Network Simplex: Part 2 Page 1
The Network Simplex Method
Reduced Costs and Node Potentials
Network Simplex: Part 2 Page 2
MCNFP Example
51
4
2
3
(0,4,5) (0,10,2)
(0,5,5)(0,4,7)
(0,10,4) (0,5,8)
(0,5,10)10
4
-4
-3
-7
(l, u, c)
Network Simplex: Part 2 Page 3
An initial BFS (Solution 1)
• Basic arcs (variables)B = {(1,3), (2,5), (3,5), (4,5)}
• Non-basic arcs at their lower boundsL = {(1,2), (1,4)}
• Non-basic arcs at their upper bounds.U = {(3,4)}
Network Simplex: Part 2 Page 4
Solving for the Basic Arcs
51
4
2
3
L = {(1,2),(1,4)}
10
4
-4
-3
-7
0
05
U = {(3,4)}
10 2
1
4
Cost = 119
Network Simplex: Part 2 Page 5
Flow on B and U
51
4
2
3
L = {(1,2),(1,4)}
10
4
-4
-3
-7
5
U = {(3,4)}
10 2
1
4
Cost = 119
Network Simplex: Part 2 Page 6
Increasing the flow on (1,2)
51
4
2
310
4
-4
-3
-7
0+
5
10- 2-
1
4+
Network Simplex: Part 2 Page 7
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.
Network Simplex: Part 2 Page 8
New BFS (Solution 2)
• Basic arcs (variables)B = {(1,2), (1,3), (2,5), (4,5)}
• Non-basic arcs at their lower boundsL = {(1,4), (3,5)}
• Non-basic arcs at their upper bounds.U = {(3,4)}
Network Simplex: Part 2 Page 9
Increasing the flow on (1,2) by 2
51
4
2
310
4
-4
-3
-7
0+2
5
10-2 2-2
1
4+2
Network Simplex: Part 2 Page 10
New Solution
51
4
2
310
4
-4
-3
-7
2
5
8
1
6
Cost = 109
L = {(1,4),(3,5)} U = {(3,4)}
Network Simplex: Part 2 Page 11
Increasing the flow on (1,4)
51
4
2
310
4
-4
-3
-7
2-
0+5
8
1+
6-
Network Simplex: Part 2 Page 12
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.
Network Simplex: Part 2 Page 13
Decreasing the flow on (3,4)
51
4
2
310
4
-4
-3
-7
2+
5-
8-
1-
6+
Network Simplex: Part 2 Page 14
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.
Network Simplex: Part 2 Page 15
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.
Network Simplex: Part 2 Page 16
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.
Network Simplex: Part 2 Page 17
Formula in the Network Context
11010001010100011001000010010000111
54321
45353425141312
A
ccccccccAcc
Network Simplex: Part 2 Page 18
Formula in the Network Context
544531132112
54321
45353425141312
,,,'
11010001010100011001000010010000111
cccc
A
ccccccccAcc
Network Simplex: Part 2 Page 19
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 where1. 1 = 02. 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
Network Simplex: Part 2 Page 20
c'12 = c12 - 1 + 2 c'12 = 5 - 0 + 2
0 = 5 + 2
Calculating Node Potentials
51
4
2
31=0
5
4
5
2
2 = -5
5 = -7
4 = -2
3 = -4
(cij)i j
Network Simplex: Part 2 Page 21
Calculating Reduced Costs
51
4
2
31=0
0
0
0
0
2 = -5
5 = -7
4 = -2
3 = -4
c'34=10+4-2
12
c'14=7+0-2
5
c'35=8+4-7
5
(c’ij)i j
Network Simplex: Part 2 Page 22
New Solution (BFS 3)
51
4
2
310
4
-4
-3
-7
3
4
7
7
Cost = 97
L = {(1,4), (3,5),(4,5)}(xij)i j
Network Simplex: Part 2 Page 23
Recalculate Node Potentials
51
4
2
31=0
5
4
2
2 = -5
5 = -7
4 = -14
3 = -4
10
(cij)i j
Network Simplex: Part 2 Page 24
Recalculate Reduced Costs
51
4
2
31=0
0
0
0
2 = -5
5 = -7
4 = -14
3 = -4
0
c'14=7-14
-7
c'35=5+4-7
2
c'45=5+14-7
12(1,4) becomes basic.It is in L and c' < 0.
(c’ij)i j
Network Simplex: Part 2 Page 25
Increasing the Flow on (1,4)
51
4
2
310
4
-4
-3
-7
3
0+4-
7-
7
Cost = 97
(xij)i j
Network Simplex: Part 2 Page 26
New Solution (BFS 4)
51
4
2
310
4
-4
-3
-7
3
4
3
7
Cost = 69
L = {(3,4), (3,5),(4,5)} (xij)i j
Network Simplex: Part 2 Page 27
Recalculate Node Potentials
51
4
2
31=0
5
4
2
2 = -5
5 = -7
4 = -7
3 = -4
7
( cij)i j
Network Simplex: Part 2 Page 28
Recalculate Reduced Costs
51
4
2
31=0
0
0
0
2 = -5
5 = -7
4 = -7
3 = -4
0
c'34 =10+4-7=7 c'35 = 8+4-7=5 c'45 = 5+7-7=5
( c’ij)i j
Network Simplex: Part 2 Page 29
Outline of Network Simplex
1. Determine a starting BFS. The n-1 basic variables correspond to a spanning tree.
2. Compute the node potentials.3. 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.
4. Select a non-basic arc (i,j) to enter the basis.5. 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.