Topic – 10 Transportation and Assignment Problems

Document Sample
Topic – 10 Transportation and Assignment Problems Powered By Docstoc
					     Topic – 10
 Transportation and
Assignment Problems
    Transportation Problem
Major Characteristics:
 Several sources (plant) with given supply
  capacity.
 Several destinations (warehouse) with given
  demand.
 Cost Matrix: unit transportation cost from each
  source to each destination.

Assumptions:
 No transshipment between any two
  destinations/two sources.
 All demand/capacity constraints should be
  attempted to meet.
    Transportation Problem
Decisions: How much should be shipped from each
  source to each destination so that total
  transportation cost is minimized?

Why Develop a New Method? (LP can certainly be
  applied)
 LP is still NP-complete to large size problems.
 Computational efficient methods are desirable to
  large size problems.

Transportation Problem Model: Tabular
  Presentation
Transportation Problem
Transportation Cost
Transportation Table
  (Seek its initial solution on board)
LP Formulation:
Let: Xad – Amount to be shipped from D to A

     Xcf – Amount to be shipped from F to C
    (9 Variables)

Objective Function:

Min: Z = 5Xad + 4Xbd + 3Xcd + 8Xae
        + 4Xbe + 3Xce + 9Xaf + 7Xbf + 5Xcf
Sub. To:

   Xad + Xbd + Xcd ≤ 100
   Xae + Xbe + Xce ≤ 300   (Supply Constraints)
   Xaf + Xbf + Xcf ≤ 300

   Xad + Xae + Xaf ≥ 300
   Xbd + Xbe + Xbf ≥ 200   (Demand Constraints)
   Xcd + Xce + Xcf ≥ 200

           (All Xij ≥ 0)
Steps in Solving
      the
Transportation
    Problem
        Solution Procedure
1. Balancing Table: Make ∑di = ∑Si
    – add a "Dummy" source when ∑ di > ∑ Si , or
    – add a "Dummy" destination when ∑ Si > ∑di.
2. Finding an initial solution: (several methods)
    – Upper-Left Corner (Northwest) Method
    – Least-Cost (Large-Profit) Method
    – VAM Method
3. Testing Optimality: (Can current solution be
   improved?)
    – Stepping-stone Procedure
    – Modified-Distribution Method
4. Improving current solution toward a better
   solution (Stepping-Stone method).
        Solution Procedure
Special cases in Transportation Problem:
  – Prohibited transportation route (take large "M"
    as cost)
  – Multiple optimal solution is possible.
  – Transshipment allowed/Route shipment
    capacity limits
      (All special cases can be easily formulated and
        solved by LP.)

Application Examples:
  – Production Planning/Scheduling/Routing
     Selection/....
             Sup. P.10-18: Problem #6
Tabular form: (by northwest method-”left corner”)

        to
              W-1         W-2        W-3       W-4
 from

Mine-1               8          13         9   500

Mine-2               11         14         5   1300

Dummy                0          0          0   600

Demand        1200        500        700       (2400)
         Transportation problem
Tabular form: (by northwest method-”left corner”)

        to
             W-1         W-2        W-3       W-4
 from

Mine-1       (1)    8          13         9   500
             500

Mine-2       (2)    11   (3)   14   (4)   5   1300
             700         500        100

Dummy               0          0    (5)   0   600
                                    600

Demand       1200        500        700       (2400)

TC=8*500+11*700+14*500+5*100+0*600=19,200
         Transportation problem
Tabular form: (by least-cost method)

        to
             W-1         W-2        W-3       W-4
 from

Mine-1       500    8          13         9   500
             700
Mine-2              11         14   600   5   1300
                         500        100
Dummy               0          0          0   600

Demand       1200        500        700       (2400)

TC=8*500+11*700+5*600+0*600=14,700 <19,200
         Problem formulation of LP
   X11-amount to be shipped from M-1to W-1
   X12-amount to be shipped from M-1to W-2
    …    …………………………………. ... …
   X23-amount to be shipped from M-2to W-3
   Min z=  Ci X ij
       =8X11+13X12+9X13+11X21+14X22+5X23
   SUB TO: X11+X12+X13≤500
           X21+14X22+5X23≤1300
           X11+X21≥1200
           X12+X22≥500
           X13+X23≥700
   (Xij≥0, i,j=1…3)
           Transportation Problem
           Maximization Objectives
Problem Description
Klein Chemicals, Inc. manufactures a product at two plants (Clifton
   Springs and Danville) and ships it to four different customers
   (denoted D1, D2, D3 and D4)
Profit per unit for shipping from each plant to each customer.
Profit Per Unit:

                                       Customers
                                       D1     D2     D3    D4
        Plant Clifton                  $32 $34 $32 $40
              Springs
              Danville                 $34 430 $28 $38
Plant Capacities and Customer Orders:

Plant        Capacities   Customers     Orders
             (Units)                    (Units)
Clifton      5000         D1            2000
Danville     3000         D2            5000
                          D3            3000
                          D4            2000
Total        8000         Total         12000
Capacity                  Orders
     The Stepping-Stone Method
1.   Select any unused square to evaluate.
2.   Begin at this square. Trace a closed path back to the
     original; square via square that are currently being
     used (only horizontal or vertical moves allowed).
3.   Place + in unused square; alternate – and + on each
     corner square of the closed path.
4.   Calculate improvement index: add together the unit
     figures found in each square containing a +; subtract
     the unit cost figure in each square containing a -.
5.   Repeat steps 1-4 for each unused square.
Stepping-Stone Method: Tracing a
Closed Path for the Des Moines to
         Cleveland route
  An Initial Solution by Northwest-Corner (VP- Corner Method)

                                Warehouse
  Plant                                                              Capacity
              Miami        Denver        Lincoln       Jackson
                       7             2             4             5
Chicago      70            30                                          100

                       3             1             5             2
Houston                    60            15                            75

                       6             9             7             4
 Buffalo                                 30            50              80

Requirem                                                                   255
                  70            90            45            50
  ent                                                                255
Total cost = 70($7) + 30($2) + 60($1) + 15($5) + 30($7) + 50($4) = $1,095
          Initial Solution by the “Least Cost Method”

                              Warehouse                                              VAM
 Plant                                                              Capacity
            Miami        Denver        Lincoln       Jackson                         Costs
                     7             2             4             5
Chicago    40            15            45                                100          $2

                     3             1             5             2
Houston                  75                                              75           $1

                     6             9             7             4
Buffalo    30                                        50                  80           $2

Require-                                                                       255
                70            90            45            50       255
 ments

 VAM
                $3            $1            $1            $2
 Costs

Total Cost = 40*7+15*2+45*4+75*1+30*6+50*4=945<(1095)
                           Optimal Solution

                                 Warehouse
   Plant                                                          Capacity
              Miami        Denver        Lincoln    Jackson
                       7             2          4             5
  Chicago    +$3           55            45         $3              100

                       3             1          5             2
 Houston     40            35            +$2        +$1             75

                       6             9          7             4
  Buffalo    30            +$5           +$1        50              80

 Requirem                                                               255
                  70            90         45            50
   ent                                                            255

Total Cost = 40($3) + 30($6) + 55($2) + 35($1) + 45($4) + 50($4) = $825
            Assignment Problem: A Special Case
   100=1 track load (can’t be split)


warehouse
             W-1       W-2       W-3       W-4       supply
  plant

  P-1              3         1         2             100
                                                 4


  P-2              5
                             9         2         1   100

  P-3              6         6         3         4   100

  P-4                                  7             100
                   9         4                   2

  demand     100       100       100       100       (400)
        Assignment Problem
Major Characteristics:
   – Finite number of Team-Job (or the like) to be assigned.
   – Assignment must be made based on one-to-one basis.
   – Payoff matrix: cost (profit) for each possible assignment.

Available Solution Techniques to Assignment Problem:
   – Linear Programming
   – Dynamic Programming
   – Integer IP (Branch-and-Bound)
   – Complete Enumeration
   – Transportation Model (special case with all Si = di = 1)
   – Hungrain Method (A "smart" method for large size
      problems)
      Assignment Problem
Special Cases in Assignment Problem:
  – Prohibited Assignment (assign a large
    "M" as cost)
  – It is possible to have multiple optimal
    solutions.
  – Special arrangement for a fixed
    assignment or One-to-More assignment.
 Special Solution Procedure
Hungrain Method: (Tabular Form)
  – Principle: Add (or subtract) a constant to all
    cells of a row (or column) in the table will not
    change final optimal solution.
  – Four-Step Procedure
  – For Max Problem: (Transforming into Min
    Problem)
      Change the sign ("+" or "-") of each cell in
       the table.
      Add the largest cell value to all cells.
     (Examples)
 Special Solution Procedure
Summary:
  – Both Transportation Problem and Assignment
    Problem are special cases of LP problem, and
    can be solved by Simplex method.
  – Tabular solution technique mainly provides a
    more efficient way for some large size practical
    problems.
  – Many practical managerial problems can be
    formulated and solved by Transportation or
    Assignment problem techniques.
    The Assignment Method Algorithm
1. Subtract the smallest number in each row from every number in that row.
    Subtract the smallest number in each column from every umber in that
    column.

2. Draw the minimum number of vertical and horizontal straight lines
    necessary to cover zeros in the table. If the number of lines equals the
    number of rows or columns, then one can make an optimal assignment
    (Step-4).

3. If the number of lines does not equal the number of rows or columns:
   Subtract the smallest number not covered by a line from every other
    uncovered number.
   Add the same number to any number lying at the intersection of any two
    lines.
    Return to step 2

4. Make optimal assignments at locations of zeros within the table.


     Service costs of different team
      assignments ($ in thousand)
Service
          zone   Z1   Z2      Z3
team


S1               20   15      31


S2               17   16      33


S3               18   19      27
     Service costs of different team
      assignments ($ in thousand)
Service
          zone   Z1       Z2       Z3
team


S1               20   5   15   0   31   16


S2               17   1   16   0   33   17

S3               18   0   19   1   27
                                        9
    Service costs of different team
     assignments ($ in thousand)
5     0     16      5     0     7


1     0     17      1     0     8


0     1     9       0     1     0
      Problem 7- Supplement, p.10-18

      mach   (M)    (N)    (O)     (P)
job           1      2      3       4
      A       6      7      5       9

      B      8      5       6      7

      C      10     8       6      6

dummy        0      0       0      0
         Problem 7- Supplement
   LP formulation
   Let x11,x22,….x34(12 variables)=job A assigned to
    machine 1
   Min: Z=6X11+7X12+5X13+9X14+…+6X34
   Sub to: X11+X12+X13+X14=1
             X21+X22+X23+X24=1
             X31+X32+X33+X34=1
             X11+X21+X31=1
             X12+X22+X32=1
             X13+X23+X33=1
             X13+X23+X33=1       all Xij=1 or 0
Written Assignment-7 is due
 on next –
Sup. P.10-22: Northeastern blood bank Case


 A) identify demand/supply     Total cost
                                If buy all
 Type: A: (6+2)=8(Pints)       8*30=240
                                6*36=216
       B: (2+4)=6(Pints)
                                5*35=175
       A: (2+3)=5(Pints)       5*40=200
                                      $831
       A: (3+2)=5(Pints)
Case I: Northeastern blood bank

 Supply:
A   :      4 (pints)
B   :      10 (pints)
O    :     11 (pints)
 AB :      7 (pints)
  Case I: Northeastern blood bank

   Identify cost matrix:
   1) allowable blood transfusion pattern
   (supply)                             B           $1000
                          A
                                  While A
                   A                            O
                  $30     AB                         $1000
(Assign a large case to prevent
                                                A    $1000
Unallowable transfusion) B
                   B              While     B
                           AB                   O    $1000
                 $36
                                                A    $1000
                                          AB     B   $1000
                  AB       AB
                                                O
                                                     $1000
         A        B        O        AB    dummy       supply

A            30   1000 1000          30           0     4

B        1000         36   1000      36           0     10

O            35       35       35    35           0     11

AB       1000     1000 1000          40           0     7

demand       8        6        5     5      (8)        (32)

                      24

				
DOCUMENT INFO
Shared By:
Stats:
views:22
posted:7/18/2010
language:English
pages:45