College of Management, NCTU Operation Research I Fall, Chap The Theory of the Simplex Method Terminology Constraint oundary equation For any constraint (functional and nonnegativity), replace its,, sign y an sign. Each forms a hyperplane (a flat geometric shape) in n-dimensional space. This hyperplane forms the constraint oundary for the corresponding constraint. Corner point feasile (CPF) solution A feasile solution that does not lie on any line segment connecting two other feasile solution. Each CPF solution lies at the intersection of n constraint oundary equations (and satisfies the other constraints). Adjacent CPF solutions Two CPF solutions are adjacent if the line segment connecting them is an edge of the feasile region. Edge of the feasile region An edge of the feasile region is a feasile line segment that lies at the intersection of n- constraint oundaries. One more property of CPF Solutions (with feasile solutions and a ounded feasile region) There are only a finite numer of CPF solutions. Defining equation Refer to the constraint oundary equations that yield (define) the indicated CPF solution. Chap-
College of Management, NCTU Operation Research I Fall, In the Wyndor example, what are the defining equations for CPF (, )? Indicating variales for constraint oundary equations Given a CPF solution, how do we tell whether a particular constraint oundary equation is one of the defining equations? Each constraint has an indicating variale that completely indicates (y whether its value is zero) whether that constraint s oundary equation is satisfied y the current solution. Type of Constraint Form of Constraint Constraint in Augmented Form Nonnegativity x x j j Constraint Boundary Equation x j x j x i n + i Functional n n ( ) a j ij x j i j + + n aij x j xn i i a j ij x j Functional n n n () a j ij x j i a + + j ij x j x n i i a j ij x j i n+ Functional n n n ( ) a j ij x j i a + + j ij x j x n i xsi i a j ij x j Indicating Variale x x i n+ i Si Indicating variale (a nonasic variale) the corresponding constraint oundary equation is satisfied is a defining equation x What are the indicating variales for the Wyndor example? CPF Solution BF solution Nonasic Variales Defining Equations (, ) (,,,, ) x x x x (, ) (,,,, ) x x x x (, ) (,,,, ) x x x x + x Chap-
College of Management, NCTU Operation Research I Fall, (, ) (,,,, ) x x x + x x (, ) (,,,, ) x x x x It is possile for a asic variale to e zero (degenerate). It implies that the CPF solution may satisfy another constraint oundary equation in addition to its n defining equations. The Simplex Method and the CPF/BF Solutions When the simplex method chooses an entering asic variale, it is choosing one of the edges emanating from the current CPF solution to move along. Deleting one constraint oundary (defining equation) from the set of n constraint oundaries defining the current solution. Increasing this variale from zero corresponds to moving along this edge. Moving away from the current solution along the intersection of the remaining n- constraint oundaries Having one of the asic variales (the leaving asic variale) decrease to zero corresponds to reaching the first new constraint oundary at the other end of this edge of the feasile region. Recall how do we perform the simplex method - x - x x + x x + x x + x + x What do we really need in a simplex iteration? Ojective row Entering column Current solution (the right hand side) If we are dealing with too many numers, can we perform the iteration in a more efficient way? Here comes the revised simplex method. Chap-
College of Management, NCTU Operation Research I Fall, The Matrix Representations for a LP Prolem Max cx S.T. Ax x Where c x,, A After adding slack variales, we have the augmented form of the prolem. x s So that the constraints ecome [A, I] x x s x and x s One of the key features of the revised simplex method involves the way in which it solves for each new BF solution after identifying its asic and nonasic variales. In a specific iteration, variales are either asic or nonasic variales. Thus, we can represent a LP prolem (in the augmented form) as c B x B + c N x N Bx B + Nx N x B, x N Chap-
College of Management, NCTU Operation Research I Fall, Example: the Wyndor Example x + x x + x x + x x + x + x x, x, x, x, x Solve the aove model x B + B - Nx N B - (x B B - B - Nx N ) The value of asic variales x B * B - c B x B + c N x N c B (B - B - Nx N ) + c N x N c B B - + (c N c B B - N) x N Move the term of nonasic variales to the left hand side to follow the same representation. + (c B B - N - c N )x N c B B - The value of the ojective function is * c B B - Coefficients of nonasic variales in the ojective row c B B - N - c N In the Wyndor Example Iteration : asic variales {x, x, x }, nonasic variales {x, x } Final iteration: asic variales {x, x, x }, nonasic variales {x, x } Chap-
College of Management, NCTU Operation Research I Fall, The Revised simplex method Recall what we really need in a simplex iteration. Determine the entering variale. Ojective row: c B B - N - c N Compute the Entering Column: only need to compute column of B - N corresponding to the entering variale. a: The corresponding interested column in N (i.e. the column of N corresponding to entering variale). d: The only needed column in B - N. d B - a (a is known since N is known) Determine the leaving variale (the minimum ratio test). That is, x B x B * x e d. (Recall x B B - B - Nx N ) x B * is the current values of asic variales. x e is the entering variale. Optimality test: Calculate c B B - N - c N (the coefficients of nonasic variales in the ojective row). If they are all non-negative, stop. Optimal is found. Update: x * B, x B, x N, B, N, c B, c N Repeat until the optimal solution is found. Chap-
College of Management, NCTU Operation Research I Fall, Example For the Revised Simplex Method -- the Wyndor Example Max x + x S.T. x + x x + x x + x + x x, x, x, x, x Chap-
College of Management, NCTU Operation Research I Fall, Chap- Some asic concepts of linear algera and matrix operations Multiply the nd row y and add to the st row. Perform the same operation on the identity matrix yields. The former operation is equivalent to. Multiply the nd row y and add to the st row, then multiple nd row y - and add to the rd row This operation is equivalent to For the original set of equations, the matrix form is cx I A c x s x Ax + Ix s After any iteration, we know that x B * B - and * c B B - Let look at the right hand side, the original vector is, now is B B c B
College of Management, NCTU Operation Research I Fall, What is the operation on the right hand side? So, we know what happen on the left hand side c A I Thus, the desired matrix form of the set of equations after any iteration is Iteration Basic Variale Eq. Coefficient of: Original Variale Slack Variales Right Side x B () (,,, m) -c A I Any () c B B - A c c B B - x B (,,, m) B - A Let s look at the second (final) iteration of the Wyndor example B - c B B - B - Chap-
College of Management, NCTU Operation Research I Fall, A Fundamental Insight Focus on the coefficients of slack variales and the information they give. After any iteration, the coefficients of the slack variales in each equation immediately reveal how that equation has een otained from the initial equations. Let s recall the Wyndor example in Taleau form. Basic Coefficient of: Variale x x x x x Right Side - - x x x - / x x / x - / x / -/ x / x -/ / For Iteration : New row old row + (/) (old row ) New row old row + () (old row ) New row (/) (old row ) New row old row + (-) (old row ) These algeraic operations amount to pre-multiplying rows to of the initial taleau y the matrix New row- / / Chap-
College of Management, NCTU Operation Research I Fall, Chap- Notice that the coefficients of the slack variales in the new taleau do indeed provide a record of the algeraic operations performed. New row [ ] [ ] [ ] / / + For Iteration New row old row + () (old row ) New row old row + (-/) (old row ) New row old row + () (old row ) New row (/) (old row ) Thus, the multiplying matrix is / / Final row - / / / / / / / / / / Final row ()(initial row ) + (/)(initial row ) + (-/)(initial row ) Final row ()(initial row ) + (/) (initial row ) + () (initial row ) Final row ()(initial row ) + (-/)(initial row ) + (/)(initial row ) Final row [ ] [ ] [ ] / / +
College of Management, NCTU Operation Research I Fall, Thus, given the initial taleau and the coefficients/parameters of the slack variales in any iteration, we could derive the rest of parameters. An example (.-) Fill the missing numers Max x + x + x S.T. x + x + /x x x /x x + x + /x x, x,, x Initial Taleau Basic Coefficient of: Variale x x x x x x Right Side x x x Final Taleau x x - x - Finally, recall that Chap-
College of Management, NCTU Operation Research I Fall, Iteration Basic Variale Eq. Coefficient of: Original Variale Slack Variales Right Side x B () (,,, m) -c A I Any x B () (,,, m) c B B - A c B - A c B B - B - c B B - B - This taleau also gives an explanation of shadow price. Chap-