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Transportation, Assignment, and Transshipment Problems. Chapter 7. Transportation, Assignment, and Transshipment Problems.

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Transportation assignment and transshipment problems

Transportation, Assignment,and Transshipment Problems

Chapter 7

MNGT 379 Operations Research


Transportation assignment and transshipment problems1
Transportation, Assignment, and Transshipment Problems

  • 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, assignment, and transshipment problems of this chapter as well as the PERT/CPM problems (in another chapter) are all examples of network problems.

  • Each of the three models of this chapter can be formulated as linear programs and solved by general purpose linear programming codes.

  • For each of the three models, 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 (including The Management Scientist) that contain separate computer codes for these models which take advantage of their network structure.

MNGT 379 Operations Research


Transportation problem
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 below.

1

d1

c11

1

c12

s1

c13

2

d2

c21

c22

2

s2

c23

3

d3

Sources

Destinations

MNGT 379 Operations Research


Transportation problem1
Transportation Problem

  • LP Formulation

    The LP formulation in terms of the amounts shipped from the origins to the destinations, xij , can be written as:

    Min cijxij

    i j

    s.t. xij<si for each origin i

    j

    xij = dj for each destination j

    i

    xij> 0 for all i and j

  • Special Cases

    The following special-case modifications to the linear programming formulation can be made:

    • Minimum shipping guarantee from i to j:

      xij>Lij

    • Maximum route capacity from i to j:

      xij<Lij

    • Unacceptable route:

      Remove the corresponding decision variable.

MNGT 379 Operations Research


Example acme block co
Example: Acme Block Co.

  • Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, and Eastwood -- 10 tons. Acme has two plants, each of which can produce 50 tons per week. Delivery cost per ton from each plant to each suburban location is shown below. How should end of week shipments be made to fill the above orders?

    NorthwoodWestwoodEastwood

    Plant 1 24 30 40

    Plant 2 30 40 42

    MIN 24x11+30x12+40x13+30x21+40x22+42x23

    S.T.

    1) 1x11+1x12+1x13<50

    2) 1x21+1x22+1x23<50

    3) 1x11+1x21=25

    4) 1x12+1x22=45

    5) 1x13+1x23=10

MNGT 379 Operations Research


Example acme block co1
Example: Acme Block Co.

Objective Function Value = 2490.000

Variable Value Reduced Costs

-------------- --------------- ------------------

x11 5.000 0.000

x12 45.000 0.000

x13 0.000 4.000

x21 20.000 0.000

x22 0.000 4.000

x23 10.000 0.000

Constraint Slack/Surplus Dual Prices

-------------- --------------- ------------------

1 0.000 6.000

2 20.000 0.000

3 0.000 -30.000

4 0.000 -36.000

5 0.000 -42.000

OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper Limit

------------ --------------- --------------- ---------------

x11 20.000 24.000 28.000

x12 No Lower Limit 30.000 34.000

x13 36.000 40.000 No Upper Limit

x21 26.000 30.000 34.000

x22 36.000 40.000 No Upper Limit

x23 No Lower Limit 42.000 46.000

RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit

------------ --------------- --------------- ---------------

1 45.000 50.000 70.000

2 30.000 50.000 No Upper Limit

3 5.000 25.000 45.000

4 25.000 45.000 50.000

5 0.000 10.000 30.000

MNGT 379 Operations Research


Assignment problem
Assignment Problem

  • An assignment problem seeks to minimize the total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is cij.

  • It assumes all workers are assigned and each job is performed.

  • An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs.

  • The network representation of an assignment problem with three workers and three jobs is shown below.

c11

1

1

c12

c13

Tasks

c21

c22

Agents

2

2

c23

c31

c32

3

3

c33

MNGT 379 Operations Research


Assignment problem1
Assignment Problem

  • LP Formulation

    Min cijxij

    i j

    s.t. xij = 1 for each agent i

    j

    xij = 1 for each task j

    i

    xij = 0 or 1 for all i and j

    • Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.

  • Special Cases

    • Number of agents exceeds the number of tasks:

      xij< 1 for each agent i

      j

    • Number of tasks exceeds the number of agents:

      Add enough dummy agents to equalize the number of agents and the number of tasks. The objective function coefficients for these new variable would be zero.

MNGT 379 Operations Research


Assignment problem2
Assignment Problem

  • Special Cases (continued)

    • The assignment alternatives are evaluated in terms of revenue or profit:

      • Solve as a maximization problem.

    • An assignment is unacceptable:

      • Remove the corresponding decision variable.

    • An agent is permitted to work a tasks:

      xij<a for each agent i

      j

MNGT 379 Operations Research


Example who does what
Example: Who Does What?

  • An electrical contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects.

    Projects

    SubcontractorABC

    Westside 50 36 16

    Federated 28 30 18

    Goliath 35 32 20

    Universal 25 25 14

    How should the contractors be assigned to minimize total mileage costs?

MNGT 379 Operations Research


Example who does what1
Example: Who Does What?

Network Representation

50

West.

A

36

16

Subcontractors

Projects

28

30

Fed.

B

18

32

35

Gol.

C

20

25

25

Univ.

14

MNGT 379 Operations Research


Example who does what2
Example: Who Does What?

  • Linear Programming Formulation

    Min 50x11+36x12+16x13+28x21+30x22+18x23

    +35x31+32x32+20x33+25x41+25x42+14x43

    s.t. x11+x12+x13 < 1

    x21+x22+x23 < 1

    x31+x32+x33 < 1

    x41+x42+x43 < 1

    x11+x21+x31+x41 = 1

    x12+x22+x32+x42 = 1

    x13+x23+x33+x43 = 1

    xij = 0 or 1 for all i and j

  • The optimal assignment is:

    SubcontractorProjectDistance

    Westside C 16

    Federated A 28

    Goliath (unassigned)

    Universal B 25

    Total Distance = 69 miles

MNGT 379 Operations Research


Transshipment problem
Transshipment Problem

  • Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes) before reaching a particular destination node.

  • Transshipment problems can be converted to larger transportation problems and solved by a special transportation program.

  • Transshipment problems can also be solved by general purpose linear programming codes.

  • The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown below.

c36

3

c13

c37

1

6

s1

d1

c14

c46

c15

4

Demand

c47

Supply

c23

c56

c24

7

2

d2

s2

c25

5

c57

Destinations

Sources

Intermediate Nodes

MNGT 379 Operations Research


Transshipment problem1
Transshipment Problem

  • Linear Programming Formulation

    xij represents the shipment from node i to node j

    Min cijxij

    i j

    s.t. xij<si for each origin i

    j

    xik - xkj= 0 for each intermediate

    ij node k

    xij = dj for each destination j

    i

    xij> 0 for all i and j

MNGT 379 Operations Research


Example zeron shelving
Example: Zeron Shelving

The Northside and Southside facilities of Zeron Industries supply three firms (Zrox, Hewes, Rockrite) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc.

Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockrite. Both Arnold and Supershelf can supply at most 75 units to its customers.

Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are:

Zeron NZeron S

Arnold 5 8

Supershelf 7 4

The costs to install the shelving at the various locations are:

ZroxHewesRockrite

Thomas 1 5 8

Washburn 3 4 4

MNGT 379 Operations Research


Example zeron shelving1
Example: Zeron Shelving

  • Network Representation

Zrox

50

1

5

Zeron

N

Arnold

75

5

8

8

Hewes

60

3

7

Super

Shelf

Zeron

S

4

75

4

4

Rock-

Rite

40

MNGT 379 Operations Research


Example zeron shelving2
Example: Zeron Shelving

  • Linear Programming Formulation

    • Decision Variables Defined

      xij = amount shipped from manufacturer i to supplier j

      xjk = amount shipped from supplier j to customer k

      where i = 1 (Arnold), 2 (Supershelf)

      j = 3 (Zeron N), 4 (Zeron S)

      k = 5 (Zrox), 6 (Hewes), 7 (Rockrite)

    • Objective Function Defined

      Minimize Overall Shipping Costs:

      Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37

      + 3x45 + 4x46 + 4x47

  • Constraints Defined

    Amount Out of Arnold: x13 + x14< 75

    Amount Out of Supershelf: x23 + x24< 75

    Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0

    Amount Through Zeron S: x14 + x24 - x45 - x46 - x47 = 0

    Amount Into Zrox: x35 + x45 = 50

    Amount Into Hewes: x36 + x46 = 60

    Amount Into Rockrite: x37 + x47 = 40

    Non-negativity of Variables: xij> 0, for all i and j.

MNGT 379 Operations Research


Example zeron shelving3
Example: Zeron Shelving

Objective Function Value = 1150.000

Variable Value Reduced Costs

X13 75.000 0.000

X14 0.000 2.000

X23 0.000 4.000

X24 75.000 0.000

X35 50.000 0.000

X36 25.000 0.000

X37 0.000 3.000

X45 0.000 3.000

X46 35.000 0.000

X47 40.000 0.000

ConstraintSlack/SurplusDual Prices

1 0.000 0.000

2 0.000 2.000

3 0.000 -5.000

4 0.000 -6.000

5 0.000 -6.000

6 0.000 -10.000

7 0.000 -10.000

MNGT 379 Operations Research


Example zeron shelving4
Example: Zeron Shelving

Optimal Solution

Zrox

50

50

75

1

5

Zeron

N

Arnold

75

5

25

8

8

Hewes

60

35

3

4

7

Super

Shelf

Zeron

S

40

75

4

4

75

Rock-

Rite

40

MNGT 379 Operations Research


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