥ 0,so \Delta c1 ≤ 8,so = 3 + \Delta c1 ≤ 11.Therefore,the allowable range to stay optimal for c1 is c1 \le 11.For changes in c2,we noted in part (a)that the fact that x2 is a basic variable requires using elementary row operations to restore proper form from Gaussian elimination.In particular,if c2 changes by \Delta c2,then \Delta c2*Row 2 needs to be added to Row 0 to obtain the new Row 0 in proper form.Therefore,the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c2 * Row 2 \ge 0.This is equivalent to Since c2 = 1 + \Delta c2, the allowable range to stay optimal for c2 is c2 \ge -3/5.For changes in c3,since x3 also is a basic variable,the same reasoning as for changes in c2 implies that the current optimal solution remains optimal (although the objective function value may change)if Row 0 + \Delta c3 * Row 1 \ge 0.This is equivalent to Since c3 = 2 + \Delta c3,the allowable range to stay optimal for c3 is c3 \ge -2/3. (c)Changes in the right-hand side only affect the final right-hand side b*.We require b* \ge 0.For b1,we need b* + S* or,equivalently, Since b1 = 20 + \Delta b1, the allowable range for b1 is b1 \ge -10.For b2,we need b* + S* or,equivalently, Since b2 = 10 + \Delta b2, the allowable range for b2 is b2 \ge -10.$ Since b₂ = 10 +

\Delta

b₂, the allowable range for b₂ is b₂

\ge

-10.

Consider the Following Problem