Two Discrete
Optimization Problems
Problem:
The Transportation
Problem
Formulating Graph Problems
You already know the following steps for formulating a graph
problem (such as the TSP and the Shortest-Path Problem):
1. Identify the vertices that represent objects in a problem.
2. Identify the edges that are lines connecting selected pairs of
vertices to indicated a relationship between the objects
associated with the two connected vertices.
3. Identify additional data by writing those values next to the
corresponding vertices and/or edges.
4. State the objective in the context of the graph and given
data.
Let’s do this for a new problem…
2
The Transportation Problem (TP)
CCC has 1000 computers at each of three plants this
month. Three customers have requested 1100, 800, and
1100 computers, respectively. These data are summarized
in the table below, along with the cost of shipping one
computer from each plant to each customer. You have been
asked to develop a least-cost shipping plan for CCC.
Plant 1
Customer 1 Customer 2 Customer 3 Supply
5
3
2
1000
Plant 2
4
7
8
1000
Plant 3
6
7
4
1000
Demand
1100
800
1100
3
Creating a Graph for the TP
Step 1: Identify the Vertices. Use one vertex for each plant
and one for each customer:
Plants
1
1
Note: An edge is
a relationship
and not a number.
2
2
Customers
3
3
Step 2: Identify the Edges. Use an edge to connect a
vertex i to a vertex j to represent the possibility of shipping
computers from Plant i to Customer j.
4
Creating a Graph for the TP
Step 3: Identify Additional Data.
(Supplies)
Plants
Customers (Demands)
5
1000
1
1
1100
2
800
3
1100
3
2
4
1000
2
7
8
6
7
1000
3
4
(Unit Shipping Costs)
5
Step 4: State the Objective
You can use words, for example, for the TSP:
Find the order in which to visit every vertex exactly once
and return to the starting vertex with least total “cost”.
OR use variables, objective function, and constraints:
Step 4(a): Identify the variables, which are those
quantities whose values, once determined, constitute
the solution to the problem.
To identify the variables, ask yourself the following questions:
1. What can you choose or control?
2. What decisions do you have to make?
3. What items affect costs or profits?
4. If you had to implement the solution, what
information would you need to know?
6
Identifying Variables for CCC
A = the number of computers to ship from P1 to C1.
OR
X1 = the number of computers to ship from P1 to C1.
X2 = the number of computers to ship from P1 to C2.
Question: What does X5 mean for this problem?
Note: Choose notation that is easy to understand.
Xij = the number of
computers to ship from
Plant i to Customer j
(i, j = 1, 2, 3)
Cust 1
Cust 2
Cust 3
Plant 1
X11
X12
X13
Plant 2
X21
X22
X23
Plant 3
X31
X32
X33
7
Identifying the Obj. Func. For CCC
Step 4(b): Identify the Objective Function (a math expression
in terms of the variables and data that reflect the goal).
Words: Minimize total transportation cost
Decompose: Total transportation cost =
Key Point: You can
 Trans. cost   Trans. cost   Trans. cost  use decomposition




 from Plant 1  from Plant 2   from Plant 3  more than once.
Decompose (again): Transportation cost from Plant 1 =
 Trans. cost   Trans. cost   Trans. cost 



  5X11 + 3X12 + 2X13
 to Cust. 1   to Cust. 2   to Cust. 3 
Math: M in 5 X 11  3 X 12  2 X 13  4 X 21  7 X 22  8 X 23  6 X 31  7 X 32  4 X 33
Plant 1
Plant 2
Plant 3
8
Identifying Constraints for CCC
Step 4(c): Identify Constraints (restrictions on the values
of the variables so that those values are acceptable).
Note: Use grouping to identify groups of constraints.
Demand Constraints ( 3 )
Words: Total number of computers received by
Customer 1 should be equal to the number requested.
Decompose: Total number of computers received by Customer 1 =
 Computers   Computers   Computers 




 from Plant 1  from Plant 2   from Plant 3 
Mathematics:
X 11  X 21  X 31  1100 (Demand for Customer 1)
X 12  X 22  X 32  800 (Demand for Customer 2)
X 13  X 23  X 33  1100 (Demand for Customer 3)
9
Identifying Constraints for CCC
Step 4(c): Identify Constraints (cont.)
Supply Constraints ( 3 )
Words: Total number of computers shipped from Plant 1
should be equal to the available supply.
Decompose: Total number of computers shipped from Plant 1 =
 Computers   Computers   Computers 




 to Cust. 1   to Cust. 2   to Cust. 3 
Mathematics:
X 11  X 12  X 13  1000 (Supply at Plant 1)
X 21  X 22  X 23  1000 (Supply at Plant 2)
X 31  X 32  X 33  1000 (Supply at Plant 3)
Logical Constraints: All X ij  0 and integer
10
Transportation Problem of CCC
Find values for the variables Xij (i = 1, 2, 3; j = 1, 2, 3) so as to
M in 5 X 11  3 X 12  2 X 13  4 X 21  7 X 22  8 X 23  6 X 31  7 X 32  4 X 33
s. t. Demand Constraints
X 11  X 21  X 31  1100 (Demand for Customer 1)
X 12  X 22  X 32  800 (Demand for Customer 2)
X 13  X 23  X 33  1100 (Demand for Customer 3)
Supply Constraints 3000
Qn: Is this a
X 11  X 12  X 13  1000 (Supply at Plant 1)
X 21  X 22  X 23  1000 (Supply at Plant 2)
X 31  X 32  X 33  1000 (Supply at Plant 3)
Logical Constraints 3000
Balanced
All X ij  0 and integer
TP
linear program?
NO…it is an
integer program.
11
Handling Too Much Demand
The algorithm for solving the transportation problem
requires that total supply = total demand.
Question: What do you do if total supply < total demand?
(Supplies) Plants
1000
Customers (Demands)
5
1
1
1100
2
800
3
2
4
1000
2
7
Means that some
customers will not
receive all of their
demand.
8
6
1000
500
3000
3500
3
7
4
3
1600
0 0
0
Dum
3500
Note: Shipping one
unit from the Dummy
Plant to Customer j
means that Customer j
will not receive one
unit of their demand.
12
Handling Too Much Supply
The algorithm for solving the transportation problem
requires that total supply = total demand.
Question: What do you do if total supply > total demand?
(Supplies) Plants
1000
Customers (Demands)
5
1
1
1100
2
800
3
2
4
1000
2
7
Means that some
plants will not ship
all of their
supplies.
8
6
1500
3
7
3
4
0
0
3500
1100
0
Dum
500
3000
3500
Note: Shipping one
unit from Plant i to the
Dummy Customer
means that Plant i will
have one unit of supply
not shipped.
13
Review of Simplex Alg. for LP
Step 0. Initialization: Find an initial extreme point, x.
Step 1. Test for Optimality: Perform a
relatively easy computation to see if the
current extreme point x is optimal. If so, stop.
x2
Step 2. Move: Use the fact that the current 150
extreme point x is not optimal to:
x’ = x + t*d
(a) DOM: Find a direction of movement
100
d pointing from x along an edge of the
feasible region that leads to improvement of the objective function.
50
(b) AOM: Determine the amount t* to
move from x in the direction d to
d
reach the next extreme point.
(c) Move: Move from x to x’ = x +
Set x = x’ and return to Step 1.
t*d.
This is a Finite
Improvement
Algorithm (FIA).
x
t*
x1
50
100
14
Finding an Initial Shipping Plan
Step 0 (Initialization). Find an initial shipping plan.
C1
5
C2
3
C3
2
P1
4
P2
6
P3
Dem.
1000
100
7
1000
Sup.
1000 0
8
1000 0
7
800
4
100
1100
800
1100
100
0
0
100
0
1000 900
12600
800
0
= cost
Note: Cells in which nothing is being shipped (cells 11, 12,
22, and 23) are called nonbasic cells. The others are called
basic cells (cells 13, 21, 31, 32, and 33). Likewise for the
variables.
15
The Test for Optimality
C1
5
P1
+1
4
3
C3
Sup.
1000
1000
2
7
8
1000
P2
6
P3
Dem.
C2
1000
7
100
1100
4
800
800
100
1100
1000
12600
= cost
TFO: First compute a reduced cost there for each
nonbasic (“empty”) cell.
The reduced costs indicates the change in the current
total costs if one unit is shipped in that nonbasic cell.
If the reduced cost of every nonbasic cell is ≥ 0, then
the current shipping plan is optimal.
16
Computing Reduced Costs
C1
5
P1
+1
4
3
2
7
C3
Sup.
10001
1000
8
1000
P2
6
P3
Dem.
C2
100 1
1100
1000
7
4
800
800
100 +1
1100
1000
12600
= cost
rc( x11 )  +5  2 + 4  6 = +1 ≥ 0
Note: To compute rc(xij), start in cell ij and, using
only the basic cells, alternate between rows and
columns to find the unique “cycle” that returns to
cell ij. Then, alternately add and subtract the unit
shipping costs around this cycle.
17
Test for Optimality
C1
5
C2
3
P1
4
P2
1000 1
6
P3
Dem.
7
7
100 +1
1100
+1
+1
800 1
800
C3
2
8
4
1000 1
+1
+1
100 1
1100
rc( x11 )  1
rc( x12 )  +3  2 + 4  7 = 2 < 0
rc( x22 )  +7  4 + 6  7 = +2 > 0
rc( x23 )  +8  4 + 6  4 = +6 > 0
Sup.
1000
1000
1000
12600
= cost
Not
optimal
18
Find the Direction of Movement
Find the cycle that starts and ends at the nonbasic cell whose
reduced cost is most negative (the rule of steepest descent).
rc( X12 )  2
C1
5
C2
3
P1
4
7
2
1000 1
6
1000
1000
7
100
1100
Sup.
8
1000
P2
P3
Dem.
+1
C3
800 1
800
4
100 +1
1100
1000
12600
= cost
19
Find the Amount of Movement
C1
5
C2
3
4
1000  1
1000
7
8
1000
P2
6
P3
Dem.
Sup.
2
+1
P1
C3
1000
7
100
1100
800  1
800
4
100 + 1
1100
1000
12600
= cost
Question: Because rc(x12) = –2, you can improve the current
plan by shipping 1 unit in cell 12. In fact, each unit shipped in
cell 12 reduces the costs by 2. So why not ship 1000 units there?
Answer: Because each unit shipped in cell 12 requires reducing
the shipment in some of the basic cells by the same amount...and
the values of these basic variables cannot be reduced below 0, so
t* = 800
 min{xij : ij is a basic cell in the cycle whose value decreases}
20
Move to the New Shipping Plan
Having determined that t* = 800 is the amount to ship in the
chosen nonbasic cell, create the new shipping plan by adjusting
the amounts shipped in each basic cell in the associated cycle.
C1
5
C2
3
+1
+800
P1
4
7
6
1000
1000
7
100
1100
Sup.
8
1000
P2
P3
Dem.
C3
2 200
1
1000800
4
0
900
800 800
100+800
1
+1
800
1100
1000
12600
11000
= cost
Note: The nonbasic cell whose value increased to t* becomes
basic (cell 12, in this example) and the basic cell in the cycle
whose value decreases to 0 becomes nonbasic (cell 32, in this
example).
21
Computing Reduced Costs Again
C1
5
C2
3
P1
4
P2
1000 1
6
P3
Dem.
800 1
7
Sup.
200 +1
1000
2
8
+1
7
100 +1
1100
C3
1000
4
800
900 1
1100
1000
11000
rc( x11 )  5  2  4  6  1  0
rc( x22 )  +7  4 + 6  4 + 2  3 = +4 > 0
rc( x23 )  8  4  6  4  6  0
rc( x32 )  7  4  2  3  2  0
= cost
optimal
22
The Stepping Stone Algorithm
Step 0: (Initialization) Find an initial shipping plan x (for
example, by using the Matrix Minimum Method).
Step 1: (Move) Compute a reduced cost for each nonbasic cell. If
all reduced costs are  0, then stop. Otherwise, choose any
nonbasic cell ij whose reduced cost is < 0, create a new shipping
plan as follows, and then return to Step 1:
(a) Find the cycle of cells that starts and ends at cell ij.
(b) Set t* = min{xij : xij is being reduced in the cycle}.
(c) Starting in cell ij, alternately increase and decrease
the values of the cells in the cycle by t*. The cell whose
value reduces to 0 becomes nonbasic.
Question: Is the final shipping plan x* obtained by the algorithm
the best, that is, is x* optimal? Yes!
23
Variations of the Trans. Prob.
•Capacitated/Uncapacitated Problems
•Prohibited Routes
•Transshipment Nodes
•Incorporating Unequal Production Costs
•Incorporating Unequal Revenue
•Lower Bounds on Supplies and Demands
24
Summary
Formulating a graph problem involves the following steps:
1. Identify the vertices by using circles to represent objects in
a problem.
2. Identify the edges by using lines to connect selected pairs
of vertices to indicated a relationship between the objects
associated with the two connected vertices.
3. Identify other data by writing those values next to the
corresponding vertices and/or edges.
4. State the objective in the context of the graph using either:
•Words, or
•Variables, an objective function, and constraints.
25
Summary
The transportation problem is a combinatorial
optimization problem.
For a balanced problem, in which the total supply
equals the total demand, the Stepping Stone
Algorithm is a finite improvement algorithm that
generates a sequence of shipping plans, each of
which has less total cost that the preceding one.
The Matrix Minimum Method is one algorithm
for finding an initial feasible solution, that is, an
initial shipping plan (see the next slide for details).
26
The Matrix Minimum Method
Repeat the following steps until there are no more cells:
Step 1. Identify a remaining cell ij whose unit shipping
cost is the least.
Step 2. Set xij = min{remaining supply at Plant i,
remaining demand for Customer j}.
Step 3. Reduce the supply at Plant i and the demand for
Customer j by xij.
Step 4. If the supply at Plant i has decreased to 0, then
cross off row i. Otherwise, the demand for Customer j is
now 0, so cross off column j.
27