Linear Programming: Transportation Problem J. Loucks, St. Edward's University (Austin, TX, USA)

1
Linear Programming Transportation ProblemJ.
Loucks, St. Edward's University (Austin, TX, USA)
Chapter 2B
2
Transportation Problem

• A network model is one which can be represented
  by a set of nodes, a set of arcs, and functions
  (e.g. costs, supplies, demands, etc.) associated
  with the arcs and/or nodes.
• Transportation problem (TP), as well as many
  other problems, are all examples of network
  problems.
• Efficient solution algorithms exist to solve
  network problems.

3
Transportation Problem

• TP can be formulated as linear programs and
  solved by general purpose linear programming
  codes.
• For the TP, if the right-hand side of the linear
  programming formulations are all integers, the
  optimal solution will be in terms of integer
  values for the decision variables.
• However, there are many computer packages, which
  contain separate computer codes for the TP which
  take advantage of its network structure.

4
Transportation Problem

• The TP problem has the following characteristics
• m sources and n destinations
• number of variables is m x n
• number of constraints is m n (constraints are
  for source capacity and destination demand)
• costs appear only in objective function
  (objective is to minimize total cost of shipping)
• coefficients of decision variables in the
  constraints are either 0 or 1

5
Transportation Problem

• The transportation problem seeks to minimize the
  total shipping costs of transporting goods from m
  origins (each with a supply si) to n destinations
  (each with a demand dj), when the unit shipping
  cost from an origin, i, to a destination, j, is
  cij.
• The network representation for a transportation
  problem with two sources and three destinations
  is given on the next slide.

6
Transportation Problem

• Network Representation

1
d1
c11
1
c12
s1
c13
2
d2
c21
c22
2
s2
c23
3
d3
SOURCES
DESTINATIONS
7
The Relationship of TP to LP

• TP is a special case of LP
• How do we formulate TP as an LP?
• Let xij quantity of product shipped from source
  i to destination j
• Let cij per unit shipping cost from source i to
  destination j
• Let si be the row i total supply
• Let dj be the column j total demand
• The LP formulation of the TP problem is

8
The Relationship of TP to LP

• LP Formulation
• The linear programming formulation in terms of
  the amounts shipped from the origins to the
  destinations, xij, can be written as
• Min SScijxij
• i j
• s.t. Sxij lt si for
  each origin i
• j
• Sxij gt dj for
  each destination j
• i
• xij gt 0 for all i
  and j

9
Transportation Problem

• LP Formulation Special Cases
• The following special-case modifications to the
  linear programming formulation can be made
• Minimum shipping guarantees from i to j
• xij gt Lij
• Maximum route capacity from i to j
• xij lt Lij
• Unacceptable routes
• delete the variable (or put a very high cost,
  called the Big-M method, on a selected pair of i
  and j)

10
Transportation Problem

• To solve the transportation problem by its
  special purpose algorithm, it is required that
  the sum of the supplies at the origins equal the
  sum of the demands at the destinations. If the
  total supply is greater than the total demand, a
  dummy destination is added with demand equal to
  the excess supply, and shipping costs from all
  origins are zero. Similarly, if total supply is
  less than total demand, a dummy origin is added.
• When solving a transportation problem by its
  special purpose algorithm, unacceptable shipping
  routes are given a cost of M (a large number).

11
Transportation Problem

• The transportation problem is solved in two
  phases
• Phase I Obtaining an initial feasible solution
• Phase II Moving toward optimality
• In Phase I, the Minimum-Cost Procedure can be
  used to establish an initial basic feasible
  solution without doing numerous iterations of the
  simplex method.
• In Phase II, the Stepping Stone Method, using the
  MODI method for evaluating the reduced costs may
  be used to move from the initial feasible
  solution to the optimal one.

12
Transportation Algorithm

• Phase I - Minimum-Cost Method (simple and
  intuitive)
• Step 1 Select the cell with the least cost.
  Assign to this cell the minimum of its remaining
  row supply or remaining column demand.
• Step 2 Decrease the row and column
  availabilities by this amount and remove from
  consideration all other cells in the row or
  column with zero availability/demand. (If both
  are simultaneously reduced to 0, assign an
  allocation of 0 to any other unoccupied cell in
  the row or column before deleting both.) GO TO
  STEP 1.

13
Transportation Algorithm

• Phase II - Stepping Stone Method
• Step 1 For each unoccupied cell, calculate the
  reduced cost by the MODI method described below.
  Select the unoccupied cell with the most
  negative reduced cost. (For maximization
  problems select the unoccupied cell with the
  largest reduced cost.) If none, STOP.
• Step 2 For this unoccupied cell generate a
  stepping stone path by forming a closed loop with
  this cell and occupied cells by drawing
  connecting alternating horizontal and vertical
  lines between them.
• Determine the minimum allocation where a
  subtraction is to be made along this path.

14
Transportation Algorithm

• Phase II - Stepping Stone Method (continued)
• Step 3 Add this allocation to all cells where
  additions are to be made, and subtract this
  allocation to all cells where subtractions are to
  be made along the stepping stone path. (Note
  An occupied cell on the stepping stone path now
  becomes 0 (unoccupied).
• If more than one cell becomes 0, make only one
  unoccupied make the others occupied with 0's.)
  GO TO STEP 1.

15
Transportation Algorithm

• MODI Method (for obtaining reduced costs)
• Associate a number, ui, with each row and vj
  with each column.
• Step 1 Set u1 0.
• Step 2 Calculate the remaining ui's and vj's by
  solving the relationship cij ui vj for
  occupied cells.
• Step 3 For unoccupied cells (i,j), the reduced
  cost cij - ui - vj.

16
Example 1 TP

• A transportation tableau is given below. Each
  cell represents a shipping route (which is an arc
  on the network and a decision variable in the LP
  formulation), and the unit shipping costs are
  given in an upper right hand box in the cell.

17
Example 1 TP

• Building Brick Company (BBC) has orders for 80
  tons of bricks at three suburban locations as
  follows Northwood 25 tons, Westwood 45
  tons, and Eastwood 10 tons. BBC has two
  plants, each of which can produce 50 tons per
  week.
• How should end of week shipments be made to
  fill the above orders given the following
  delivery cost per ton
• Northwood Westwood
  Eastwood
• Plant 1 24
  30 40
• Plant 2 30 40
  42

18
Example 1 TP

• Initial Transportation Tableau
• Since total supply 100 and total demand 80,
  a dummy destination is created with demand of 20
  and 0 unit costs.

19
Example 1 TP

• Least Cost Starting Procedure
• Iteration 1 Tie for least cost (0), arbitrarily
  select x14. Allocate 20. Reduce s1 by 20 to 30
  and delete the Dummy column.
• Iteration 2 Of the remaining cells the least
  cost is 24 for x11. Allocate 25. Reduce s1 by
  25 to 5 and eliminate the Northwood column.
• Iteration 3 Of the remaining cells the least
  cost is 30 for x12. Allocate 5. Reduce the
  Westwood column to 40 and eliminate the Plant 1
  row.
• Iteration 4 Since there is only one row with
  two cells left, make the final allocations of 40
  and 10 to x22 and x23, respectively.

20
Example 1 TP

• Iteration 1
• MODI Method
• 1. Set u1 0
• 2. Since u1 vj c1j for occupied cells in
  row 1, then
• v1 24, v2 30, v4 0.
• 3. Since ui v2 ci2 for occupied cells in
  column 2, then u2 30 40, hence u2 10.
• 4. Since u2 vj c2j for occupied cells in
  row 2, then
• 10 v3 42, hence v3 32.

21
Example 1 TP

• Iteration 1
• MODI Method (continued)
• Calculate the reduced costs (circled numbers on
  the next slide) by cij - ui vj.
• Unoccupied Cell Reduced
  Cost
• (1,3) 40 - 0 -
  32 8
• (2,1) 30 - 24
  -10 -4
• (2,4) 0 -
  10 - 0 -10

22
Example 1 TP

• Iteration 1 Tableau

23
Example 1 TP

• Iteration 1
• Stepping Stone Method
• The stepping stone path for cell (2,4) is
  (2,4), (1,4), (1,2), (2,2). The allocations in
  the subtraction cells are 20 and 40,
  respectively. The minimum is 20, and hence
  reallocate 20 along this path. Thus for the next
  tableau
• x24 0 20 20 (0 is its current
  allocation)
• x14 20 - 20 0 (blank for the
  next tableau)
• x12 5 20 25
• x22 40 - 20 20
• The other occupied cells remain the
  same.

24
Example 1 TP

• Iteration 2
• MODI Method
• The reduced costs are found by calculating
  the ui's and vj's for this tableau.
• 1. Set u1 0.
• 2. Since u1 vj cij for occupied cells in
  row 1, then
• v1 24, v2 30.
• 3. Since ui v2 ci2 for occupied cells in
  column 2, then u2 30 40, or u2 10.
• 4. Since u2 vj c2j for occupied cells in
  row 2, then
• 10 v3 42 or v3 32 and, 10
  v4 0 or v4 -10.

25
Example 1 TP

• Iteration 2
• MODI Method (continued)
• Calculate the reduced costs (circled numbers on
  the next slide) by cij - ui vj.
• Unoccupied Cell Reduced Cost
• (1,3) 40 - 0 - 32
  8
• (1,4) 0 - 0 -
  (-10) 10
• (2,1) 30 - 10 - 24
  -4

26
Example 1 TP

• Iteration 2 Tableau

27
Example 1 TP

• Iteration 2
• Stepping Stone Method
• The most negative reduced cost is -4
  determined by x21. The stepping stone path for
  this cell is (2,1),(1,1),(1,2),(2,2). The
  allocations in the subtraction cells are 25 and
  20 respectively. Thus the new solution is
  obtained by reallocating 20 on the stepping stone
  path. Thus for the next tableau
• x21 0 20 20 (0 is its
  current allocation)
• x11 25 - 20 5
• x12 25 20 45
• x22 20 - 20 0 (blank for the
  next tableau)
• The other occupied cells remain the
  same.

28
Example 1 TP

• Iteration 3
• MODI Method
• The reduced costs are found by calculating
  the ui's and vj's for this tableau.
• 1. Set u1 0
• 2. Since u1 vj c1j for occupied cells in
  row 1, then
• v1 24 and v2 30.
• 3. Since ui v1 ci1 for occupied cells in
  column 2, then u2 24 30 or u2 6.
• 4. Since u2 vj c2j for occupied cells in
  row 2, then
• 6 v3 42 or v3 36, and 6 v4
  0 or v4 -6.

29
Example 1 TP

• Iteration 3
• MODI Method (continued)
• Calculate the reduced costs (circled numbers on
  the next slide) by cij - ui vj.
• Unoccupied Cell Reduced Cost
• (1,3) 40 - 0 - 36
  4
• (1,4) 0 - 0 -
  (-6) 6
• (2,2) 40 - 6 -
  30 4

30
Example 1 TP

• Iteration 3 Tableau
• Since all the reduced costs are non-negative,
  this is the optimal tableau.

31
Example 1 TP

• Optimal Solution
• From To
  Amount Cost
• Plant 1 Northwood 5 120
• Plant 1 Westwood 45
  1,350
• Plant 2 Northwood 20
  600
• Plant 2 Eastwood 10
  420
• Total Cost 2,490

32
Example 2 TP
Des Moines (100 units capacity)
Cleveland (200 units Req.)
Boston (200 units req.)
Albuquerque (300 units Req.)
Evansville (300 units capacity)
Fort Lauderdale (300 units capacity)
33
Example 2 TP
34
Example 2 TP
35
Example 2 TP

• What is the initial solution?
• Since we want to minimize the cost, it is
  reasonable to be greedy and start at the
  cheapest cell (3/unit) while shipping as much as
  possible. This method of getting the initial
  feasible solution is called the least-cost rule
• Any ties are broken arbitrarily. (Des Moines to
  Cleveland - 3)(Evansville to Cleveland -
  3)Lets start ad cell 1-3 (Des Moines to
  Cleveland)
• In making the flow assignments make sure not to
  exceed the capacities and demands on the margins!

36
Example 2 TP

Start
100
0
100
37
Example 2 TP

100 100
0
200
0
100
38
Example 2 TP

100 200 100
0
0
200
0
0
100
39
Example 2 TP

100 200 100 300
0
0
200
0
0
0
0
100
40
Example 2 TP

• The initial feasible solution via the least-cost
  rule is
• Ship 100 units from Des Moines to Cleveland
• Ship 100 units from Evansville to Cleveland
• Ship 200 units from Evansville to Boston
• Ship 300 units from Fort Lauderdale to
  Albuquerque
• The total cost of this solution is
• 100(3) 100(3) 200(4) 300(9) 4,100
• Is the current solution optimal?
• Perhaps, but we are not sure. Hence, some more
  work is needed.

41
Example 2 TP

• The cells that have a number in them are referred
  to as basic cells (or basic variables)
• In general, all basic variables are strictly
  positive
• When at least one basic variable is equal to
  zero, the corresponding solution is called
  degenerate
• When a solution is degenerate, exercise caution
  in using sensitivity reports

42
Example 2 TP

• How many basic variables (cells) should there be
  in the transportation tableau?
• Answer (RC-1), where R number of rows and C
  number of columns. Here, (RC-1) 33-1 5
• What are they (there are 4 of them)?
• Ship 100 units from Des Moines to Cleveland
• Ship 100 units from Evansville to Cleveland
• Ship 200 units from Evansville to Boston
• Ship 300 units from Fort Lauderdale to
  Albuquerque

43
Example 2 TP

• Is this solution degenerate?
• Answer Yes, because (RC-1) 33-1 5 gt 4
• Hence, we can arbitrarily make (5-4) 1 variable
  which is at level zero basic.
• Let us make Des Moines to Albuquerque a basic
  variable at level zero
• How do we determine if the current solution is
  optimal?
• Let us price currently nonbasic variables
  (cells), which are all at level zero using the
  stepping stone algorithm

44
Example 2 TP
0 100
200 100 300
45
Example 2 TP

• Evansville Albuquerque
• 8 3 3 5 3 gt 0

46
Example 2 TP
0 100
200 100 300
47
Example 2 TP

• Des Moines - Boston
• 4 3 3 - 4 0

48
Example 2 TP
0 100
200 100 300
49
Example 2 TP

• Fort Lauderdale - Boston
• 7 9 5 - 3 3 4 -1 lt 0

50
Example 2 TP
0 100
200 100 300
51
Example 2 TP

• Fort Lauderdale - Cleveland
• 5 29 5 - 3 -2 lt 0

52
Example 1 TP

• The objective function improvement potentials for
  the nonbasic variables are
• Evansville Albuquerque 3 gt 0
• Des Moines - Boston 0 0
• Fort Lauderdale - Boston -1 lt 0
• Fort Lauderdale - Cleveland -2 lt 0
• The largest improvement (here reduction in cost)
  in the objective function can be achieved by
  increasing shipments from Fort Lauderdale to
  Cleveland
• What i2 the limit on the increase?

53
Example 2 TP
0 K 100 -K
200 100 300 -K K

54
Example 2 TP

• If we increase shipments from Fort Lauderdale to
  Cleveland by K, we must reduce shipments in the
  basic cells in the same row and column so that
  the totals on the margins remain the same
• Clearly, because negative quantities cannot be
  shipped, we must impose that
• K gt 0 (no problem)
• 300 K gt 0 (K lt 300)
• 0 K gt 0 (no problem)
• 100 K gt0 (K lt100)
• Hence, let K 100
• The new basis becomes

55
Example 2 TP
100
200 100 200 100
56
Example 2 TP

• Observe that the number of basic variables
  (cells) is equal to (RC-1) 5. Hence, the
  current basic solution is feasible and not
  degenerate
• Also observe that Des Moines Cleveland left the
  basis while Fort Lauderdale Cleveland entered
  the basis (one for one interchange)
• Is the current solution optimal?
• Let us price the non-basic variables (cells)

57
Example 2 TP
100
200 100 200 100
58
Example 2 TP

• Evansville - Albuquerque
• 8 3 5 9 1 gt 0

59
Example 2 TP
100
200 100 200 100
60
Example 2 TP

• Des Moines - Boston
• 4 5 9 5 3 4 2 gt 0

61
Example 2 TP
100
200 100 200 100
62
Example 2 TP

• Fort Lauderdale - Boston
• 7 5 3 - 4 1 gt 0

63
Example 2 TP
100
200 100 200 100
64
Example 2 TP

• Des Moines - Cleveland
• 3 5 9 - 5 2 gt 0

65
Example 2 TP

• The objective function improvement potentials for
  the nonbasic variables are
• Evansville - Albuquerque 1 gt 0
• Des Moines - Boston 2 gt 0
• Fort Lauderdale - Boston 1 gt 0
• Des Moines - Cleveland 2 gt 0
• Since all improvement potentials are positive, it
  is not possible to achieve further improvement
  (here reduction in cost) in the objective
  function
• Hence, the current solution is optimal

66
Example 2 TP

• The final optimal (and not degenerate) solution
  is
• Ship 100 units from Des Moines to Albuquerque
• Ship 200 units from Fort Lauderdale to
  Albuquerque
• Ship 200 units from Evansville to Boston
• Ship 100 units from Evansville to Cleveland
• Ship 100 units from Fort Lauderdale to Cleveland
• The total cost of this solution is
• 100(5) 200(9) 200(4) 100(3) 100(5) 3,900

67
Example 2 TP via LP

• In our example the LP formulation is
• Min z 5 x11 4x12 3x13 8 x21 4x22 3x23
  9 x31 7x32 5x33
• s.t.
• x11 x12 x13 lt 100 (row 1)
• x21 x22 x23 lt 300 (row 2)
• x31 x32 x33 lt 300 (row 3)
• x11 x21 x31 gt 300 (column 1)
• x12 x22 x32 gt 200 (column 2)
• x31 x32 x33 gt 200 (column 3)
• xij gt 0 for i 1, 2, 3 and j 1, 2, 3
  (nonnegativity)

68
Example 2 TP via LP

• The solver formulation and solution is

69
Example 2 TP via LP

• The solver answer report is

70
Example 2 TP via LP

• The solver sensitivity report is

71
Example 2 TP via LP

• The solver limits report is

72
Example 3 Tropicsun (TP via LP)

• Tropicsun is a grower of oranges with locations
  in the cities of Mt. Dora, Eustis and Clermont.
  Tropicsun currently has 275,000 bushels of citrus
  at the grove in Mt. Dora 400,000 bushels in
  Eustis and 300,000 bushels in Clermont. The
  citrus processing plants are located in Ocala
  (capacity of 200,000 bushels), Orlando (600,000
  bushels) and Leesburg (225,000 bushels).
  Tropicsun contracts with a trucking company to
  transport its fruit, which charges a flat rate
  for every mile that each bushel of fruit must be
  transported. Observe that the problem is not
  balanced, since the total of the fruit supplies
  (275,000 400,000 300,000 975,000) is not
  equal to the total of the capacities (200,000
  600,000 225,000 1,025,000). The difference of
  50,000 bushels will be assigned to a dummy
  source, which represents unutilized capacity. The
  distances (in miles) between the groves and
  processing plants are as follows

73
Example 3 Tropicsun (TP via LP)

• Distances (in miles) between groves and plants
• Grove Ocala Orlando Leesburg
• Mt. Dora 21 50 40
• Eustis 35 30 22
• Clermont 55 20 25

74
Example 3 TP via LP Tropicsun
Processing
Plants
Groves
Distances (in miles)
Supply
Capacity
21
Mt. Dora
Ocala
200,000
275,000
1
4
50
40
35
30
Eustis
Orlando
600,000
400,000
2
5
22
55
20
Clermont
Leesburg
225,000
300,000
3
6
25
75
Example 3 Defining the Decision Variables
Xij of bushels shipped from node i to node
j Specifically, the nine decision variables
are X14 of bushels shipped from Mt. Dora
(node 1) to Ocala (node 4) X15 of bushels
shipped from Mt. Dora (node 1) to Orlando (node
5) X16 of bushels shipped from Mt. Dora (node
1) to Leesburg (node 6) X24 of bushels
shipped from Eustis (node 2) to Ocala (node
4) X25 of bushels shipped from Eustis (node
2) to Orlando (node 5) X26 of bushels shipped
from Eustis (node 2) to Leesburg (node 6) X34
of bushels shipped from Clermont (node 3) to
Ocala (node 4) X35 of bushels shipped from
Clermont (node 3) to Orlando (node 5) X36 of
bushels shipped from Clermont (node 3) to
Leesburg (node 6)
76
Example 3 Defining the Objective Function
Minimize the total number of bushel-miles. MIN
21X14 50X15 40X16 35X24 30X25 22X26
55X34 20X35 25X36
77
Example 3 Defining the Constraints

• Capacity constraints
• X14 X24 X34 lt 200,000 Ocala
• X15 X25 X35 lt 600,000 Orlando
• X16 X26 X36 lt 225,000 Leesburg
• Supply constraints
• X14 X15 X16 275,000 Mt. Dora
• X24 X25 X26 400,000 Eustis
• X34 X35 X36 300,000 Clermont
• Nonnegativity conditions
• Xij gt 0 for all i and j

78
Example 3 Solver Formulation
79
Example 3 Solver Solution
80
Example 4 The Sentry Lock Corporation
81
Example 4 Solver Formulation
82
The End of Chapter 2B