Chapter 3
Transportation Problems
QEM - Chapter 3
Design of Transport Networks
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 2 7 4 25
F3 9 1 4 8 50
Demand 15 20 30 35 100
Transportation Problem: Model & LP
(c) Prof. Richard F. Hartl Kapitel 3 / 2
(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 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 1 4 8
Demand 15 20 30 35
15
20 30
35 15 + 35 = 50
20 + 30 = 50
Total cost = 15 * 1 + 20 * 1 + 30 * 4 + 35 * 4 = 295
(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/4
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 3Kapitel 3/5
Capacity constraints
Total capacity = total demand
Customer
Factory V1 V2 V3 V4 capacity
F1 10 5 6 11 25
F2 1 2 7 4 25
F3 9 1 4 8 50
Demand 15 20 30 35 100
100
100
(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/6
Constructive Heuristics (Basic Feasible Solution)
Heuristics: 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 3Kapitel 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,
ORif row resource depleted delete row i.
4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantityOtherwise continue with 1.
EXAMPLE
(c) Prof. Richard F. Hartl QEM - Chapter 3 Kapitel 3/8
Column Minimum Algorithm
i / j 1 2 3 4 Capacity
1 10 5 6 11 25
2 1 2 7 4 25
3 9 1 4 8 50
Demand 15 20 30 35 100
15
20 30 0
10
25
30
10
0
0
0
Total cost = 15 * 1 + 20 * 1 + 30 * 4 + 25 * 11 + 10 * 4 + 0 * 8 = 470
Just one colums not deleted
ALGORITHM
SS 2011 EK Produktion & LogistikKapitel 3/9
Vogel Approximation
0. Initialization: empty transport tableau. No rows or columns are deleted
1. In each row and column (not yet deleted) compute opportunity cost = difference between the smallest and the second smallest cij (not yet deleted).
2. Where this opportunity cost is largest choose smallest cij and make xij maximal.
3. If column resource depleted delete column j,
ORif row resource depleted delete row i.
4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantityOtherwise continue with 1.
SS 2011 EK Produktion & Logistik Kapitel 3/10
Vogel Approximation
i / j 1 2 3 4 capacity
1 10 5 6 11 25
2 1 2 7 4 25
3 9 1 4 8 50
Nachfrage 15 20 30 35 100
15
20 5 25
10
25
8 21
3
1
1
4
10 2
25
4 3
30 4
5
5
Total cost = 15 * 1 + 20 * 1 + 25 * 6 + 5 * 4 + 10 * 4 + 25 * 8
= 445
Column deleted recompute row differences
Row deleted recompute column differences
Jut one row not deleted
(besser als 470 zuvor)
QEM - Chapter 3
General 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
variable: Transportation quantity xij from i to j
LP-Formulation:
Transportation cost
Supply
Demand
Non negativity
min1 1
m
iij
n
jijxcK
n
jiji xs
1i = 1, …, m
m
iijj xd
1j = 1, …, n
0ijx i = 1, …, m; j = 1, …, n
Mdn
jj
1
m
1=iisData must satisfy Total demand = Total supply
(c) Prof. Richard F. Hartl Kapitel 3 / 11
QEM - Chapter 3
In example:
Supply: x11 + x12 + x13 + x14 = 25 (i=1)
x21 + x22 + x23 + x24 = 25 (i=2)
x31 + x3 2+ x33 + x34 = 50 (i=3) :
Damand: x11 + x21 + x31 = 15
(j=1) x12 + x22 + x32 = 20
(j=2) x13 + x23 + x33 = 30 (j=3)
x14 + x24 + x34 = 35 (j=4) Non negativity:
xij 0 für i = 1, … , 3; j = 1, … , 4
K = (10x11+5x12+6x13+11x14) + (x21+2x22+7x23+4x24) + (9x31+x32+4x33+8x34) min
(c) Prof. Richard F. Hartl Kapitel 3 / 12
QEM - Chapter 3
Solution: As LP (o.k. but less efficient)
In each column exactly 2 of the m + n elements are ≠ 0
Make use of special structure:
1 . . . 1
1 . . . 1
. . .
1 . . . 1
1 ..
.
1
1 ..
.
1
. . . 1 ..
.
1
Transportation simplex or network simplex
Starting solution
Iteration (stepping stone)
(c) Prof. Richard F. Hartl Kapitel 3 / 13
QEM - Chapter 3
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
1c11 c12
…c1n
s1 u1
2c21 c22
…c2n
s2 u2
… … … … … …
mcm1 cm2
…cmn
sm um
dj d1 d2 … dn
vj v1 v2 … vn
(c) Prof. Richard F. Hartl Kapitel 3 / 14
QEM - Chapter 3
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.
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)
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 Kapitel 3 / 15
QEM - Chapter 3
Example
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
15 10
10 15
15 35
-3-4
-7-6
25
10 5 10 14
-3
-6
0
For didactical raesons we choose poor starting solution, because after Vogel only few steps (if any) can be demonstrated
(c) Prof. Richard F. Hartl 16
QEM - Chapter 3
Total cost = 10*15 + 5*10 + 2*10 + 7*15 + 4*15 + 8*35 = 665.
Check equality of primal and dual objective (optional):
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.
j
n
jj
m
iii
m
iij
n
jij dvsuxcK
111 1
K = 25*0 + 25*(-3) + 50*(-6) + 15*10 + 20*5 + 30*10 + 35*14 = 665
(c) Prof. Richard F. Hartl Kapitel 3 / 17
QEM - Chapter 3
Chain reaction:
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
15+
15-
35-
15 10
10
-3-4
-7-6
25
10 5 10 14
-3
-6
0
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.
+
K = 665 – 7 *
= 665 – 7*15 = 560
(c) Prof. Richard F. Hartl Kapitel 3 / 18
QEM - Chapter 3
New BV x24 gets value = 15 → chain reaction x34 = 35-15 = 20
x33 = 15+15 = 30
x23 is no BV anymore, all other BV remain same
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
43
7-6
-5-2
10+ 15-
10-
10 5 3 7
-3
1
0
30 20
15
Next iteration:
K = 560 – 6 *
= 560 – 6*10 = 500
(c) Prof. Richard F. Hartl Kapitel 3 / 19
QEM - Chapter 3
Next iteration:
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
-2-3
76
14
20
10 5 9 13
-9
-5
05-
30-
15-
20+
10+
K = 500 – 3 *
= 500 – 3*5 = 485
(c) Prof. Richard F. Hartl Kapitel 3 / 20
QEM - Chapter 3
Next iteration:
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
-2
13
73
4
20- 5+
7 5 6 10
-6
-2
0
25-
10
25
15
K = 485 – 2 *
= 485 – 2*20 = 445
(c) Prof. Richard F. Hartl Kapitel 3 / 21
QEM - Chapter 3
Next iteration:
i\j 1 2 3 4 si ui
110 5 6 11
25
21 2 7 4
25
39 1 4 8
50
dj 15 20 30 35
vj
2 13
75
4
25
5 1 4 8
-4
0
2
20 5
10
25
15
All coefficients non-negative → optimal solution found.
Basic variables: x13 = 25
x21 = 15
x24 = 10
x32 = 20
x33 = 5
x34 = 25
Total cost: K = 445
(c) Prof. Richard F. Hartl Kapitel 3 / 22
QEM - Chapter 3
TP is a LP-Problem with equality constraints dual variables ui azw. vj can also be negative (free variables).
Sensitivity analysis
If small enough, then the dual variables ui and vj do not change cost change by (ui + vj):
From duality theory follows
Clearly some si AND soem dj must change at the same time;Otherwise total demand ≠ total capacity
si si + for some i anddj dj + for some j
K K + (ui + vj)
(c) Prof. Richard F. Hartl Kapitel 3 / 23
QEM - Chapter 3
Example:
i\j 1 2 3 4 si ui
11 4 2 5
22
26 1 5 6
24
37 5 3 5
16
dj 10 13 22 17
vj
10
1 -1 2 4
2
1
0
10 6
11
12
13
Change of data?
s1 s1 + and
d2 d2 +
Optimal cost change to K = 173 + (u1 + v2) = 173 - ; i.e. cost decrease! (this paradoxon can occur if some ui and/or vj are negative. In most cases cost will increase.)
22 +
13 +
(c) Prof. Richard F. Hartl Kapitel 3 / 24
QEM - Chapter 3
How large can be before basis changes: similar to stepping stone
i\j 1 2 3 4 si ui
11 4 2 5 22
+
26 1 5 6
24
37 5 3 5
16
dj 10 13 + 22 17
vj
10
1 -1 2 4
2
1
0
6+10-
11-
12+
13+
K = 173 -
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.
Check: K = 1*10+2*(12+ ) + 1*(13+ ) +6*(11- ) +3*(10- ) +5*(6 + ) = 173 -
(c) Prof. Richard F. Hartl Kapitel 3 / 25
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