ESE403. Formula Sheets. Following is a list of the formulae that will be supplied to your during the final.

Transcription

1 ESE403 Formula Sheets Following is a list of the formulae that will be supplied to your during the final. Chapter 5 Formula for Revised Simplex Method: x k = entering basic variable a ik = coefficient of x k in current Eq.(i), for I = 1, 2,, m (identified in step 2 of iteration) r = number of equation containing the leaving basic variable = 1 = In matrix form, the formula becomes: = Where matrix E is an identity matrix except that its rth column is replaced by the vector =,h = 1 = Formula for Matrix Form Simplex Method: Initial matrix 0 0 Final matrix = = Chapter 8 General Procedure for Constructing an Initial BF Solution. To begin, all source rows and destination columns of the transportation simplex tableau are initially under consideration for providing a basic variable (allocation).

2 1. From the rows and columns still under consideration, select the next basic variable (allocation) according to some criterion. 2. Make that allocation large enough to exactly use up the remaining supply in its row or the remaining demand in its column (whichever is smaller). 3. Eliminate that row or column (whichever had the smaller remaining supply or demand) from further consideration. (If the row and column have the same remaining supply and demand, then arbitrarily select the row as the one to be eliminated. The column will be used later to provide a degenerate basic variable, i.e., a circled allocation of zero.) 4. If only one row or only one column remains under consideration, then the procedure is completed by selecting every remaining variable (i.e., those variables that were neither previously selected to be basic nor eliminated from consideration by eliminating their row or column) associated with that row or column to be basic with the only feasible allocation. Otherwise, return to step 1. Northwest corner rule: Begin by selecting x 11 (that is, start in the northwest corner of the transportation simplex tableau). Thereafter, if x ij was the last basic variable selected, then next select x i,j+1 (that is, move one column to the right) if source i has any supply remaining. Otherwise, next select x i+1,j (that is, move one row down). Vogel s approximation method: For each row and column remaining under consideration, calculate its difference, which is defined as the arithmetic difference between the smallest and next-to-the-smallest unit cost c ij still remaining in that row or column. (If two unit costs tie for being the smallest remaining in a row or column, then the difference is 0.) In that row or column having the largest difference, select the variable having the smallest remaining unit cost. (Ties for the largest difference, or for the smallest remaining unit cost, may be broken arbitrarily). Russell s approximation method: For each source row ii remaining under consideration, determine its u i, which is the largest unit cost c ij still remaining in that row. For each destination column j remaining under consideration, determine its v j, which is the largest unit cost c ij still remaining in that column. For each variable x ij not previously selected in these rows and columns, calculate ij = c ij u i v j. Select the variable having the largest (in absolute terms) negative value of ij, (Tie may be broken arbitrarily.) Summary of the Transportation Simplex Method: Initialization: Construct an initial BF solution. Go to optimality test. Optimality test: Derive u i and v j by selecting the row having the largest number of allocations, setting its u i = 0, and then solving the set of equations c ij = u i + v j for each (i, j) such that x ij is basic. If c ij u i v j 0 for every (i, j) such that x ij is nonbasic, then the current solution is optimal, so stop. Otherwise, go to an iteration.

3 Iteration: 1. Determine the entering basic variable: Select the nonbasic variable x ij having the largest (in absolute terms) negative value of c ij -u i -v j. 2. Determine the leaving basic variable: Identify the chain reaction required to retain feasibility when the entering basic variable is increased. From the donor cells, select the basic variable having the smallest value. 3. Determine the new BF solution: Add the value of the leaving basic variable to the allocation for each recipient cell. Subtract this value form the allocation for each donor cell. Summary of the Hungarian Algorithm 1. Subtract the smallest number in each row from every number in the row. (This is called row reduction.) Enter the results in a new table. 2. Subtract the smallest number in each column of the new table from every number in the column. (This is called column reduction.) Enter the results in another table. 3. Test whether an optimal set of assignments can be made. You do this by determining the minimum number of lines needed to cover (i.e., cross out) all zeros. Since this minimum number of lines equals the maximum number of assignments that can be make to zero element positions, if the minimum number of lines equals the number of rows, an optimal set of assignment is possible. (If you find that a complete set of assignments to zero element positions is not possible, this means that you did not reduce the number of lines covering all zeros down to the minimum number.) In that case, go to step 6. Otherwise go on to step If the number of lines is less than the number of rows, modify the table in the following way: a. Subtract the smallest uncovered number from every uncovered number in the table. b. Add the smallest uncovered number to the numbers at intersections of covering lines. c. Numbers crossed out but not at the intersections of cross-out lines carry over unchanged to the next table. 5. Repeat steps 3 and 4 until an optimal set of assignments is possible 6. Make the assignments one at a time in positions that have zero elements. Begin with rows or columns that have only one zero. Since each row and each column needs to receive exactly one assignment, cross out both the row and the column involved after each assignment is made. Then move on to the rows and columns that are not yet crossed out to select the next assignment, with preference again given to any such row or column that has only one zero that is not crossed out. Continue until every row and every column has exactly one assignment and so has been crossed out. The complete set of assignments made in this way is an optimal solution for the problem.

4 Chapter 9 Algorithm for the Shortest-Path Problem Object of nth iteration: Find the nth nearest node to the origin (to be repeated for n = 1, 2, until the nth nearest node is the destination. Input for nth iteration: n-1 nearest nodes to the origin (solved for at the previous iterations), including their shortest path and distance from the origin. (These nodes, plus the origin, will be called solved nodes; the others are unsolved nodes.) Candidates for nth nearest node: Each solved node that is directly connected by a link to one or more unsolved node provides one candidate = the unsolved node with the shortest connecting link. (Ties provide additional candidates.) Calculation of nth nearest node: For each such solved node and its candidate, add the distance between them and the distance of the shortest path from the origin to this solved node. The candidate with the smallest such total distance is the nth nearest node (ties provide additional solved nodes), and its shortest path is the one generating this distance. Algorithm for the Minimum Spanning Tree Problem 1. Select any node arbitrarily, and then connect it (i.e., add a link) to the nearest distinct node. 2. Identify the unconnected node that is closest to a connected node, and then connect these two nodes (i.e., add a link between them). Repeat this step until all nodes have been connected. 3. Tie breaking: Ties for the nearest distinct node (step 1) or the closest unconnected node (step 2) may be broken arbitrarily, and the algorithm must still yield an optimal solution. However, such ties are a signal that there may be (but need not be) multiple optimal solutions. All such optimal solutions can be identified by pursuing all ways of breaking ties to their conclusion. The Augmenting Path Algorithm for the Maximum Flow Problem 1. Identify an augmenting path by finding some directed path from the source to the sink in the residual network such that every arc on this path has strictly positive residual capacity. (If no augmenting path exists, the net flows already assigned constitute an optimal flow pattern.)

5 2. Identify the residual capacity c* of this augmenting path by finding the minimum of the residual capacities of the arcs on this path. Increase the flow in this path by c* 3. Decrease by c* the residual capacity of each arc on this augmenting path. Increase by C* the residual capacity of each arc in the opposite direction on this augmenting path. Return to step 1. Chapter 11 Summary of the BIP Branch-and-Bound Algorithm Initialization: Set Z*= -. Apply the bounding step, fathoming step, and optimality test described below to the whole problem. If not fathomed, classify this problem as the one remaining subproblem for performing the first full iteration below. Steps for each iteration: 4. Branching: Among the remaining (unfathomed) subproblem, select the one that was created most recently. (Break ties according to which has the larger bound.) Branch from the node for this subproblem to create two new subproblems by fixing the next variable (the branching variable) at either 0 or Bounding: For each new subproblem, apply the simplex method to its LP relaxation to obtain an optimal solution, including the value of Z, for this LP relaxation. If this value of Z is not an integer, round it down to an integer. (If it was already an integer, no change is needed) This integer value of Z is the bound for the subproblem. 6. Fathoming: For each new subproblem, apply the three fathoming tests, and discard those subproblems that are fathomed by any of the tests. Optimality test: Stop when there are no remaining subproblems; the current incumbent is optima. Otherwise, return to perform another iteration. Summary of Fathoming Criteria. A subproblem is fathomed if an analysis of its relaxation reveals that Criterion 1: Feasible solution of the subproblem must have Z Z*, or Criterion 2: The subproblem has no feasible solutions, or Criterion 3: An optimal solution of the subproblem has been found. Summary of the MIP Branch-and-Bound Algorithm

6 Initialization: Set Z* = -. Apply the bounding step, fathoming step, and optimality test descriabed below to the whole problem. If not fathomed, classify this problem as the one remaining subproblem for performing the first full iteration below. Steps for each iteration: 1. Branching: Among the remaining (unfathomed) subproblems, select the one that was created most recently. (Break ties according to which has the larger bound.) Among the integer-restricted variables that have a noninteger value in the optimal solution for the LP relaxation of the subproblem, choose the first one in the natural ordering of the variables to be the branching variable. Let x j be this variable and xj* its value in this solution. Branch from the node for the subproblem to create two new subproblems by adding the respective constraints x j [x j *] and x j [x j *] Bounding: For each new subproblem, obtain its bound by applying the simplex method (or the dual simplex method when reoptimizing) to its LP relaxation and using the value of Z for the resulting optimal solution. 3. Fathoming: For each new subproblem, apply the three fathoming tests given below, and discard those subproblems that are fathomed by any of the tests. Test 1: Its bound Z*, where Z* is the value of Z for the current incumbent Test 2: Its LP relaxation has no feasible solutions Test 3: The optimal solution for its LP relaxation has integer values for the integer restricted variables. (If this solution is better than the incumbent, it becomes the new incumbent and test 1 is reapplied to all unfathomed subproblems with the new larger Z*.) Optimality test: Stop when there are no remaining subproblems: the current incumbent is optimal. Otherwise, perform another iteration. Procedure for Tightening a Constraint Denote the constraint by a 1 x 1 +a 2 x 2 + +a n x n b. 1. Calculate S = sum of the positive a j. 2. Identify any a j 0 such that S < b + a j a. If none, stop; the constraint cannot be tightened further. b. If a j > 0, go to step 3. c. If a j < 0, go to step (a j >0) Calculate a j = S-b and b = S-a j. Reset a j = a j and b = b. Return to step (a j <0) Increase a j to a j = b-s. Return to step 1.