Systems Analysis in Construction

Systems Analysis in Construction CB312 Construction & Building Engineering Department- AASTMT by A h m e d E l h a k e e m & M o h a m e d S a i e d

3. Linear Programming Optimization Simplex Method 135

Linear Programming: the Simplex Method 136 An Overview of the Simplex Method Standard Form Tableau Form Setting Up the Initial Simplex Tableau Improving the Solution Calculating the Next Tableau Solving a Minimization Problem Special Cases

Introduction to Simplex Method The Simplex Method: It is the general method to solve any LP (Fleet Assignment at Delta Air Lines deals with 60,000 variables and 40,000 constraints); It is an analytical method at which a set of simultaneous linear Equations is solved together; The number of variables (n) is greater that the number of equations (m), accordingly (n-m) variables should be assumed as zeros to have a solution; The variables assumed to be zeros are called Non-Basic and the ones obtained from solving the equations are called Basic; Simplex is an iterative procedure method for moving from one Basic Feasible Solution (extreme point) to another until the optimal solution is reached; 137

Introduction to Simplex Method The Simplex Method: Steps Leading to the Simplex Method Formulate Problem as LP Put In Standard Form Put In Tableau Form Execute Simplex Method 138

Simplex Method: Standard Form Convert from Inequalities to Equations ( > or < = ) + ve RHS + ve Variables, i.e., (x1, x2,. > 0) Max or Min 2x 1-3x 2-4x 3 s. t. x 1 + x 2 + x 3 30 2x 1 + x 2 + 3x 3 60 x 1 - x 2 + 2x 3 = -20 x 1 0 x 2 0 139

Simplex Method: Standard Form An LP is in standard form when: RHS of constraints are non-negative All constraints are equalities (Equations) All variables are non-negative. Putting an LP formulation into standard form involves: Use of Mathematical Modeling Transformation Adding slack variables to < constraints Subtracting surplus variables from > constraints Using substitution rules. 140

Simplex Method: Standard Form Max or Min 2x 1-3x 2-4x 3 s. t. x 1 + x 2 + x 3 30 2x 1 + x 2 + 3x 3 60 x 1 - x 2 + 2x 3 = -20 x 1 0 x 2 0 x 3?? 2-Add slack 2-Subtrack surplus 1- Multiply by -1 3- Substitution 4- Substitution 141

Simplex Method: Standard Form Max or Min 2x 1-3x 2-4x 3 s. t. x 1 + x 2 + x 3 30 2x 1 + x 2 + 3x 3 60 x 1 - x 2 + 2x 3 = -20 x 1 0 Max or Min 2x 1-3x 2-4x 3 s. t. x 1 + x 2 + x 3 + s 1 = 30 2x 1 + x 2 + 3x 3 - s 2 = 60 -x 1 + x 2-2x 3 = 20 142 x 1, s 1, s 2 > 0

Simplex Method: Standard Form Max or Min 2x 1-3x 2-4x 3 s. t. x 1 + x 2 + x 3 + s 1 = 30 2x 1 + x 2 + 3x 3 - s 2 = 60 -x 1 + x 2-2x 3 = 20 x 1, s 1, s 2 > 0 x 2 < 0, Assume x 2 = -y 2 and y 2 > 0 143 Max or Min 2x 1 + 3y 2-4x 3 s. t. x 1 - y 2 + x 3 + s 1 = 30 2x 1 - y 2 + 3x 3 - s 2 = 60 -x 1 - y 2-2x 3 = 20 x 1, y 2 s 1, s 2 > 0

Simplex Method: Standard Form Max or Min 2x 1 + 3y 2-4x 3 s. t. x 1 - y 2 + x 3 + s 1 = 30 2x 1 - y 2 + 3x 3 - s 2 = 60 -x 1 - y 2-2x 3 = 20 x 1, y 2 s 1, s 2 > 0 Assume x 3 = w 3 -y 3 w 3,y 3 > 0 Max or Min 2x 1 + 3y 2 4w 3 + 4y 3 s. t. x 1 - y 2 + w 3 - y 3 + s 1 = 30 2x 1 - y 2 + 3w 3-3y 3 - s 2 = 60 - x 1 - y 2-2w 3 + 2y 3 = 20 144 x 1, y 2, w 3, y 3, s 1, s 2 > 0

Simplex Method: Standard Form Max or Min 2x 1 + 3y 2 4w 3 + 4y 3 s. t. x 1 - y 2 + w 3 - y 3 + s 1 = 30 2x 1 - y 2 + 3w 3-3y 3 - s 2 = 60 - x 1 - y 2-2w 3 + 2y 3 = 20 x 1, y 2, w 3, y 3, s 1, s 2 > 0 145

Simplex Method: Solving Equations 3x - y + z = 9 x + y - 3z = -9 2x - y + 3z = 15 3-1 1 1 1-3 2-1 3 x y z = 9-9 15 146

Simplex Method: 3-1 1 1 1-3 2-1 3 x y z = 9-9 15 3-1 1 9 1 1-3 -9 2-1 3 15 R1 R2 R3 R2 + R3, R2 x (-3) +R1 3-1 1 9 1 1-3 -9 2-1 3 15 1 0 0 v 1 0 1 0 v 2 0 0 1 v 3 x y z = v 1 v 2 v 3 147

Simplex Method: 3-1 1 9 1 1-3 -9 2-1 3 15 1 0 0 v 1 0 1 0 v 2 0 0 1 v 3 x y z = v 1 v 2 v 3 0 1 0 v 2 1 0 0 v 1 0 0 1 v 3 x y z = v 1 v 2 v 3 Select pivot element, make it 1 and the rest in the column = 0 148

Simplex Method: 3-1 1 9 1 1-3 -9 2-1 3 15 1 0 0 v 1 0 1 0 v 2 0 0 1 v 3 3 1 2 Selected as pivot element make it 1 and the rest zeros 1 0 0 149 3 1 2 0 0 1 How?

Simplex Method: a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 b 1 R 1 b 2 R 2 b 3 R 3 Step 1 R 1 /a 11 a 12 b 1 Step 2 a 13 1 a 11 a 11 a 21 a 22 a 23 a 31 a 32 a 33 a 11 R b 2 R 2 b 3 R 3 To make a 21 = 0 (-1xa 21 ) x R* + R 2 To make a 31 = 0 (-1xa 31 ) x R* + R 3 To make a i1 = 0 (-1xa i1 ) x R* + R i 150

Simplex Method: 3-1 1 9 R 1 1 1-3 -9 R 2 2-1 3 15 R 3 1-1/3 1/3 9/3 R* =R 1 /3 1 1-3 -9 R 2 2-1 3 15 R 3 + -1 1/3-1/3-3 (-1)xR* -2 2/3-2/3-6 (-2)xR* 1-1/3 1/3 3 R* 0 4/3-10/3-12 (-1)xR* +R 2 0-1/3 7/3 9 (-2)xR* +R 3 151

Simplex Method: 1-1/3 1/3 3 R 1 0 4/3-10/3-12 R 2 0-1/3 7/3 9 R 3 1-1/3 1/3 3 R 1 0 1-5/2-9 R*=R 2 /(4/3) 0-1/3 7/3 9 R 3 + 0 1/3-5/6-3 (- -1/3)xR* 0 1/3-5/6-3 (- -1/3)xR* 1 0-1/2 0 (1/3)xR*+R 1 0 1-5/2-9 R* 0 0 3/2 6 (1/3)xR*+R 3 152

Simplex Method: 1 0-1/2 0 R 1 0 1-5/2-9 R 2 0 0 3/2 6 R 3 1 0 0 2 (1/2)R*+R 1 0 1 0 1 (5/2)R*+R 2 0 0 1 4 R*=R 3 /(3/2) + 0 0 1/2 2 (1/2)xR* 0 0 5/2 10 (5/2)xR* 3-1 1 9 1 1-3 -9 2-1 3 15 1 0 0 2 0 1 0 1 0 0 1 4 x y z = 2 1 4 153

Simplex Method: Standard Form Max or Min 2x 1 + 3y 2 4w 3 + 4y 3 s. t. x 1 - y 2 + w 3 - y 3 + s 1 = 30 2x 1 - y 2 + 3w 3-3y 3 - s 2 = 60 - x 1 - y 2-2w 3 + 2y 3 = 20 x 1, y 2, w 3, y 3, s 1, s 2 > 0 6 variables & 3 Equations 154

Simplex Method: Tableau Form 6 variables & 3 Equations x1 y2 w3 y3 s1 s2 Basis C 2 3-4 4 RHS 1-1 1-1 1 30 2-1 3-3 -1 60-1 -1-2 2 20 155

Simplex Method: Tableau Form 6 variables & 3 Equations x1 y2 w3 y3 s1 s2 Basis C 2 3-4 4 0 0 RHS 1-1 1-1 1 0 30 2-1 3-3 0-1 60-1 -1-2 2 0 0 20 z c-z 156

Simplex Method: Tableau Form The Simplex Method: It is the general method to solve any LP (Fleet Assignment at Delta Air Lines deals with 60,000 variables and 40,000 constraints); It is an analytical method at which a set of simultaneous linear Equations is solved together; The number of variables (n) is greater that the number of equations (m), accordingly (n-m) variables should be assumed as zeros to have a solution; The variables assumed to be zeros are called Non-Basic and the ones obtained from solving the equations are called Basic; Simplex is an iterative procedure method for moving from one Basic Feasible Solution (extreme point) to another until the optimal solution is reached; 157

Example 1 Max 5x 1 + 7x 2 s.t. x 1 < 6 2x 1 + 3x 2 < 19 x 1 + x 2 < 8 x 1, x 2 > 0 158

Example 1: Standard Form Max 5x 1 + 7x 2 + 0s 1 + 0s 2 + 0s 3 s. t. 1x 1 + 0x 2 + 1s 1 + 0s 2 + 0s 3 = 6 2x 1 + 3x 2 + 0s 1 + 1s 2 + 0s 3 = 19 1x 1 + 1x 2 + 0s 1 + 0s 2 + 1s 3 = 8 x 1, x 2, s 1, s 2, s 3 > 0 5 variables & 3 Equations 159 2 =(5-3) variables are Non-Basic with values = 0 and 3 variables are Basic with values > 0 (can get by solving the 3 equations)

Example 1: Max (z) 5x 1 + 7x 2 1x 1 + 1s 1 = 6 2x 1 + 3x 2 + 1s 2 = 19 1x 1 + 1x 2 + 1s 3 = 8 160 If we assumed that s 1 and s 3 = 0 then x 1 = (6-1*0) = 6 x 2 = (8-1*6-1*0) = 2 s 2 = (19-2*6-3*2) = 1 z = 5*6 +7*2 = 44 Basic-Feasible Solution

Example 1: Max (z) 5x 1 + 7x 2 1x 1 + 1s 1 = 6 2x 1 + 3x 2 + 1s 2 = 19 1x 1 + 1x 2 + 1s 3 = 8 161 If we assumed that s 1 and s 2 = 0 then x 1 = (6-1*0) = 6 x 2 = (19-1*0-2*6)/3 = 7/3 Non-Feasible s 3 = (8-1*6-1*7/3) = -1/3 < 0 Solution z = 5*6 +7*7/3 = 46.333

Example 1: Simplex is an iterative procedure method for moving from one Basic- Feasible - Solution (extreme point) to another until the optimal solution is reached. Basic-Feasible Solutions 162 Assume x 1 & x 2 = 0

Simplex Method: Tableau Form How to assume Basic-Feasible variables or determine the Non-Basic variables for an Initial Tableau 163

Simplex Method: Tableau Form The simplex tableau is a convenient means for performing the calculations required by the simplex method. A set of equations is in tableau form if for each equation: Its right hand side (RHS) is non-negative, and There is a basic variable. (A basic variable for an equation is a variable whose coefficient in the equation is +1 and whose coefficient in all other equations of the problem is 0.) 164 To generate an initial tableau form: An artificial variable must be added to each constraint that does not have a basic variable.

Example 1: Standard Form Max 5x 1 + 7x 2 + 0s 1 + 0s 2 + 0s 3 s. t. 1x 1 + 0x 2 + 1s 1 + 0s 2 + 0s 3 = 6 2x 1 + 3x 2 + 0s 1 + 1s 2 + 0s 3 = 19 1x 1 + 1x 2 + 0s 1 + 0s 2 + 1s 3 = 8 x 1, x 2, s 1, s 2, s 3 > 0 5 variables & 3 Equations 165 3 variables are Basic with values > 0 (s 1, s 2, & s 3 for the initial table) 2 variables are Non-Basic with values = 0 (Rest of variables x 1 & x 2 ).

Simplex Method: Tableau Form x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z c-z s 1, s 2, s 3 Basic & x 1, x 2 Non-Basic 166

Simplex Method: Tableau Form x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z c-z x 1 = 0 & x 2 = 0 s 1, s 2 & s 3 = 6, 19 & 8 167

Simplex Method: Initial Tableau x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 z 1 = 0 x 1 + 0 x 2 + 0 x 1 = 0 168

Simplex Method: Initial Tableau x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 169

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 The complete solution is: s 1 = 6, s 2 = 19, s 3 = 8, x 1 = 0, x 2 = 0 & Z = 0 170 Is this solution optimum? Why?

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 x 2 is assumed to be 0 (non-basic) if moved to basic (each increase in x 2 by 1 unit will affect Z by +7 units, i.e., add to Z maximization). 171 x 1 is assumed to be 0 (non-basic) if moved to basic (each increase in x 1 by 1 unit will affect Z by +5 units, i.e., add to Z maximization).

Simplex Method: Improving the Solution Pivot Element make it 1 and the reset of the column 0s x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 19/3 8/1 172 Entering Column at maximum +ve (c-z) make x 2 Basic variable Leaving Row at minimum RHS/EC make s 2 Non-Basic variable

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 s 2 0 2 3 0 1 0 19 s 3 0 1 1 0 0 1 8 z 0 0 0 0 0 0 c-z 5 7 0 0 0 173 x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 x 2 7 2/3 1 0 1/3 0 19/3 s 3 0 1 1 0 0 1 8 z c-z R* R 1 R 3

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 x 2 7 2/3 1 0 1/3 0 19/3 s 3 0 1 1 0 0 1 8 z c-z R 1 R* R 3-2/3-1 0-1/3 0-19/3-1 x R* 174 x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 x 2 7 2/3 1 0 1/3 0 19/3 s 3 0 1/3 0 0-1/3 1 5/3 z 14/3 7 0 7/3 0 44.33 c-z 1/3 0 0-7/3 0 The complete solution: s 1 = 6, s 2 = 0, s 3 = 5/3, x 1 = 0, x 2 = 19/3 & Z = 44.33-1xR*+ R 3

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 1 0 1 0 0 6 x 2 7 2/3 1 0 1/3 0 19/3 s 3 0 1/3 0 0-1/3 1 5/3 z 14/3 7 0 7/3 0 44.33 c-z 1/3 0 0-7/3 0 6/1 19/2 5 x 1 IN & s 3 OUT 175 x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 0 0 1 1-3 1 x 2 7 0 1 0 1-2 3 x 1 5 1 0 0-1 3 5 z 5 7 0 2 1 46 c-z 0 0 0-2 -1 -R* + R 1-2/3 R* + R 2 R* No more improvements then optimum solution reached

Simplex Method: Improving the Solution x 1 x 2 s 1 s 2 s 3 Basis C 5 7 0 0 0 RHS s 1 0 0 0 1 1-3 1 x 2 7 0 1 0 1-2 3 x 1 5 1 0 0-1 3 5 z 5 7 0 2 1 46 c-z 0 0 0-2 -1 The complete solution: s 1 = 1, s 2 = 0, s 3 = 0, x 1 = 5, x 2 = 3 & Z = 46 176

Simplex Method: missing basic variables Max 2x 1 + 3y 2 4w 3 + 4y 3 + 0s 1 + 0s 2 s. t. x 1 - y 2 + w 3 - y 3 + s 1 = 30 2x 1 - y 2 + 3w 3-3y 3 - s 2 = 60 - x 1 - y 2-2w 3 + 2y 3 = 20 x 1, y 2, w 3, y 3, s 1, s 2 > 0 s 1 is a basic variable for equation 1, What about equations 2 and 3? 177

Simplex Method: missing basic variables Max 2x 1 + 3y 2 4w 3 + 4y 3 + 0s 1 + 0s 2 Ma 2 Ma 3 s. t. x 1 - y 2 + w 3 - y 3 + s 1 = 30 2x 1 - y 2 + 3w 3-3y 3 - s 2 + a 2 = 60 - x 1 - y 2-2w 3 + 2y 3 + a 3 = 20 x 1, y 2, w 3, y 3, s 1, s 2, a 2, a 3 > 0 178 a 2 and a 3 are artificial variables with no meaning Penalized the objective function by subtracting M (big number) multiplied by the artificial variable, accordingly it will not affect the objective function if a = 0.

Simplex Method: Tableau Form x 1 y 2 w 3 y 3 s 1 s 2 a 2 a 3 Basis C 2 3-4 4 0 0 -M -M RHS s 1 0 1-1 1-1 1 0 0 0 30 a 2 -M 2-1 3-3 0-1 1 0 60 a 3 -M -1-1 -2 2 0 0 0 1 20 z -M 2M -M M 0 M -M -M -80M c-z M+2 3-2M M-4 4-M 0 -M 0 0 179

Simplex Method: Tableau Form x 1 y 2 w 3 y 3 s 1 s 2 a 2 a 3 Basis C 2 3-4 4 0 0 -M -M RHS s 1 0 1-1 1-1 1 0 0 0 30 a 2 -M 2-1 3-3 0-1 1 0 60 a 3 -M -1-1 -2 2 0 0 0 1 20 z -M 2M -M M 0 M -M -M -80M c-z M+2 3-2M M-4 4-M 0 -M 0 0 a 2 column can be removed in the next table as it equals 0 180

Simplex Method: Artificial Variables Example Max 50x 1 + 40x 2 s. t. 3x 1 + 5x 2 < 150 (Assembly time) 1x 2 < 20 (Ultra Portable display) 8x 1 + 5x 2 < 300 (Warehouse space) 1x 1 + 1x 2 > 25 (Minimum total production) x 1, x 2 > 0 181

Simplex Method: Artificial Variables Example Max 50x 1 + 40x 2 + 0s 1 + 0s 2 + 0s 3 + 0s 4 Ma 4 s. t. 3x 1 + 5x 2 + 1s 1 + 0s 2 + 0s 3 + 0s 4 + 0a 4 = 150 0x 1 + 1x 2 + 0s 1 + 1s 2 + 0s 3 + 0s 4 + 0a 4 = 20 8x 1 + 5x 2 + 0s 1 + 0s 2 + 1s 3 + 0s 4 + 0a 4 = 300 1x 1 + 1x 2 + 0s 1 + 0s 2 + 0s 3-1s 4 + 1a 4 = 25 x 1, x 2, s 1, s 2, s 3, s 4, a 4 > 0 182

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 a 4 Basis C 50 40 0 0 0 0 -M RHS s 1 0 3 5 1 0 0 0 0 150 s 2 0 0 1 0 1 0 0 0 20 s 3 0 8 5 0 0 1 0 0 300 a 4 -M 1 1 0 0 0-1 1 25 z -M -M 0 0 0 M -M -25M c-z 50+M 40+M 0 0 0 -M 0 183

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 a 4 Basis C 50 40 0 0 0 0 -M RHS s 1 0 3 5 1 0 0 0 0 150 150/3 s 2 0 0 1 0 1 0 0 0 20 s 3 0 8 5 0 0 1 0 0 300 a 4 -M 1 1 0 0 0-1 1 25 300/8 25/1 z -M -M 0 0 0 M -M -25M c-z 50+M 40+M 0 0 0 -M 0 184 a 4 is assumed to have a value (basic variable) however as being a leaving variable, it should be moved to non-basic variables with value of 0. But as an artificial variable (no meaning for it), there is no longer need for it.

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 Basis C 50 40 0 0 0 0 RHS s 1 0 0 2 1 0 0 3 75 s 2 0 0 1 0 1 0 0 20 s 3 0 0-3 0 0 1 8 100 x 1 50 1 1 0 0 0-1 25-3R*+R1 No change -8R*+R3 R* z 50 50 0 0 0-50 1250 c-z 0-10 0 0 0 50 The complete solution: s 1 = 75, s 2 = 20, s 3 = 100, x 1 = 25, x 2 = 0, s 4 = 0 & Z = 1250 185

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 Basis C 50 40 0 0 0 0 RHS s 1 0 0 2 1 0 0 3 75 75/3 s 2 0 0 1 0 1 0 0 20 s 3 0 0-3 0 0 1 8 100 100/8 x 1 50 1 1 0 0 0-1 25 z 50 50 0 0 0-50 1250 c-z 0-10 0 0 0 50 186

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 Basis C 50 40 0 0 0 0 RHS s 1 0 0 25/8 1 0-3/8 0 37.5 s 2 0 0 1 0 1 0 0 20 s 4 0 0-3/8 0 0 1/8 1 12.5 x 1 50 1 5/8 0 0 1/8 0 37.5-3R*+R1 No change R* R*+R4 z 50 125/4 0 0 50/8 0 1875 c-z 0 35/4 0 0-50/8 0 The complete solution: s 1 = 37.5, s 2 = 20, s 3 = 0, x 1 = 37.5, x 2 = 0, s 4 = 12.5 & Z = 1875 187

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 Basis C 50 40 0 0 0 0 RHS s 1 0 0 25/8 1 0-3/8 0 37.5 s 2 0 0 1 0 1 0 0 20 37.5*8/25 20/1 s 4 0 0-3/8 0 0 1/8 1 12.5 x 1 50 1 5/8 0 0 1/8 0 37.5 37.5*8/5 z 50 125/4 0 0 50/8 0 1875 c-z 0 35/4 0 0-50/8 0 188

Simplex Method: Artificial Variables Example x 1 x 2 s 1 s 2 s 3 s 4 Basis C 50 40 0 0 0 0 RHS x 2 40 0 1 8/25 0-3/25 0 12 s 2 0 0 0-8/25 1 3/25 0 8 s 4 0 0 0 3/25 0 2/25 1 17 x 1 50 1 0-1/5 0 1/5 0 30 R* -R*+R2 3/8R*+R3-5/8R*+R4 z 50 40 2.8 0 5.2 0 1980 c-z 0 0-2.8 0-5.2 0 The complete solution: s 1 = 0, s 2 = 8, s 3 = 0, x 1 = 30, x 2 = 12, s 4 = 17 & Z = 1980 189

Simplex Method: Minimization To solve a minimization problem multiply the objective function by (-1) and solve it as a maximization problem Min 5x 1 + 7x 2 = Max -5x 1-7x 2 After solving the problem using the Simplex procedure the values of the variable will be the same however the value of (Z) in the last table should be multiplied by (-1). 190

Simplex Method: Special Cases Infeasibility Unboundedness Alternative Optimal Solution Sensitivity Analysis 191

Simplex Method: Infeasibility Infeasibility is detected in the simplex method when an artificial variable remains positive in the final tableau. LP Formulation Final Tableau MAX 2x 1 + 6x 2 s. t. 4x 1 + 3x 2 < 12 2x 1 + x 2 > 8 x 1, x 2 > 0 x 1 x 2 s 1 s 2 a 2 Basis C 2 6 0 0 -M RHS x 1 2 1 3/4 1/4 0 0 3 a 2 -M 0-1/2-1/2-1 1 2 z 2 M/2+3/2 M/2+1/2 M -M 6-2M c-z 0 -M/2+9/2 -M/2-1/2 -M 0 192

Simplex Method: Unboundedness A linear program has an unbounded solution if all entries in an entering column are non-positive. LP Formulation Final Tableau MAX 2x 1 + 6x 2 s. t. 4x 1 + 3x 2 > 12 2x 1 + x 2 > 8 x 1, x 2 > 0 x 1 x 2 s 1 s 2 Basis C 3 4 0 0 RHS x 2 4 3 1 0-1 8 s 1 0 2 0 1-1 3 z 12 4 0-4 32 c-z -9 0 0 4 193

Simplex Method: Alternative Opt. Sol. A linear program has alternate optimal solutions if the final tableau has a c j - z j value equal to 0 for a non-basic variable. Final Tableau 194 x 1 x 2 x 3 s 1 s 2 s 3 s 4 Basis C 2 4 6 0 0 0 0 RHS s 3 0 0 0 2 4-2 1 0 8 x 2 4 0 1 2 2-1 0 0 6 x 1 2 1 0-1 1 2 0 0 4 s 4 0 0 0 1 3 2 0 1 12 z 2 4 6 10 0 0 0 32 c-z 0 0 0-10 0 0 0

Simplex Method: Alternative Opt. Sol. In the previous slide we see that the optimal solution is: x 1 = 4, x 2 = 6, x 3 = 0, s 1 = 0, s 2 = 0, s 3 = 8, s 4 = 12 and z = 32 Note that x 3 is non-basic and its c 3 - z 3 = 0. This 0 indicates that if x 3 were increased, the value of the objective function would not change. Another optimal solution can be found by choosing x 3 as the entering variable and performing one iteration of the simplex method. The new tableau on the next slide shows an alternative optimal solution is: x 1 = 7, x 2 = 0, x 3 = 3, s 1 = 0, s 2 = 0, s 3 = 2, s 4 = 9 and z = 32 195

Simplex Method: Alternative Opt. Sol. Final Tableau 196 x 1 x 2 x 3 s 1 s 2 s 3 s 4 Basis C 2 4 6 0 0 0 0 RHS s 3 0 0-1 0 2-1 1 0 2 x 3 6 0 0.5 1 1-0.5 0 0 3 x 1 2 1 0.5 0 2 1.5 0 0 7 s 4 0 0-0.5 0 2 2.5 0 1 9 z 2 4 6 10 0 0 0 32 c-z 0 0 0-10 0 0 0

Simplex Method: Summary Initial Tableau Step 1: If the problem is a minimization problem, multiply the objective function by -1. Step 2: If the problem formulation contains any constraints with negative right-hand sides, multiply each constraint by -1. Step 3: Add a slack variable to each < constraint. Step 4: Subtract a surplus variable and add an artificial variable to each > constraint. 197

Simplex Method: Summary Initial Tableau Step 5: Add an artificial variable to each = constraint. Step 6: Set each slack and surplus variable's coefficient in the objective function equal to zero. Step 7: Set each artificial variable's coefficient in the objective function equal to -M, where M is a very large number. Step 8: Each slack and artificial variable becomes one of the basic variables in the initial basic feasible solution. 198

Simplex Method: Summary Procedure Step 1: Determine Entering Variable Identify the variable with the most positive value in the c j - z j row. (The entering column is called the pivot column.) Step 2: Determine Leaving Variable For each positive number in the entering column, compute the ratio of the right-hand side values divided by these entering column values. If there are no positive values in the entering column, STOP; the problem is unbounded. Otherwise, select the variable with the minimal ratio. (The leaving row is called the pivot row.) 199

Simplex Method: Summary Procedure Step 3: Generate Next Tableau Divide the pivot row by the pivot element (the entry at the intersection of the pivot row and pivot column) to get a new row. We denote this new row as (row *). Replace each non-pivot row i with: [new row i] = [current row i] - [(a ij ) x (row *)], where a ij is the value in entering column j of row i Step 4: Calculate zj Row for New Tableau For each column j, multiply the objective function coefficients of the basic variables by the corresponding numbers in column j and sum them. 200

Simplex Method: Summary Procedure Step 5: Calculate c j - z j Row for New Tableau For each column j, subtract the z j row from the c j row. If none of the values in the c j - z j row are positive, GO TO STEP 1. If there is an artificial variable in the basis with a positive value, the problem is infeasible. STOP. Otherwise, an optimal solution has been found. The current values of the basic variables are optimal. The optimal values of the non-basic variables are all zero. If any non-basic variable's c j - z j value is 0, alternate optimal solutions might exist. STOP. 201

END of Simplex Method