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4.4 The Simplex Method and the Standard Minimization Problem Question 1: What is a standard minimization problem? Question 2: How is the standard minimizati on problem related to the dual standard maximization problem? Question 3: How do you apply the Simp lex Method to a standard minimization problem? In Section 4.3, the Simplex Method was used to solve the standard maximization problem. With some modifications, it can also be used to solve the standard minimization problem. These problems shar e characteristics and are called the dual of the other. In this section, we learn what a standard minimization problem is and how it is connected to the standard maximization probl em. Utilizing the connection between the dual problems, we will solve the standard minimization problem with the Simplex Method.

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Question 1: What is a standard minimization problem? In Section 4.3, we learned that some types of linear programming problems, where the objective function is maximized, are call ed standard maximization problems. A similar form exists for another for li near programming problems where the objective function is minimized. A standard minimization problem is a type of linear programming problem in which the objective function is to be minimized and has the form 11 2 2 nn wdy dy dy where ,, dd are real numbers and ,, are decision variables. The decision variables must represent non- negative values. The other constraints for the standard minimization problem have the form 11 2 2 nn ey ey ey f t where ,, ee and are real numbers and . The standard minimization problem is written with the decision variables ,, , but any letters could be used as long as t he standard minimization problem and the corresponding dual maximization problem do not share the same variable names. Often a problem can be rewritten to put it into standard minimization form. In particular, constraints are often manipu lated algebraically so the ea ch constraint has the form 11 2 2 nn ey ey ey f t . Example 1 demonstrates how a constraint can be changed to put it in the proper form.

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For the problems in this section, we will requ ire the coefficients of the objective function be positive. Although this is not a requirement of the Simplex Method, it simplifies the presentation in this section. Example 1 Write As A Sta ndard Minimization Problem In section 4.2, we solved the linear programming problem 12 21 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt using a graph. Rewrite this linear programming problem as a standard minimization problem. Solution In a standard minimization probl em, the objective function must have the form 11 2 2 nn wdy dy dy where ,, dd are real number constants and ,, are the decision vari ables. The objective function matches this form with . Each constraint must have the form 11 2 2 nn ey ey ey f t where ,, ee and are real number constants. Additionally, the constant f must be non-negative. The second constraint, 12 74 32 yy , fits this form perfectly. The first constraint appears to hav e the correct type of terms, but variable terms are on both sides of t he inequality. To put in the proper format, add to both sides of the inequality: 12 yy

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With this change, we can write the problem as a standard minimization problem, 12 12 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt In addition to adding and subtracting terms to a constraint, we can also multiply or divide the terms in a constraint by nonzero real numbers. However, remember that the direction of the inequality changes when you mu ltiply or divide by a negative number. This can complicate or even prevent a li near programming problem from being changed to standard minimization form.

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Question 2: How is the standard minimizati on problem related to the dual standard maximization problem? At this point, the connection between t he standard minimization problem and the standard maximization problem is not clear. Let’s look at an example of a standard minimization problem and another rela ted standard maximization problem. The linear programming problem 12 12 12 12 Minimize 10 20 subject to 416 34 24 0, 0 wy y yy yy yy t t tt is a standard minimization problem. The related dual maximization problem is found by forming a matrix before the objec tive function is modified or slack variables are added to the constraints. The entries in this matrix are formed from the coe fficients and constants in the constraints and objective function: To find the coefficients and constants in the dual problem, switch the rows and columns. In other words, make the ro ws in the matrix above become the columns in a new matrix, 1310 4420 16 20 0 1416 3424 10 20 0 Coefficients from the first constraint Coefficients from the second constraint Coefficients from the objective function Constant from the first constraint Constant from the second constraint No constants in the objective function

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The values in the new matrix help us to fo rm the constraints and obj ective function in a standard maximization problem: Notice the inequalities have switched direct ions since the dual problem is a standard maximization problem and the nam es of the variables are di fferent from the original minimization problem. Putting these details t ogether with non-negativity constraints, we get the standard maximization problem 12 12 12 12 Maximize 16 24 subject to 310 44 20 0, 0 zx x xx xx xx d d tt This strategy works in general to find the dual problem. Example 2 Find the Dual Maximization Problem In Example 1, we rewrote a linear programming problem as a standard minimization problem, 1310 4420 16 24 0 12 310 xx 12 44 20 xx 12 Maximize 16 24 zx x

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12 12 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt Find the dual maximization probl em associated with this standard minimization problem. Solution The dual maximization problem can be formed by examining a matrix where the first tw o rows are the coefficients and constants of the constraints and the last row contains the coefficients on the right side of the objective function. In the case of this standard ma ximization problem, we get the x matrix 12 7432 41 0 The vertical line separates the coefficients from the constants, and the horizontal line separates the entries corresponding to the constraints from the entries corresponding to the obj ective function. Notice that the entries are written befor e introducing slack variables or rearranging the objective function. The zero in the last column corresponding to the objective function comes from the fact that the objective function has no constants in it. The coefficients and constants for t he dual maximization problem are formed when the rows and columns of this matrix are interchanged. The new matrix,

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74 141 2320 is utilized to find the dual problem . The first row corresponds to the constraint 12 74 xx d . The second row corresponds to the constraint 12 41 xx d . Notice that each constraint incl udes a less than or equal to ) to insure it fits the format of a standard ma ximization problem. The last row corresponds to the objective function 12 232 zx x . These inequalities and equations ar e combined to yield the standard maximization problem 12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt

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Question 3: How do you apply the Simp lex Method to a standard minimization problem? Example 2 illustrates how to convert a standard minimization problem into a standard maximization problem. These problems are call ed the dual of each other. The solutions of the dual problems are related and c an be exploited to solve both problems simultaneously. Let’s look at the solution of each linear programming problem graphically. For each problem, let’s look at a graph of the feasible region and a table of corner points with corresponding objective function values. From t he table, we see that the solutions share the same objective function value at their respective solutions. 12 yy Minimize 12 10 20 wy y 16,0 160 4,3 100 0,6 120 12 xx Maximize 12 16 24 zx x 5, 0 90 2.5, 2.5 100 10 0, 80

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Although the corner points yielding the ma ximum or minimum are not the same, the value of the objective function at t he optimal corner point is the same, 100 . In other words, 10 4 20 3 100 yields the same value as 16 2.5 24 2.5 100 Another connection between the dual problems is evident if we apply the Simplex Method to the dual maximization problem 12 12 12 12 Maximize 16 24 subject to 310 44 20 0, 0 zx x xx xx xx d d tt If we rearrange the objective function and add sl ack variables to the constraints, we get the system of equations 121 12 2 12 31 44 20 16 24 xxs xx s xx z This system corresponds to the initial simple x tableau shown below. The pivot column is the second column and the quotients can be formed to yield 1212 10 20 1 1 0 0 3.3 4010 5 16 24 0 0 1 10 20 xxssz The pivot for this tableau is the in the first row, second column. 10

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If we multiply the first row by , the pivot becomes a one an d results in the tableau The first simplex iteration is completed by cr eating zeros in the rest of the pivot column. To change these entries, multiply the first row by -4 and add it to the second row. Then multiply the first row by 24 and add it to the third row. Now that the pivot is a one and the rest of the pivot column are zeros, look at the indicator row to see if another Simplex Method iteration is needed. Si nce the entry in the first column of the indicator row, -8, is negative, we make the first column the new pivot column. The quotients for each row of the tableau are formed below: 12 1 2 11 33 88 10 10 33 20 2 33 100 10 010 2.5 80 8 0180 xx s sz y y The smallest ratio is in the second row. The pivot, , must be changed to a one by multiplying the second row by , 12 ecomes 40 44 33 3 82 33 3 400 4: 4401020 010 24 : 82480080 16 24 0 0 1 0 8 0 80180 121 2 10 11 33 3 820 33 3 100 010 80 8 0180 xx s sz 13 24 ecomes ecomes 10 11 33 3 1310010 100 1212 10 11 33 3 100 4401020 16 24 0 0 1 0 xxssz 11

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Once the pivot is a one, row operations are used to change the rest of the pivot column to zeros. The entry in the first row of the pivot column is changed to a zero by placing the sum of times the second row and the first row in the first row. The entry in the third row of the pivot column is changed to a zero by placing the sum of 8 times the second ro w and the third row in the third row, Since the indicator row no longer contains any negative entries, we have reached the final tableau. If we examine the final simp lex tableau carefully, we can see the solution to the standard maximization problem and the standard minimization problem: 21 ecomes 111 368 6 10 11 33 3 11 28 2 00 100 01 0 8: 80 43020 80 8 0180 0 0 4 3 1 100 12 1 2 11 28 2 35 28 2 01 0 10 0 0 0 4 3 1 100 xx s s z 23 ecomes ecomes 82 33 3 35 28 2 010 10 0 1212 10 11 33 3 35 28 2 100 10 0 80 8 0180 xxssz 12

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The final simplex tableau gives the soluti on to the standard maximization problem and the solution to the correspond ing dual standard minimization problem. This means that as long as we can solve the standard maximization problem with the Simplex Method, we get the solution to the dual standard minimi zation problem for free. This suggests a strategy for solving standar d minimization problems. How to Solve a Standard Minimization Problem with the Dual Problem 1. Make sure the minimization pr oblem is in standard form. If it is not in standard form, modi fy the problem to put it in standard form. 2. Find the dual standard maximization problem. 3. Apply the Simplex Method to solve the dual maximization problem. 4. Once the final simplex tableau has been calculated, the minimum value of the standard minimization problem’s objective function is the same as the maximum value of the standard maximization problem ’s objective function. 121 2 11 28 2 35 28 2 01 0 10 0 00 4 3 1100 xx s s z The solution to the standard maximization problem is 55 12 22 ,, xx . The solution to the dual minimization problem is 12 ,4,3 yy . The value for the objective function in the dual problems is 100 zw . 13

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5. The solution to the standar d minimization problem is found in the bottom row of the final simplex tableau in the columns corresponding to the slack variables. Example 3 Find the Optimal Solution In section 4.2, we solved the linear programming problem 12 21 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt using a graph. In 1.1Question 1Exa mple 2, we found the associated dual maximization problem, 12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt Apply the Simplex Method to this dual problem to solve the minimization problem. Solution In Example 1 and Example 2 we wrote this problem as a standard minimization problem an d found the dual maximization problem. In this example, we’ll take the dual problem, 14

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12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt and apply the Simplex Method. The initial simplex tableau is fo rmed from the syst em of equations 121 12 2 12 74 41 232 0 xxs xx s xx z Notice that the slack variables and are included in the equations corresponding to the constraints, and the objective function has been rearranged appropriately. The initial tableau is 1212 71004 140101 2 320010 xxssz . The most negative entry in the indicator row is -32 , so the second column is the pivot column. Now calc ulate the quotients to find the pivot row, 1212 100 0.57 1 0 1 0 0.25 23200 0 xxssz 15

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The smallest quotient co rresponds to putting the pivot in the second row, second column. To change the entry in this position to a one, multiply the second row by : To put zeros in the rest of the pivot column, we utilize more row operations. Since the indicator row is non-negativ e, this tableau corresponds to the optimal solution. The solution is found in the indicator row under the columns for the slack variabl es. The lowest value for is and occurs at 12 ,0,8 yy . This strategy works for st andard minimization problems in volving more variables or more constraints. Example 4 has two decisi on variables, but three constraints. This changes the sizes of the matrices involved, but not the process of applying the Simplex Method to the dual standard maximization problem. 21 ecomes 777 444 679 444 7: 70 0 :71004 01 0 32 : 8320808 2 320010 600818 12 679 444 111 444 01 0 10 0 600818 xyss z 23 32 ecomes ecomes 111 444 140101 10 0 1212 111 444 71004 10 0 2320010 xxssz 16

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Example 4 Find the Minimum Cost In Section 4.2, we found the cost of contracting barrels of American ale from contract brewery 1 and barrels of America ale from contract brewery 2. The lin ear programming problem for this application is 12 12 21 21 12 Minimize 100 125 subject to 10,000 0.25 0, 0 CQQ QQ QQ QQ QQ t tt Follow the parts a through c to solv e this linear programming problem. a. Rewrite this problem so that it is a standard minimization problem. Solution The objective function must have the form 11 2 2 nn wdy dy dy where ,, are the decision variables, and ,, dd are constants. In this ca se the decision variables are and , and is used instead of . A different name for the variable is acceptable as long as the terms on the right side each contain a constant times a variable. The constraints must have the form 11 2 2 nn ey ey ey f t , where ,, ee and f are constants. The first constraint, 12 10,000 QQ t , has the proper format, but wit h the decision variables and instead of and . 17

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The second and third constraints mu st be modified to match the form 11 2 2 nn yf t . Subtract 0.25 from both sides of the constraint 21 0.25 QQ to yield 12 0.25 0 QQ t The third constraint is converted to the proper form by rearranging the inequality 12 QQ . Subtract from both sides to yield 12 QQ These changes lead to a standard minimization problem, 12 12 12 12 12 Minimize 100 125 subject to 10,000 0.25 0 0, 0 CQQ QQ QQ QQ QQ t t t tt b. Find the dual problem for t he standard minimization problem. Solution The dual maximization problem is found by forming a matrix from the constraints and objective function. The coefficients and constants in the constraints co mpose the first three rows. The coefficients from the objective functi on are placed the fourth row of the matrix. 18

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12 12 12 12 12 Minimize 100 125 subject to 10,000 0.25 0 0, 0 CQQ QQ QQ QQ QQ t t t tt 1 1 10,000 0.25 1 0 110 100 125 0 The dual maximization problem’s coe fficients and constant are found by switching the rows and columns of the matrix 1110,000 0.25 1 0 110 100 125 0 1 0.25 1 100 1 1 1 125 10,000 0 0 0 We’ll use the decision variables , , and and write the corresponding dual maximization problem, 123 123 123 Maximize 10,000 subject to 0.25 100 125 0, 0, 0 zx xxx xxx xxx d d ttt c. Apply the Simplex Method to the dual problem to find the solution to the standard minimization problem. Solution The standard minimization problem is solved by applying the Simplex Method to the dual maximization problem. The first step is to write out the system of equations we ’ll work with including the slack variables: switch rows and columns 19

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1231 1232 0.25 100 125 10,000 xxxs xxxs xz This system of equations corresponds to the initial simplex tableau 12 1 0.25 1 1 0 0 100 1 1 1 0 1 0 125 10,000 0 0 0 0 1 0 xxxssz The only negative number in the indicator row is - 10,000 , so the pivot column is the first column. We choose the pivot row by forming quotients from the last co lumn and the pivot column: 12 100 125 0.25 1 1 0 0 100 11010 125 10,000 0 0 0 0 1 0 100 125 xxxssz The smallest quotient is 100 so the first row is the pivot row. Conveniently, the pivot is already a one and we can use row operations to change the rest of the column to zeros. 12 ecomes 12 3 12 10.25 1 1 00 100 01.25 2 1 10 25 0 2500 10,000 10,000 0 1 1,000,000 xx x ssz 13 10,000 ecomes 1: 10.25 1 100 100 1 1 1 0 1 0 125 01.25 2 110 25 10,000 : 10,000 2500 10,000 10,000 0 0 1,000,000 10,000 0 0 0 0 1 0 0 2500 10,000 10,000 0 1 1,000,000 20

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The new matrix still has a negative number in the indicator row, so we must choose a new pivot. The second column is the pivot column. The second row is the pivot row since it contains the only admissible quotient, 25 1.25 . The pivot, 1.25 , is changed to a one by multiplying the second row by 1.25 : With a one in the pivot, we can use row operations to put zeros in the rest of the pivot column. Multiply the pivot row by 0.25 and add it to the first row to put a zero at the top of the pivot column. Multiply the pivot row by 2500 and add it to the third row to pu t a zero at the bottom of the pivot row: No entry in the indicator row is negat ive, so we know that this tableau corresponds to the solution. The solu tion to the minimization problem lies in the indicator row in the columns corresponding to the slack variables, 1.25 ecomes 1.25 01.25 2 1 1 025 0 1 1.6 0.8 0.8 0 20 12 3 1 2 1 0.25 1 1 0 0 100 0 1 1.6 0.8 0.8 0 20 0 2500 10,000 10,000 0 1 1,000,000 xx x s sz 0.25 : 0 0.25 0.4 0.2 0.2 0 5 1 0.25 1 1 0 0 100 1 0 0.6 0.8 0.2 0 105 2500 : 0 2500 4000 2000 2000 0 50,000 0 2500 10,000 10,000 0 1 1,000,000 0 0 6,000 8,000 2000 1 1,050,000 21 0.25 ecomes 12 3 1 0 0.6 0.8 0.2 0 105 0 1 1.6 0.8 0.8 0 20 0 0 6000 8000 2000 1 1,050,000 xx x s s z 23 2500 ecomes 21

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12 3 1 0 0.6 0.8 0.2 0 105 0 1 1.6 0.8 0.8 0 20 0 0 6000 8000 2000 1 1,050,000 xx x s s z The lowest cost is $ 1,050,000 and occurs when 12 , 8000, 2000 QQ . This means the lowest cost occurs when 8000 barrels are contracted from brewery 1 and 2000 barrels are contracted from brewery 2. Solution to the minimization roblem Lowest cost 22

4 The Simplex Method and the Standard Minimization Problem Question 1 What is a standard minimization problem Question 2 How is the standard minimizati on problem related to the dual standard maximization problem ID: 22460

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4.4 The Simplex Method and the Standard Minimization Problem Question 1: What is a standard minimization problem? Question 2: How is the standard minimizati on problem related to the dual standard maximization problem? Question 3: How do you apply the Simp lex Method to a standard minimization problem? In Section 4.3, the Simplex Method was used to solve the standard maximization problem. With some modifications, it can also be used to solve the standard minimization problem. These problems shar e characteristics and are called the dual of the other. In this section, we learn what a standard minimization problem is and how it is connected to the standard maximization probl em. Utilizing the connection between the dual problems, we will solve the standard minimization problem with the Simplex Method.

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Question 1: What is a standard minimization problem? In Section 4.3, we learned that some types of linear programming problems, where the objective function is maximized, are call ed standard maximization problems. A similar form exists for another for li near programming problems where the objective function is minimized. A standard minimization problem is a type of linear programming problem in which the objective function is to be minimized and has the form 11 2 2 nn wdy dy dy where ,, dd are real numbers and ,, are decision variables. The decision variables must represent non- negative values. The other constraints for the standard minimization problem have the form 11 2 2 nn ey ey ey f t where ,, ee and are real numbers and . The standard minimization problem is written with the decision variables ,, , but any letters could be used as long as t he standard minimization problem and the corresponding dual maximization problem do not share the same variable names. Often a problem can be rewritten to put it into standard minimization form. In particular, constraints are often manipu lated algebraically so the ea ch constraint has the form 11 2 2 nn ey ey ey f t . Example 1 demonstrates how a constraint can be changed to put it in the proper form.

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For the problems in this section, we will requ ire the coefficients of the objective function be positive. Although this is not a requirement of the Simplex Method, it simplifies the presentation in this section. Example 1 Write As A Sta ndard Minimization Problem In section 4.2, we solved the linear programming problem 12 21 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt using a graph. Rewrite this linear programming problem as a standard minimization problem. Solution In a standard minimization probl em, the objective function must have the form 11 2 2 nn wdy dy dy where ,, dd are real number constants and ,, are the decision vari ables. The objective function matches this form with . Each constraint must have the form 11 2 2 nn ey ey ey f t where ,, ee and are real number constants. Additionally, the constant f must be non-negative. The second constraint, 12 74 32 yy , fits this form perfectly. The first constraint appears to hav e the correct type of terms, but variable terms are on both sides of t he inequality. To put in the proper format, add to both sides of the inequality: 12 yy

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With this change, we can write the problem as a standard minimization problem, 12 12 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt In addition to adding and subtracting terms to a constraint, we can also multiply or divide the terms in a constraint by nonzero real numbers. However, remember that the direction of the inequality changes when you mu ltiply or divide by a negative number. This can complicate or even prevent a li near programming problem from being changed to standard minimization form.

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Question 2: How is the standard minimizati on problem related to the dual standard maximization problem? At this point, the connection between t he standard minimization problem and the standard maximization problem is not clear. Let’s look at an example of a standard minimization problem and another rela ted standard maximization problem. The linear programming problem 12 12 12 12 Minimize 10 20 subject to 416 34 24 0, 0 wy y yy yy yy t t tt is a standard minimization problem. The related dual maximization problem is found by forming a matrix before the objec tive function is modified or slack variables are added to the constraints. The entries in this matrix are formed from the coe fficients and constants in the constraints and objective function: To find the coefficients and constants in the dual problem, switch the rows and columns. In other words, make the ro ws in the matrix above become the columns in a new matrix, 1310 4420 16 20 0 1416 3424 10 20 0 Coefficients from the first constraint Coefficients from the second constraint Coefficients from the objective function Constant from the first constraint Constant from the second constraint No constants in the objective function

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The values in the new matrix help us to fo rm the constraints and obj ective function in a standard maximization problem: Notice the inequalities have switched direct ions since the dual problem is a standard maximization problem and the nam es of the variables are di fferent from the original minimization problem. Putting these details t ogether with non-negativity constraints, we get the standard maximization problem 12 12 12 12 Maximize 16 24 subject to 310 44 20 0, 0 zx x xx xx xx d d tt This strategy works in general to find the dual problem. Example 2 Find the Dual Maximization Problem In Example 1, we rewrote a linear programming problem as a standard minimization problem, 1310 4420 16 24 0 12 310 xx 12 44 20 xx 12 Maximize 16 24 zx x

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12 12 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt Find the dual maximization probl em associated with this standard minimization problem. Solution The dual maximization problem can be formed by examining a matrix where the first tw o rows are the coefficients and constants of the constraints and the last row contains the coefficients on the right side of the objective function. In the case of this standard ma ximization problem, we get the x matrix 12 7432 41 0 The vertical line separates the coefficients from the constants, and the horizontal line separates the entries corresponding to the constraints from the entries corresponding to the obj ective function. Notice that the entries are written befor e introducing slack variables or rearranging the objective function. The zero in the last column corresponding to the objective function comes from the fact that the objective function has no constants in it. The coefficients and constants for t he dual maximization problem are formed when the rows and columns of this matrix are interchanged. The new matrix,

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74 141 2320 is utilized to find the dual problem . The first row corresponds to the constraint 12 74 xx d . The second row corresponds to the constraint 12 41 xx d . Notice that each constraint incl udes a less than or equal to ) to insure it fits the format of a standard ma ximization problem. The last row corresponds to the objective function 12 232 zx x . These inequalities and equations ar e combined to yield the standard maximization problem 12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt

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Question 3: How do you apply the Simp lex Method to a standard minimization problem? Example 2 illustrates how to convert a standard minimization problem into a standard maximization problem. These problems are call ed the dual of each other. The solutions of the dual problems are related and c an be exploited to solve both problems simultaneously. Let’s look at the solution of each linear programming problem graphically. For each problem, let’s look at a graph of the feasible region and a table of corner points with corresponding objective function values. From t he table, we see that the solutions share the same objective function value at their respective solutions. 12 yy Minimize 12 10 20 wy y 16,0 160 4,3 100 0,6 120 12 xx Maximize 12 16 24 zx x 5, 0 90 2.5, 2.5 100 10 0, 80

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Although the corner points yielding the ma ximum or minimum are not the same, the value of the objective function at t he optimal corner point is the same, 100 . In other words, 10 4 20 3 100 yields the same value as 16 2.5 24 2.5 100 Another connection between the dual problems is evident if we apply the Simplex Method to the dual maximization problem 12 12 12 12 Maximize 16 24 subject to 310 44 20 0, 0 zx x xx xx xx d d tt If we rearrange the objective function and add sl ack variables to the constraints, we get the system of equations 121 12 2 12 31 44 20 16 24 xxs xx s xx z This system corresponds to the initial simple x tableau shown below. The pivot column is the second column and the quotients can be formed to yield 1212 10 20 1 1 0 0 3.3 4010 5 16 24 0 0 1 10 20 xxssz The pivot for this tableau is the in the first row, second column. 10

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If we multiply the first row by , the pivot becomes a one an d results in the tableau The first simplex iteration is completed by cr eating zeros in the rest of the pivot column. To change these entries, multiply the first row by -4 and add it to the second row. Then multiply the first row by 24 and add it to the third row. Now that the pivot is a one and the rest of the pivot column are zeros, look at the indicator row to see if another Simplex Method iteration is needed. Si nce the entry in the first column of the indicator row, -8, is negative, we make the first column the new pivot column. The quotients for each row of the tableau are formed below: 12 1 2 11 33 88 10 10 33 20 2 33 100 10 010 2.5 80 8 0180 xx s sz y y The smallest ratio is in the second row. The pivot, , must be changed to a one by multiplying the second row by , 12 ecomes 40 44 33 3 82 33 3 400 4: 4401020 010 24 : 82480080 16 24 0 0 1 0 8 0 80180 121 2 10 11 33 3 820 33 3 100 010 80 8 0180 xx s sz 13 24 ecomes ecomes 10 11 33 3 1310010 100 1212 10 11 33 3 100 4401020 16 24 0 0 1 0 xxssz 11

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Once the pivot is a one, row operations are used to change the rest of the pivot column to zeros. The entry in the first row of the pivot column is changed to a zero by placing the sum of times the second row and the first row in the first row. The entry in the third row of the pivot column is changed to a zero by placing the sum of 8 times the second ro w and the third row in the third row, Since the indicator row no longer contains any negative entries, we have reached the final tableau. If we examine the final simp lex tableau carefully, we can see the solution to the standard maximization problem and the standard minimization problem: 21 ecomes 111 368 6 10 11 33 3 11 28 2 00 100 01 0 8: 80 43020 80 8 0180 0 0 4 3 1 100 12 1 2 11 28 2 35 28 2 01 0 10 0 0 0 4 3 1 100 xx s s z 23 ecomes ecomes 82 33 3 35 28 2 010 10 0 1212 10 11 33 3 35 28 2 100 10 0 80 8 0180 xxssz 12

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The final simplex tableau gives the soluti on to the standard maximization problem and the solution to the correspond ing dual standard minimization problem. This means that as long as we can solve the standard maximization problem with the Simplex Method, we get the solution to the dual standard minimi zation problem for free. This suggests a strategy for solving standar d minimization problems. How to Solve a Standard Minimization Problem with the Dual Problem 1. Make sure the minimization pr oblem is in standard form. If it is not in standard form, modi fy the problem to put it in standard form. 2. Find the dual standard maximization problem. 3. Apply the Simplex Method to solve the dual maximization problem. 4. Once the final simplex tableau has been calculated, the minimum value of the standard minimization problem’s objective function is the same as the maximum value of the standard maximization problem ’s objective function. 121 2 11 28 2 35 28 2 01 0 10 0 00 4 3 1100 xx s s z The solution to the standard maximization problem is 55 12 22 ,, xx . The solution to the dual minimization problem is 12 ,4,3 yy . The value for the objective function in the dual problems is 100 zw . 13

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5. The solution to the standar d minimization problem is found in the bottom row of the final simplex tableau in the columns corresponding to the slack variables. Example 3 Find the Optimal Solution In section 4.2, we solved the linear programming problem 12 21 12 12 Minimize 4 subject to 74 32 0, 0 wyy yy yy yy t t tt using a graph. In 1.1Question 1Exa mple 2, we found the associated dual maximization problem, 12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt Apply the Simplex Method to this dual problem to solve the minimization problem. Solution In Example 1 and Example 2 we wrote this problem as a standard minimization problem an d found the dual maximization problem. In this example, we’ll take the dual problem, 14

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12 12 12 12 Maximize z 2 32 subject to 74 41 0, 0 xx xx xx d d tt and apply the Simplex Method. The initial simplex tableau is fo rmed from the syst em of equations 121 12 2 12 74 41 232 0 xxs xx s xx z Notice that the slack variables and are included in the equations corresponding to the constraints, and the objective function has been rearranged appropriately. The initial tableau is 1212 71004 140101 2 320010 xxssz . The most negative entry in the indicator row is -32 , so the second column is the pivot column. Now calc ulate the quotients to find the pivot row, 1212 100 0.57 1 0 1 0 0.25 23200 0 xxssz 15

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The smallest quotient co rresponds to putting the pivot in the second row, second column. To change the entry in this position to a one, multiply the second row by : To put zeros in the rest of the pivot column, we utilize more row operations. Since the indicator row is non-negativ e, this tableau corresponds to the optimal solution. The solution is found in the indicator row under the columns for the slack variabl es. The lowest value for is and occurs at 12 ,0,8 yy . This strategy works for st andard minimization problems in volving more variables or more constraints. Example 4 has two decisi on variables, but three constraints. This changes the sizes of the matrices involved, but not the process of applying the Simplex Method to the dual standard maximization problem. 21 ecomes 777 444 679 444 7: 70 0 :71004 01 0 32 : 8320808 2 320010 600818 12 679 444 111 444 01 0 10 0 600818 xyss z 23 32 ecomes ecomes 111 444 140101 10 0 1212 111 444 71004 10 0 2320010 xxssz 16

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Example 4 Find the Minimum Cost In Section 4.2, we found the cost of contracting barrels of American ale from contract brewery 1 and barrels of America ale from contract brewery 2. The lin ear programming problem for this application is 12 12 21 21 12 Minimize 100 125 subject to 10,000 0.25 0, 0 CQQ QQ QQ QQ QQ t tt Follow the parts a through c to solv e this linear programming problem. a. Rewrite this problem so that it is a standard minimization problem. Solution The objective function must have the form 11 2 2 nn wdy dy dy where ,, are the decision variables, and ,, dd are constants. In this ca se the decision variables are and , and is used instead of . A different name for the variable is acceptable as long as the terms on the right side each contain a constant times a variable. The constraints must have the form 11 2 2 nn ey ey ey f t , where ,, ee and f are constants. The first constraint, 12 10,000 QQ t , has the proper format, but wit h the decision variables and instead of and . 17

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The second and third constraints mu st be modified to match the form 11 2 2 nn yf t . Subtract 0.25 from both sides of the constraint 21 0.25 QQ to yield 12 0.25 0 QQ t The third constraint is converted to the proper form by rearranging the inequality 12 QQ . Subtract from both sides to yield 12 QQ These changes lead to a standard minimization problem, 12 12 12 12 12 Minimize 100 125 subject to 10,000 0.25 0 0, 0 CQQ QQ QQ QQ QQ t t t tt b. Find the dual problem for t he standard minimization problem. Solution The dual maximization problem is found by forming a matrix from the constraints and objective function. The coefficients and constants in the constraints co mpose the first three rows. The coefficients from the objective functi on are placed the fourth row of the matrix. 18

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12 12 12 12 12 Minimize 100 125 subject to 10,000 0.25 0 0, 0 CQQ QQ QQ QQ QQ t t t tt 1 1 10,000 0.25 1 0 110 100 125 0 The dual maximization problem’s coe fficients and constant are found by switching the rows and columns of the matrix 1110,000 0.25 1 0 110 100 125 0 1 0.25 1 100 1 1 1 125 10,000 0 0 0 We’ll use the decision variables , , and and write the corresponding dual maximization problem, 123 123 123 Maximize 10,000 subject to 0.25 100 125 0, 0, 0 zx xxx xxx xxx d d ttt c. Apply the Simplex Method to the dual problem to find the solution to the standard minimization problem. Solution The standard minimization problem is solved by applying the Simplex Method to the dual maximization problem. The first step is to write out the system of equations we ’ll work with including the slack variables: switch rows and columns 19

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1231 1232 0.25 100 125 10,000 xxxs xxxs xz This system of equations corresponds to the initial simplex tableau 12 1 0.25 1 1 0 0 100 1 1 1 0 1 0 125 10,000 0 0 0 0 1 0 xxxssz The only negative number in the indicator row is - 10,000 , so the pivot column is the first column. We choose the pivot row by forming quotients from the last co lumn and the pivot column: 12 100 125 0.25 1 1 0 0 100 11010 125 10,000 0 0 0 0 1 0 100 125 xxxssz The smallest quotient is 100 so the first row is the pivot row. Conveniently, the pivot is already a one and we can use row operations to change the rest of the column to zeros. 12 ecomes 12 3 12 10.25 1 1 00 100 01.25 2 1 10 25 0 2500 10,000 10,000 0 1 1,000,000 xx x ssz 13 10,000 ecomes 1: 10.25 1 100 100 1 1 1 0 1 0 125 01.25 2 110 25 10,000 : 10,000 2500 10,000 10,000 0 0 1,000,000 10,000 0 0 0 0 1 0 0 2500 10,000 10,000 0 1 1,000,000 20

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The new matrix still has a negative number in the indicator row, so we must choose a new pivot. The second column is the pivot column. The second row is the pivot row since it contains the only admissible quotient, 25 1.25 . The pivot, 1.25 , is changed to a one by multiplying the second row by 1.25 : With a one in the pivot, we can use row operations to put zeros in the rest of the pivot column. Multiply the pivot row by 0.25 and add it to the first row to put a zero at the top of the pivot column. Multiply the pivot row by 2500 and add it to the third row to pu t a zero at the bottom of the pivot row: No entry in the indicator row is negat ive, so we know that this tableau corresponds to the solution. The solu tion to the minimization problem lies in the indicator row in the columns corresponding to the slack variables, 1.25 ecomes 1.25 01.25 2 1 1 025 0 1 1.6 0.8 0.8 0 20 12 3 1 2 1 0.25 1 1 0 0 100 0 1 1.6 0.8 0.8 0 20 0 2500 10,000 10,000 0 1 1,000,000 xx x s sz 0.25 : 0 0.25 0.4 0.2 0.2 0 5 1 0.25 1 1 0 0 100 1 0 0.6 0.8 0.2 0 105 2500 : 0 2500 4000 2000 2000 0 50,000 0 2500 10,000 10,000 0 1 1,000,000 0 0 6,000 8,000 2000 1 1,050,000 21 0.25 ecomes 12 3 1 0 0.6 0.8 0.2 0 105 0 1 1.6 0.8 0.8 0 20 0 0 6000 8000 2000 1 1,050,000 xx x s s z 23 2500 ecomes 21

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12 3 1 0 0.6 0.8 0.2 0 105 0 1 1.6 0.8 0.8 0 20 0 0 6000 8000 2000 1 1,050,000 xx x s s z The lowest cost is $ 1,050,000 and occurs when 12 , 8000, 2000 QQ . This means the lowest cost occurs when 8000 barrels are contracted from brewery 1 and 2000 barrels are contracted from brewery 2. Solution to the minimization roblem Lowest cost 22

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