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The Network Simplex Method. Reduced Costs and Node Potentials. 4. 2. (0,4,5). (0,10,2). -3. 1. 3. 5. (0,10,4). (0,5,8). -7. 10. (0,5,10). 4. (0,5,5). (0,4,7). -4. MCNFP Example. (l, u, c). An initial BFS (Solution 1). Basic arcs (variables) B = {(1,3), (2,5), (3,5), (4,5)}

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## The Network Simplex Method

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**The Network Simplex Method**Reduced Costs and Node Potentials**4**2 (0,4,5) (0,10,2) -3 1 3 5 (0,10,4) (0,5,8) -7 10 (0,5,10) 4 (0,5,5) (0,4,7) -4 MCNFP Example (l, u, c)**An initial BFS (Solution 1)**• Basic arcs (variables) B = {(1,3), (2,5), (3,5), (4,5)} • Non-basic arcs at their lower bounds L = {(1,2), (1,4)} • Non-basic arcs at their upper bounds. U = {(3,4)}**Solving for the Basic Arcs**4 L = {(1,2),(1,4)} U = {(3,4)} 2 4 0 -3 10 2 1 3 5 -7 10 5 0 1 Cost = 119 4 -4**Flow on B and U**4 L = {(1,2),(1,4)} U = {(3,4)} 2 4 -3 10 2 1 3 5 -7 10 5 1 Cost = 119 4 -4**Increasing the flow on (1,2)**4 2 4+ 0+ -3 10- 2- 1 3 5 -7 10 5 1 4 -4**Increasing flow on (1,2)**• Consider increasing the flow on (1,2) by • Cost = 119+(c12 + c25 – c13 – c35) =119 + (5+2-4-8) = 119 + (-5) • Maximum possible increase is = 2. • Flow on arc (3,5) will go to zero.**New BFS (Solution 2)**• Basic arcs (variables) B = {(1,2), (1,3), (2,5), (4,5)} • Non-basic arcs at their lower bounds L = {(1,4), (3,5)} • Non-basic arcs at their upper bounds. U = {(3,4)}**Increasing the flow on (1,2) by 2**4 2 4+2 0+2 -3 10-2 2-2 1 3 5 -7 10 5 1 4 -4**New Solution**4 L = {(1,4),(3,5)} U = {(3,4)} 2 6 2 -3 8 1 3 5 -7 10 5 1 4 Cost = 109 -4**Increasing the flow on (1,4)**4 2 6- 2- -3 8 1 3 5 -7 10 5 0+ 1+ 4 -4**Increasing the flow on (1,4)**• Consider increasing the flow on (1,4) by • Cost = 109+(c14 + c45 – c12 – c25) =109 + (7+5-5-2) = 109 + 5 • Increasing the flow on (1,4) will make the solution worse.**Decreasing the flow on (3,4)**4 2 6+ 2+ -3 8- 1 3 5 -7 10 5- 1- 4 -4**Decreasing the flow on (3,4)**• Consider decreasing the flow on (d,4) by • Cost = 109 + (-c13 -c34 - c45 + c12 + c25) =109 + (-4-10-5+5+2) = 109 - 12 • Maximum decrease is = 1. Flow on (4,5) goes to 0.**Reduced Costs: Generic Simplex Method**• The row 0 coefficients of the Simplex Tableau are known as reduced costs. • The reduced cost of a basic variable is zero. • Non-basic variables with negative reduced cost are eligible to enter the basis. • If all the reduced costs are non-negative, then the Simplex method terminates.**Formula to Compute Reduced Costs**• Let B be the set of basic variables. • Let AB be the submatrix of A comprised of the columns corresponding to the basic variables. • Let c’ be the vector of reduced costs (i.e. row 0 of the tableau). • c’ = c – A where = cB(AB)-1. • are referred to as the dual multipliers.**Reduced Costs: Network Simplex**• Let c’ij be the reduced cost of arc (i,j) and i be the potential at node i. • c’ij = cij - i+j where • 1 = 0 • c’ij = 0 for all basic arcs • Eligible non-basic arcs • Arcs in L are eligible if c’ij < 0 • Arcs in U are eligible if c’ij > 0**(cij)**i j Calculating Node Potentials c'12 = c12 - 1 + 2 c'12 = 5 - 0 + 2 0 = 5 + 2 2 = -5 2 2 5 3 = -4 4 1 3 5 5 = -7 1=0 5 4 4 = -2**(c’ij)**i j Calculating Reduced Costs 2 = -5 c'34=10+4-2 c'14=7+0-2 2 0 0 c'35=8+4-7 3 = -4 5 0 1 3 5 5 = -7 1=0 12 5 0 4 4 = -2**(xij)**i j New Solution (BFS 3) 4 L = {(1,4), (3,5),(4,5)} 2 7 3 -3 7 1 3 5 -7 10 4 4 Cost = 97 -4**(cij)**i j Recalculate Node Potentials 2 = -5 2 2 5 3 = -4 4 1 3 5 5 = -7 1=0 10 4 4 = -14**(c’ij)**i j Recalculate Reduced Costs 2 = -5 c'14=7-14 2 c'35=5+4-7 0 0 c'45=5+14-7 3 = -4 2 0 1 3 5 5 = -7 1=0 0 -7 12 4 (1,4) becomes basic. It is in L and c' < 0. 4 = -14**(xij)**i j Increasing the Flow on (1,4) 4 2 7 3 -3 7- 1 3 5 -7 10 4- 0+ 4 Cost = 97 -4**(xij)**i j New Solution (BFS 4) 4 L = {(3,4), (3,5),(4,5)} 2 7 3 -3 3 1 3 5 -7 10 4 4 Cost = 69 -4**( cij)**i j Recalculate Node Potentials 2 = -5 2 2 5 3 = -4 4 1 3 5 5 = -7 1=0 7 4 4 = -7**( c’ij)**i j Recalculate Reduced Costs 2 = -5 c'34 =10+4-7=7 2 c'35 = 8+4-7=5 0 0 c'45 = 5+7-7=5 3 = -4 0 1 3 5 5 = -7 1=0 0 4 4 = -7**Outline of Network Simplex**• Determine a starting BFS. The n-1 basic variables correspond to a spanning tree. • Compute the node potentials. • Compute the reduced costs for the non-basic arcs. • The current solution is optimal if all arcs in L have non-negative reduced cost and all arcs in U have non-positive reduced cost. • Select a non-basic arc (i,j) to enter the basis. • Identify the cycle created by adding (i,j) to the tree. Use flow conservation to determine the new flows on the arcs in the cycle. The arc that first hits its lower or upper limit leaves the basis.

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