1. Transportation Problems
Chapter 3
Transportation Problems

2. Design of Transport Networks
Transportation Problem: Model & LP
Situation:
Location of warehouses and customers are given
Supply and demand given
Example:
3 warehouses and 4 customers
Transportation cost per unit from i to j
Total demand must be = total supply
Factory or
customer
warehouse V1 V2 V3 V4
Production F1 10 5 6 11 25 F2 2 7 4 F3 9 1 8 50
Demand 15 20 30 35
100
(c) Prof. Richard F. Hartl
QEM - Chapter 3

3. Solution in the Uncapacitated case
No capacity constraint  solution using column minimum procedure
customer
Factory V1 V2 V3 V4
Capacity used F1 10 5 6 11 F2 1 2 7 4 F3 9 8
Demand 15 20 30 35 15 35
= 50 20 30
= 50
Total cost =
15 * * * * 4
= 295
(c) Prof. Richard F. Hartl
QEM - Chapter 3

4. (c) Prof. Richard F. Hartl
Capacity constraints
Not all customers will be delivered from their closest factory!
Solution as a special LP problem - starting (basic feasible) solution - iteration (modi, stepping stone)
(c) Prof. Richard F. Hartl
QEM - Chapter 3

5. (c) Prof. Richard F. Hartl
Capacity constraints
Total capacity = total demand
Customer
Factory V1 V2 V3 V4
capacity F1 10 5 6 11 25 F2 1 2 7 4 F3 9 8 50
Demand 15 20 30 35
100
100
100
(c) Prof. Richard F. Hartl
QEM - Chapter 3

6. Constructive Heuristics (Basic Feasible Solution)
Simple to apply, Simple to understand
„Reasonably“ good solutions
Optimality not guaranteed
Examples:
Column minimum („Greedy“)
Vogel approximation („Regret“)
(c) Prof. Richard F. Hartl
QEM - Chapter 3

7. Column Minimum Algorithm
0. Initialization: empty transport tableau. No rows or columns are deleted
1. Proceed with the first (not yet deleted) column (from left to right)
2. In choose the cell with the smallest unit cost cij and choose transportation quantity xij maximal.
3. If column resource depleted  delete column j,
OR if row resource depleted  delete row i.
Just one row or columd not deleted  all cells in this row or colum get maximal transportation quantity Otherwise continue with 1.
EXAMPLE
(c) Prof. Richard F. Hartl
QEM - Chapter 3

8. Column Minimum Algorithm
i / j 1 2 3 4
Capacity 10 5 6 11 25 7 9 8 50
Demand 15 20 30 35
100
Just one colums not deleted 25 10 15 10 30 20 30
Total cost =
= 470
15 * * * * * * 8
ALGORITHM
(c) Prof. Richard F. Hartl
QEM - Chapter 3

9. Vogel Approximation
0. Initialization: empty transport tableau. No rows or columns are deleted
In each row and column (not yet deleted) compute opportunity cost = difference between the smallest and the second smallest cij (not yet deleted).
Where this opportunity cost is largest choose smallest cij and make xij maximal.
If column resource depleted  delete column j,
OR if row resource depleted  delete row i.
Just one row or columd not deleted  all cells in this row or colum get maximal transportation quantity Otherwise continue with 1.
SS 2011
EK Produktion & Logistik

10. EK Produktion & Logistik
Vogel Approximation
i / j 1 2 3 4
capacity 10 5 6 11 25 7 9 8 50
Nachfrage 15 20 30 35
100
Column deleted  recompute row differences 25 1 5 15 10 10 1 2 20 5 25 30 3 4 5 25 8 1 4 2 4 3
Row deleted  recompute column differences
Jut one row not deleted
Total cost =
15 * * * * * * 8
= 445
(besser als 470 zuvor)
SS 2011
EK Produktion & Logistik

11. General formulation: LP-Formulation:
m supplyers with supply si, i = 1, …, m
n customers with demand dj, j = 1, …, n
Transportation cost cij per unit from i to j, i = 1, …, m; j = 1, …, n
LP-Formulation:
variable: Transportation quantity xij from i to j
Transportation cost
i = 1, …, m
Supply
j = 1, …, n
Demand
Non negativity
i = 1, …, m; j = 1, …, n
Data must satisfy
Total demand = Total supply
(c) Prof. Richard F. Hartl
QEM - Chapter 3

12. In example:
K = (10x11+5x12+6x13+11x14) + (x21+2x22+7x23+4x24) + (9x31+x32+4x33+8x34)  min
Supply:
x11 + x12 + x13 + x = 25 (i=1)
x21 + x22 + x23 + x24 = 25 (i=2)
x31 + x3 2+ x33 + x34 = 50 (i=3) :
Damand:
x x x = 15 (j=1)
x x x = 20 (j=2)
x x x = 30 (j=3)
x x x34 = 35 (j=4)
Non negativity:
xij  0 für i = 1, … , 3; j = 1, … , 4
(c) Prof. Richard F. Hartl
QEM - Chapter 3

13. Solution: As LP (o.k. but less efficient) 1 . . . 1 . . .
Make use of special structure:
. . . 1 .
In each column exactly 2 of the m + n elements are ≠ 0
Transportation simplex or network simplex
Starting solution
Iteration (stepping stone)
(c) Prof. Richard F. Hartl
QEM - Chapter 3

14. Exakt Algorithm: MODI, Stepping Stone
Equivalent to simplex method
Start with basic feasible solution (m+n-1 basic variables)
Iteration step:
Initalize the tableau
i\j 1 2 … n si ui
c11
c12
c1n s1 u1
c21
c22
c2n s2 u2 m
cm1
cm2
cmn sm um dj d1 d2 dn vj v1 v2 vn
(c) Prof. Richard F. Hartl
QEM - Chapter 3

15. cij = ui + vj if xij is basic variable
Compute dual varables ui und vj [MODI]
cij = ui + vj if xij is basic variable
Since ui and vj are not unique, normalize one of these dual varables = 0. (Choose row or colums with most basic varaibles  easier computation of ui and vj)
For all non basic variables compute cij – ui – vj (reduced cost coefficient in objective function). New (entering) BV where this coefficient is most negative. If all coefficients non-negative  optimal solution.
Increase new BV and perform chain reaction (donor cell, recipient cell). Transportation quantities in each row and column must remain same. The BV, which first becomes zero, leaves the basis. [stepping stone]
Compute new basic solution (perform chain reaction).
(c) Prof. Richard F. Hartl
QEM - Chapter 3

16. Example
For didactical raesons we choose poor starting solution, because after Vogel only few steps (if any) can be demonstrated
i\j 1 2 3 4 si ui 10 5 6 11 25 7 9 8 50 dj 15 20 30 35 vj -3 -4 15 10 35 -7 -6 -3 2 5 -6 10 5 10 14
(c) Prof. Richard F. Hartl
QEM - Chapter 3

17. Total cost = 10*15 + 5*10 + 2*10 + 7*15 + 4*15 + 8*35 = 665.
Check equality of primal and dual objective (optional):
K = 25*0 + 25*(-3) + 50*(-6) + 15* *5 + 30* *14 = 665
The most negative cost ceofficient cij – ui – vj of a NBV is -7 for x24  new basic variable x24.
Chain rection: increase new BV (starting from 0) by  investigate consequences for other BV (in some row and some column  must be subtracted) The BV, that constrains  most is the leaving BV.
(c) Prof. Richard F. Hartl
QEM - Chapter 3

18. New BV x24 increases from 0 to . Add +  odor -  for some other BV.
Chain reaction:
New BV x24 increases from 0 to . Add +  odor -  for some other BV.
If x24 increases by , x23 and x34 will decrease by , and x33 increases by . For  = 15 BV x23 becomes 0 → BV x23 leaves.
i\j 1 2 3 4 si ui 10 5 6 11 25 7 9 8 50 dj 15 20 30 35 vj
K = 665 – 7 * 
= 665 – 7*15 = 560 -3 -4 -7 -6 2 5 10 14 15 10
15-
+ 
15+ 
35- 
(c) Prof. Richard F. Hartl
QEM - Chapter 3

19. -3  1 New BV x24 gets value  = 15 → chain reaction x34 = 35-15 = 20
x23 is no BV anymore, all other BV remain same
Next iteration:
i\j 1 2 3 4 si ui 10 5 6 11 25 7 9 8 50 dj 15 20 30 35 vj
K = 560 – 6 * 
= 560 – 6*10 = 500 4 3 7 -6 -5 -2
15- 
10- 
10+  -3  30 20 15 1 10 5 3 7
(c) Prof. Richard F. Hartl
QEM - Chapter 3

20. Next iteration: 5-   -9 10+  -5 i\j 1 2 3 4 si ui 10 5 6 11 25 7 98 50 dj 15 20 30 35 vj
K = 500 – 3 * 
= 500 – 3*5 = 485 -2 -3 7 6 1 4
5- 
30-  20  -9
20+ 
10+ 
15-  -5 10 5 9 13
(c) Prof. Richard F. Hartl
QEM - Chapter 3

21. Next iteration: -6 15 -2  i\j 1 2 3 4 si ui 10 5 6 11 25 7 9 8 50 dj20 30 35 vj
K = 485 – 2 * 
= 485 – 2*20 = 445 -2 1 3 7 4
20-
5+  -6 15 10 -2 
25-  25 7 5 6 10
(c) Prof. Richard F. Hartl
QEM - Chapter 3

22. All coefficients non-negative → optimal solution found.
Next iteration:
i\j 1 2 3 4 si ui 10 5 6 11 25 7 9 8 50 dj 15 20 30 35 vj
All coefficients non-negative → optimal solution found. 2 1 3 7 5 4 2 25
Basic variables:
x13 = 25
x21 = 15
x24 = 10
x32 = 20
x33 = 5
x34 = 25 -4 15 10 20 5 25 5 1 4 8
Total cost: K = 445
(c) Prof. Richard F. Hartl
QEM - Chapter 3

23. si  si +  for some i and dj  dj +  for some j
Sensitivity analysis
TP is a LP-Problem with equality constraints  dual variables ui azw. vj can also be negative (free variables).
From duality theory follows
si  si +  for some i and dj  dj +  for some j
If  small enough, then the dual variables ui and vj do not change  cost change by (ui + vj):
K  K + (ui + vj)
Clearly some si AND soem dj must change at the same time; Otherwise total demand ≠ total capacity
(c) Prof. Richard F. Hartl
QEM - Chapter 3

24. Example: Change of data? s1  s1 +  and d2  d2 + 
i\j 1 2 3 4 si ui 5 22 6 24 7 16 dj 10 13 17 vj
22 + 
13 + 
Change of data?
s1  s1 +  and
d2  d2 +  10 12 2 13 11 1 10 6 1 -1 2 4
Optimal cost change to K = (u1 + v2) = ; i.e. cost decrease! (this paradoxon can occur if some ui and/or vj are negative. In most cases cost will increase.)
(c) Prof. Richard F. Hartl
QEM - Chapter 3

25. How large can  be before basis changes: similar to stepping stone
i\j 1 2 3 4 si ui 5 22
+  6 24 7 16 dj 10
13 +  17 vj
K = 
Check: K = 1*10+2*(12+ ) + 1*(13+ ) +6*(11- ) +3*(10- ) +5*(6 + ) =  10
12+
13+ 2
10-
11- 1
6+ 1 -1 2 4
Clearly x33 first becomes 0, if  increases. Hence upper bound  10. Similar for negative . Here x34 first becomes 0, if  decreases. Hence lower bound  -6.
(c) Prof. Richard F. Hartl
QEM - Chapter 3