# MSci 331- Integer Programming by zXQ66G

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```									MSci 331- Transportation Problem
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
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
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
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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