Linear Programming: Chapter 14
Network Flows: Applications
Robert J. Vanderbei
October 22, 2007
Operations Research and Financial Engineering
Princeton University
Princeton, NJ 08544
http://www.princeton.edu/∼rvdb
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Linear Programming Network Flows Applications - Essay - Mathematics, Essays (high school) of Mathematics
Equal number of supply and demand nodes. Every supply node has a supply of one. Every demand node has a demand for one. Each supply node is connected to every demand node (called a complete bipartite graph). Solution is required to be all integers.
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Linear Programming: Chapter 14
Network Flows: Applications
Robert J. Vanderbei
October 22, 2007
Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 http://www.princeton.edu/∼rvdb
Transportation Problem
Each node is one of two types:
- source (supply) node
- destination (demand) node Every arc has:
- its tail at a supply node
- its head at a demand node Such a graph is called bipartite.
Tree Solution
Leaving arc: (a,b) Entering arc: (i,h) Etc., etc., etc.
Assignment Problem
Transportation problem in which
- Equal number of supply and demand nodes.
- Every supply node has a supply of one.
- Every demand node has a demand for one.
- Each supply node is connected to every demand node (called a complete bipartite graph).
- Solution is required to be all integers.
Notes:
- These problems are very common.
- They are notoriously degenerate ( 2 n constraints but only n nonzero flows).
Network Flow Formulation
- Put bi =
1 i 6 = r −(m − 1) i = r
- Solve min-cost network flow problem.
- Shortest path from i to r: follow tree arcs.
- Length (of time) of shortest path = y r∗ − y i∗.
Notation Used in Following Algorithms
- Put vi = min. time from i to r
- Called label in networks literature.
- Called value in dynamic programming literature.
Label Correcting Algorithm
Dynamic Programming
- Bellman’s Equation = Principle of Dynamic Programming
vr = 0 (1) vi = min{cij + vj : (i, j) ∈ A} (2) T = {(i, j) ∈ A : vi = cij + vj } – not necessarily a tree (3)
- Method of Successive Approximation
- Initialize: v i(0) =
0 i = r ∞ i 6 = r
- Iterate: v i( k+1) =
0 i = r min{cij + v( jk ) : (i, j) ∈ A} i 6 = r
- Stop: when a pass leaves vi’s unchanged.
Label Setting Algorithm
Dijkstra’s Algorithm
Notations:
- F = set of finished nodes (labels are set).
- hi, i ∈ N = next node to visit after i (heading).
Dijkstra’s Algorithm:
- Initialize: F = ∅, vj =
0 j = r ∞ j 6 = r
- Iterate:
- While unfinished nodes remain, select the one with smallest vk. Call it j. Add it to set of finished nodes F.
- For each unfinished node i having an arc connecting it to j: ∗ If cij + vj < vi, then set
vi = cij + vj (4) hi = j (5)
Dijkstra’s Algorithm—Complexity
- Each iteration finishes one node: m iterations
- Work per iteration:
- Selecting an unfinished node: ∗ Naively, m compar- isons. ∗ Using appropriate data structures, a heap, log m compar- isons.
- Update adjacent arcs.
- Overall: m log m + n.