MSci 331- Integer Programming by zXQ66G

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									MSci 331- Transportation Problem
                      Selvaprabu Nadarajah
               Department of Management Sciences,
                      University of Waterloo




                       G.B. Dantzig
       “most practical planning problems could be
     reformulated as a system of linear inequalities.”
    “ground rules for selecting good plans by general
                   objective functions.”
              Expectations

• What skills should you expect to learn?

  – Understand the practical relevance of the
    transportation problem

  – Formulate a LP model for the problem

  – Solve the resulting model
Transportation Problem

     a1
                      b1
     a2               b2

 Supplies        Demand
 Available       requirements



                      bn
     a3
   Snow Shovelling

   S1
                     D1

   S2
                     D2


Snow affected   Snow dumping sites
areas



                     D3

   S3
         Distribution

    P1
                             C1

    P2
                             C2


Plants                  Customers




                             C3

    P3
              LP Formulation
• Decision Variables
  – Number of units shipped from i warehouse to
    customer j


• Objective
  – Minimize total transportation cost


• Constraints
  – Supply is not exceeded
  – Demand is satisfied
                                    LP contd...

Let X ij be the number of units shipped from plant i
  to customer j

 Min           C
              i   j
                        ij   X ij

  s.t.   X
          j
                  ij    Si           i      (1)

         X
          i
                  ij    Dj           j      (2)

         X ij  0                     i, j   (3)
                  Assumptions
• Demand and Supply
  Each supplier (source) has a fixed supply that must be
  distributed and each customer (sink) has a demand that
  must be satisfied.

• Feasibility Property
  A feasible solution exists only if total supply equals
  demand (i.e. the problem is “Balanced”)

• Objective/Cost
  The cost is the unit distribution cost times the number of
  units shipped (i.e. linear function)
           Balanced Problem

• A problem is balanced when total demand
  equals total supply

                     S  D
                      i
                          i
                              j
                                  j




• In real-life problems supply is usually not equal
  to demand. Solution?
                  Balancing
• If total supply exceeds demand, then add a
  dummy/fictitious demand point, with its demand
  equal to the surplus supply

• If total demand exceeds supply, then add a
  dummy/fictitious supply point, with its supply
  equal to the surplus demand.

• Cost associated with dummy points is zero
                Balancing Example
                         Excess Demand


                Customer 1   Customer 2   Customer 3   Supply
Plant 1            $75           $60          $69          50
Plant 2            $79           $73          $68          100
Plant 3            $85           $76          $70          50
Dummy plant 4       $0           $0           $0           70
Demand              80           90           100          270
         Transportation Tableau
                v1               v2                v3         Supply
                                                                50
    u1   x11               x12               x13
                     75               60                 69
                                                               100
   u2    x21               x22               x23
                     79               73                 68
                                                                50
   u3    x31               x32               x33
                     85               76                 70
                                                                70
   u4    x41               x42               x43
                     0                   0               0
Demand          80               90                100

               xij    Flow from i to j

               ui     Dual value for supply i

               vj     Dual value for supply j
             Transportation Simplex
1. If the problem is unbalanced, balance it. Setup the
   transportation tableau

2. Find a basic feasible solution

3. Set vlast  0 and determine u i ’s and v ’s such that
                                            j
   ui  v j  cij for all basic variables.


4. If the reduced cost cij  ui  v j  0 for all non-basic
   variables (minimization problem), then the current BFS
   is optimal. Stop! Else, enter variable with most negative
   reduced cost

5. Using the new BFS, repeat steps 2 and 3
  Step 2: Northwest corner rule

                                              v3        Supply
               v1             v2
                                                          50
    u1   50
                    75             60              69
   u2                    70                              100     70
         30
                    79             73              68

   u3                    20             30                50     30
                    85             76              70

   u4                                   70                70
                    0              0               0
Demand        80              90             100
              30              20             70
         Northwest Solution

                                             v3        Supply
              v1             v2
                                                         50
    u1   50
                   75             60              69
   u2                                                   100
         30             70
                   79             73              68

                                       30                50
   u3                   20
                   85             76              70

   u4                                  70                70
                   0              0               0
Demand        80             90             100
                    Question 1

Assume there are m supply points and n demand
points, how many allocations (non-zero entries)
will be present in a basic feasible solution (BFS)?

   – Is it m + n?

   – Correct answer: m + n – 1, since the problem is
     balanced
                   Question 2

How can units take non-integer values?

  –   If the demand and supply are integer, the solution to
      the transportation problem is always integer feasible

  –   why? The constraint matrix is Unimodular
     Transportation Simplex Roadmap
1. If the problem is unbalanced, balance it. Setup the
   transportation tableau

2. Find a basic feasible solution : Northwest Corner Rule

3. Set vlast  0 and determine u i ’s and v ’s such that
                                            j
   ui  v j  cij for all basic variables.


4. If the reduced cost cij  ui  v j  0 for all non-basic
   variables, then the current BFS is optimal. Else enter
   variable with most negative reduced cost

5. Using the new BFS, repeat steps 2 and 3
       Step 3: Finding Dual Values
• We know that the number of equations (m +n - 1) is
  one less than the number of dual variables (m + n)

•       Set a variable to zero,vlast  0 and solve to determine
    u i ’s and v ’s such that ui  v j  cij for all basic
                 j

        variables
                     Example

           v1  12        v2  6             v3  0     Supply

                                                          50
u1  63   50
                    75             60              69
                                                         100
u 2  67 30              70
                    79             73              68

                                        30                50
u3  70                  20
                    85             76              70

                                        70                70
u4  0
                    0              0               0
 Demand        80             90             100
     Transportation Simplex Roadmap
1. If the problem is unbalanced, balance it. Setup the
   transportation tableau

2. Find a basic feasible solution : Northwest Corner Rule

3. Set vlast  0 and determine u i ’s and v ’s such that
                                            j
   ui  v j  cij for all basic variables.


4. If the reduced cost cij  ui  v j  0 for all non-basic
   variables, then the current BFS is optimal. Else enter
   variable with most negative reduced cost

5. Using the new BFS, repeat steps 2 and 3
     Step 4: Entering Variable

1. Compute the reduced cost of non-basic
   variables cij  cij  (ui  v j )

2. Enter variable with the most negative
   reduced cost
            v1  12            v2  6          v3  0     Supply


                          -9              6                 50
u1  63    50
                     75              60              69

                          70              1                100
u 2  67   30
                     79              73              68

u3  70 3                 20              30                50

                     85              76              70

u 4  0 -12               -6              70                70

                     0               0               0
 Demand         80              90             100
   Pivoting by finding a loop: Rules
Definition: An ordered sequence of at least four
 different cells is called a loop if:
  1. It starts at the entering-variable cell

  2. Any two consecutive cells lie in either the same row or
     the same column.

  3. No three consecutive cells lie in the same row or column.

  4. The last cell in the sequence has a row or a column in
     common with the first cell in the sequence.

  5. Except for the entering-variable cell all corners of the
     loops must coincide with a basic variable
                     Finding a loop
            v1  12            v2  6          v3  0     Supply


                          -9              6                 50
u1  63    50
                     75              60              69

u 2  67   30        -   70         +   1                100

                     79              73              68

u3  70 3                 20         -   30      +        50

                     85              76              70

u 4  0 -12          +   -6              70         -     70

                      0              0               0
 Demand         80              90             100

                       = 20
                        New BFS
                                                        Supply
              v1             v2              v3
                                                          50
   u1    50
                   75              60              69

  u2     10             90                               100

                   79              73              68
                                                          50
  u3                                    50
                   85              76              70
  u4     20                             50                70

                   0               0               0
Demand        80              90             100
     Transportation Simplex Roadmap
1. If the problem is unbalanced, balance it. Setup the
   transportation tableau

2. Find a basic feasible solution : Northwest Corner Rule

3. Set vlast  0 and determine u i ’s and v ’s such that
                                            j
   ui  v j  cij for all basic variables.


4. If the reduced cost cij  ui  v j  0 for all non-basic
   variables, then the current BFS is optimal. Else enter
   variable with most negative reduced cost

5. Using the new BFS, repeat steps 2 and 3
                          Iteration 2
             v1  0           v2  6          v3  0      Supply


                          -9              -6                 50
u1  75    50
                     75              60               69

u 2  79   10        -   90              -11     +        100

                     79              73               68

u3  70 15                12              50                 50

                     85              76               70

u4  0     20        +   6               50          -     70

                      0              0                0
 Demand         80              90              100

                       = 10
                         Iteration 3
              v1  0          v2  5          v3  0     Supply


u1  75   50        -   -20        +   -6                50

                    75              60              69

u 2  68 11              90         -   10      +       100

                    79              73              68

u3  70 15               6               50                50

                    85              76              70

u4  0    30        +   -5              40         -     70

                     0               0              0
 Demand        80              90             100

                      = 40
                           Iteration 4
             v1  20           v2  5          v3  0      Supply


                                          14                 50
u1  55     10             40
                      75             60               69

 u 2  68 -9               50             50                100

                      79             73               68

u3  70 -5                 1              50                 50

                      85             76               70

u 4  20   70             15             -20                70

                      0              0                0
  Demand         80             90              100
                            Iteration 5
              v1  11           v2  5          v3  0     Supply

                                                             50
  u1  55 9                 50             14
                       75             60              69

  u 2  68   10             40             50               100

                       79             73              68

  u3  70 4                 1              50                50

                       85             76              70

 u 4  11   70             15             11                70

                       0              0               0
   Demand         80             90             100

All reduced costs are positive. Optimal solution
reached. Stop!!
Assignment Problem

   M1
                J1

   M2
                J2


Machines
                Jobs




                J3

   M3
Assignment and Transportation
• Assignment problem is a special case of the
  transportation problem

• Each demand and supply equals 1

    Min     C i   j
                             ij   X ij

    s.t.
           X
            j
                        ij   1          i      (1)

           X
            i
                        ij   1          j      (2)

           X ij  0                      i, j   (3)

								
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