# Transportation and Network Flow Problems by wulinqing

VIEWS: 24 PAGES: 32

• pg 1
```									    Transportation
and
Network Flow
Problems
From Linear & Nonlinear Programming ch.5
Presented by Wei Shiou
Information Management Dept.
National Taiwan University
2002.11.7
Introduction
   Part A
 The Transportation Problem
 The Assignment Problem

   Part B
 Minimum Cost Flow
 Maximal Flow

 There are m origins that contain
various amounts of a commodity
that must be shipped to n
destinations to meet demand
requirements
 It is assumed that the system is
balanced    m       n

 ai   bj
i 1   j 1
For Example
   A specific transportation problem
with four origins and five
destinations is defined by
a    30,80,10,60
b    10,50,20,80,20 
3   4 6 8 9
2   2 4 5 5
C             
2   2 2 3 2
           
3   3 2 4 2
Standard Form
x11  x12  ....x1n                                                                   a1
x 21  x 22  ....x 2 n                                      a2
..........
..........                 ..........
..........        ..........
..........
..........        ..........
..........                 ..
..........
x m1  x m 2  ....x mn  a m

x11      x 21        x m1      b1
x12                     x 21                               x m1               b2
..........
..........                 ..........
..........        ..........
..........
..........        ..........
..........                 ..
..........
x1n                     x2n                                x mn  bn

 We can let xij = aibj / S
for i=1 to m and j=1 to n then it is
clearly a feasible solution
 相當於將supply依照比例分配給每個
destination

   Theorem.
   A transportation problem always has a
solution , but there is exactly one
redundant equality constraint . When any
one of the equality constraints is dropped ,
the remaining system of n+m-1 equality
constraints is linearly independent .
   完整證明於 page.120
Simplex Method的應用
   Step 1
用西北角法找出 initial basic feasible solution
   Step 2
計算 simplex multipliers and the relative cost
coefficients , if all relative cost coefficients are
nonnegative then stop. otherwise, go to
Step.3
   Step 3
選擇一個nonbasic variable corresponding to
enter the basis。Compute the cycle of change
and update the solution. Go to Step.2
   請參照page.130
The Northwest Corner Rule
   Solution array：

10   20                  30
30   20   30        80
10        10
40   20   60
10   50   20   80   20
The Northwest Corner Rule
   Step 1
   Step 2
Allocate the maximum feasible amount
consistent with row and column sum
requirements involving that cell
   Step 3
Move one cell to the right if there is any
remaining row requirement. Otherwise move
one cell down till the end.

   A nonsingular square matrix M is said to
be triangular if by a permutation of its
rows and columns it can be put in the
form of a lower triangular matrix
   判斷 M 是否 triangular 的過程叫做 back
substitution, 可參照page.123
Basis Triangularity
   Basis Triangularity Theorem:
Every basis of the transportation
problem is triangular. 完整證明於
page.124
   Corollary:
If the row and column sums of a
transportation problem are integers,
then the basic variables in any basic
solution are integers.
Simplex Multipliers
   We partition the vector of
multipliers as   (u, v)
Where ui represents the multiplier
associated with the ith row sum
constraint
and vj represents the multiplier
associated with the jth column sum
constraint
Simplex Multipliers
   In transportation problem, the
relative cost coefficients are
rij  cij  ui  v j   for i  1,2,....,m
j  1,2,....,n
   任意給定一個multiplier值，可算出所有其他值，
由此計算rij，找出下一個enter variable
Cycle of Change
   Corollary:
If the unit costs cij of a transportation problem
are all integers, then the simplex multipliers
associated with iany mbasis are integers
 1,2,....,

   Theorem:
Let B be a basis from A (ignoring one row), and
let d be another column. Then the
components of the vector y=B-1d are either
0,+1,-1
Cycle of Change
   From the Theorem mentioned before, we
know that if a new enter variable has a
value S, then S is equal to the value
subtracted from the old basic variable
   We will set S equal to the smallest
magnitude of these basic variables in the
stepping-stone path
Example
   如page.131

3    4    6   8   9      5   10 20                30
2    2    4   5   5      3      30 20 30          80
2    2    2   3   2      1            10          10
3    3    2   4   2      2          + 40 20       60
-2   -1   1   2   0          10 50 20 80 20
Coefficient array        Transportation tableau
Degeneracy
   In a transportation tableau with m
rows and n columns, there must be
m+n-1 cells with allocations;
otherwise, it is degenerate.
   必須引入特殊符號來代替basic variable的
「零」，否則可能找不到stepping-stone path
The Assignment Problem
   問題描述:
Assigning n workers to n jobs, there is a
benefit(or cost) for each assignment. Each
worker must be assigned to exactly one job
and each job must have one assigned worker
   General formulation:
n            n

Minimize   c x
j 1    i 1
ij   ij

n
Subject to  x  1 for i  1 to
ij                   n
j 1
n

x
i 1
ij    1 for j  1 to n

xij  0 for i  1 to n , j  1 to n
Minimum Cost Flow

 Consider a network having n nodes.
Corresponding to each node I, there is a number
bi representing the available supply at the
node(using bi<0 to represent a required
demand)
 To determine flows xij>=0 in each arc of the
network so that the net flow into each node I is
bi while minimizing the total cost
 Still assume that the net work is balanced
Problem Structure (MCF)
n         n

Minimize       c
j 1        i 1
ij   x ij

n              n

Subject to     x x
j 1
ij
k 1
ki    bi for i  1 to n

xij  0 for i  1 to n , j  1 to n
   The transportation problem is a special
case of this problem!
   The more general problem is often
termed the transshipment problem
Simplex Method (MCF)
   Step 1
   Step 2
Compute simplex multipliers and relative cost
coefficients. If all rij are nonnegative, stop;
Otherwise, go Step 3
   Step 3
Select a nonbasic flow with negative relative
cost coefficient to enter the basis. And move
out the flow from the cycle, go Step 2
Maximal Flow
   問題描述: (以下簡稱MF)
To determine the maximal flow possible
from one given source node to a sink
node under arc capacity constraints.
   Reachable
自行參考page.141或Ford-Fulkerson’s Algorithm
Capacitated Networks
   A capacitated network is a network in
which some arcs are assigned
nonnegative capacities
   The capacities of an arc(I,j) is denoted kij
and this capacity is indicated on the
graph by placing the number kij adjacent
to the arc
General Form (MF)
maxinize            f
n                    n
subject to   x
j 1
1j     x j1  f  0
j 1
n                  n

xj 1
ij     x ji
j 1
0     i  1, m

n                    n

xj 1
mj      x jm  f  0
j 1

i, j           0  xij  k ij
Min Cut Theorem
 In a network the maximal flow
between a source and a sink is equal
to the minimal cut capacity of all
cuts separating the source and sink
 The capacity of the cut is the sum
of the capacities of the arcs in the
cut
 證明於page.147
Primal-Dual Algorithm
n             n

Minimize     c
j 1            i 1
ij       x ij

n                  n

Subject to x x
j 1
ij
k 1
ki        bi for i  1 to n

xij  0 for i  1 to n , j  1 to n

m                       n

Maximize     u a  v b
i    i                     j    j
i 1                    j 1

Subject to   ui + vj <= cij
Primal-Dual Algorithm
m                 n
minimize          yi 1
i   zj
j 1
n
subject to      x
j1
ij        y i  ai   i  1 to m

m

x
i 1
ij     zi  b j       j  1 to n

for all i , j   xij  0 y i  0 z j  0
for i , j  S xij  0
Primal-Dual Algorithm (MF)
m         n

   Since  y i        z in any solution
j
i 1         j 1

   Alternatively, the problem can be
cast as a maximal flow problem. The
objective can be rewritten as
m          n       n
       m

    ai   xij     b j   xij 

i 1       j 1   j 1       i 1  
Dual Feasible Solution
   With Primal-Dual Algorithm , it is
unnecessary to construct the network
explicitly!
Hungarian’s Method
This is The End
Thank You All

```
To top