Transcript pps

Chapter 2B
Linear Programming: Transportation Problem
J. Loucks, St. Edward's University (Austin, TX, USA)
Slide 1 of 82
Transportation Problem



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 problem (TP), as well as many other
problems, are all examples of network problems.
Efficient solution algorithms exist to solve network
problems.
Slide 2 of 82
Transportation Problem



TP can be formulated as linear programs and solved
by general purpose linear programming codes.
For the TP, 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, which
contain separate computer codes for the TP which
take advantage of its network structure.
Slide 3 of 82
Transportation Problem

The TP problem has the following characteristics:
• m sources and n destinations
• number of variables is m x n
• number of constraints is m + n (constraints are for
source capacity and destination demand)
• costs appear only in objective function (objective is
to minimize total cost of shipping)
• coefficients of decision variables in the constraints
are either 0 or 1
Slide 4 of 82
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 on the next slide.
Slide 5 of 82
Transportation Problem

Network Representation
1
d1
2
d2
3
d3
c11
s1
c12
1
c13
c21 c
22
s2
2
c23
SOURCES
DESTINATIONS
Slide 6 of 82
The Relationship of TP to LP



TP is a special case of LP
How do we formulate TP as an LP?
• Let xij = quantity of product shipped from source i to
destination j
• Let cij = per unit shipping cost from source i to
destination j
• Let si be the row i total supply
• Let dj be the column j total demand
The LP formulation of the TP problem is:
Slide 7 of 82
The Relationship of TP to LP

LP Formulation
The linear programming formulation in terms of the
amounts shipped from the origins to the destinations, xij, can be
written as:
Min SScijxij
ij
s.t. Sxij < si for each origin i
j
Sxij > dj for each destination j
i
xij > 0 for all i and j
Slide 8 of 82
Transportation Problem

LP Formulation Special Cases
The following special-case modifications to the linear
programming formulation can be made:
• Minimum shipping guarantees from i to j:
xij > Lij
• Maximum route capacity from i to j:
xij < Lij
• Unacceptable routes:
delete the variable (or put a very high cost, called
the Big-M method, on a selected pair of i and j)
Slide 9 of 82
Transportation Problem


To solve the transportation problem by its special
purpose algorithm, it is required that the sum of the
supplies at the origins equal the sum of the demands
at the destinations. If the total supply is greater than
the total demand, a dummy destination is added
with demand equal to the excess supply, and
shipping costs from all origins are zero. Similarly, if
total supply is less than total demand, a dummy
origin is added.
When solving a transportation problem by its special
purpose algorithm, unacceptable shipping routes are
given a cost of +M (a large number).
Slide 10 of 82
Transportation Problem



The transportation problem is solved in two phases:
• Phase I — Obtaining an initial feasible solution
• Phase II — Moving toward optimality
In Phase I, the Minimum-Cost Procedure can be used
to establish an initial basic feasible solution without
doing numerous iterations of the simplex method.
In Phase II, the Stepping Stone Method, using the
MODI method for evaluating the reduced costs may
be used to move from the initial feasible solution to
the optimal one.
Slide 11 of 82
Transportation Algorithm

Phase I - Minimum-Cost Method (simple and intuitive)
• Step 1: Select the cell with the least cost. Assign to
this cell the minimum of its remaining row supply or
remaining column demand.
• Step 2: Decrease the row and column availabilities
by this amount and remove from consideration all
other cells in the row or column with zero
availability/demand. (If both are simultaneously
reduced to 0, assign an allocation of 0 to any other
unoccupied cell in the row or column before deleting
both.) GO TO STEP 1.
Slide 12 of 82
Transportation Algorithm

Phase II - Stepping Stone Method
• Step 1: For each unoccupied cell, calculate the
reduced cost by the MODI method described below.
Select the unoccupied cell with the most
negative reduced cost. (For maximization problems
select the unoccupied cell with the largest reduced
cost.) If none, STOP.
• Step 2: For this unoccupied cell generate a stepping
stone path by forming a closed loop with this cell
and occupied cells by drawing connecting
alternating horizontal and vertical lines between
them.
Determine the minimum allocation where a
subtraction is to be made along this path.
Slide 13 of 82
Transportation Algorithm

Phase II - Stepping Stone Method (continued)
• Step 3: Add this allocation to all cells where
additions are to be made, and subtract this allocation
to all cells where subtractions are to be made along
the stepping stone path. (Note: An occupied cell on
the stepping stone path now becomes 0
(unoccupied).
If more than one cell becomes 0, make only
one unoccupied; make the others occupied with 0's.)
GO TO STEP 1.
Slide 14 of 82
Transportation Algorithm

MODI Method (for obtaining reduced costs)
Associate a number, ui, with each row and vj with
each column.
• Step 1: Set u1 = 0.
• Step 2: Calculate the remaining ui's and vj's by
solving the relationship cij = ui + vj for occupied cells.
• Step 3: For unoccupied cells (i,j), the reduced cost =
cij - ui - vj.
Slide 15 of 82
Example 1: TP

A transportation tableau is given below. Each cell
represents a shipping route (which is an arc on the
network and a decision variable in the LP
formulation), and the unit shipping costs are given in
an upper right hand box in the cell.
D1
S1
S2
Demand
25
D2
D3
Supply
15
30
20
30
40
35
45
50
30
10
Slide 16 of 82
Example 1: TP
Building Brick Company (BBC) has orders for 80
tons of bricks at three suburban locations as follows:
Northwood — 25 tons, Westwood — 45 tons, and
Eastwood — 10 tons. BBC has two plants, each of
which can produce 50 tons per week.
How should end of week shipments be made to fill
the above orders given the following delivery cost per
ton:
Northwood Westwood Eastwood
Plant 1
24
30
40
Plant 2
30
40
42
Slide 17 of 82
Example 1: TP

Initial Transportation Tableau
Since total supply = 100 and total demand = 80, a
dummy destination is created with demand of 20 and 0
unit costs.
Northwood Westwood
Eastwood
Dummy
24
30
40
0
30
40
42
0
Plant 1
Plant 2
Demand
25
45
10
Supply
50
50
20
Slide 18 of 82
Example 1: TP

Least Cost Starting Procedure
• Iteration 1: Tie for least cost (0), arbitrarily select x14.
Allocate 20. Reduce s1 by 20 to 30 and delete the
Dummy column.
• Iteration 2: Of the remaining cells the least cost is 24
for x11. Allocate 25. Reduce s1 by 25 to 5 and
eliminate the Northwood column.
• Iteration 3: Of the remaining cells the least cost is 30
for x12. Allocate 5. Reduce the Westwood column to
40 and eliminate the Plant 1 row.
• Iteration 4: Since there is only one row with two
cells left, make the final allocations of 40 and 10 to x22
and x23, respectively.
Slide 19 of 82
Example 1: TP

Iteration 1
• MODI Method
1. Set u1 = 0
2. Since u1 + vj = c1j for occupied cells in row 1, then
v1 = 24, v2 = 30, v4 = 0.
3. Since ui + v2 = ci2 for occupied cells in column 2,
then u2 + 30 = 40, hence u2 = 10.
4. Since u2 + vj = c2j for occupied cells in row 2, then
10 + v3 = 42, hence v3 = 32.
Slide 20 of 82
Example 1: TP

Iteration 1
• MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(2,1)
(2,4)
Reduced Cost
40 - 0 - 32 = 8
30 - 24 -10 = -4
0 - 10 - 0 = -10
Slide 21 of 82
Example 1: TP

Iteration 1 Tableau
Plant 1
Plant 2
vj
Northwood Westwood
Eastwood
Dummy
5
30
+8
40
20
0
40
40
10
42
-10
0
25
24
-4
30
24
30
32
ui
0
10
0
Slide 22 of 82
Example 1: TP

Iteration 1
• Stepping Stone Method
The stepping stone path for cell (2,4) is (2,4), (1,4),
(1,2), (2,2). The allocations in the subtraction cells are 20
and 40, respectively. The minimum is 20, and hence
reallocate 20 along this path. Thus for the next tableau:
x24 = 0 + 20 = 20 (0 is its current allocation)
x14 = 20 - 20 = 0 (blank for the next tableau)
x12 = 5 + 20 = 25
x22 = 40 - 20 = 20
The other occupied cells remain the same.
Slide 23 of 82
Example 1: TP

Iteration 2
• MODI Method
The reduced costs are found by calculating
the ui's and vj's for this tableau.
1. Set u1 = 0.
2. Since u1 + vj = cij for occupied cells in row 1, then
v1 = 24, v2 = 30.
3. Since ui + v2 = ci2 for occupied cells in column 2,
then u2 + 30 = 40, or u2 = 10.
4. Since u2 + vj = c2j for occupied cells in row 2, then
10 + v3 = 42 or v3 = 32; and, 10 + v4 = 0 or v4 = -10.
Slide 24 of 82
Example 1: TP

Iteration 2
• MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(1,4)
(2,1)
Reduced Cost
40 - 0 - 32 = 8
0 - 0 - (-10) = 10
30 - 10 - 24 = -4
Slide 25 of 82
Example 1: TP

Iteration 2 Tableau
Plant 1
Plant 2
vj
Northwood Westwood
Eastwood
Dummy
25
30
+8
40
+10
0
20
40
10
42
20
0
25
24
-4
30
24
30
36
ui
0
10
-6
Slide 26 of 82
Example 1: TP

Iteration 2
• Stepping Stone Method
The most negative reduced cost is = -4 determined
by x21. The stepping stone path for this cell is
(2,1),(1,1),(1,2),(2,2). The allocations in the subtraction
cells are 25 and 20 respectively. Thus the new solution
is obtained by reallocating 20 on the stepping stone
path. Thus for the next tableau:
x21 = 0 + 20 = 20 (0 is its current allocation)
x11 = 25 - 20 = 5
x12 = 25 + 20 = 45
x22 = 20 - 20 = 0 (blank for the next tableau)
The other occupied cells remain the same.
Slide 27 of 82
Example 1: TP

Iteration 3
• MODI Method
The reduced costs are found by calculating
the ui's and vj's for this tableau.
1. Set u1 = 0
2. Since u1 + vj = c1j for occupied cells in row 1, then
v1 = 24 and v2 = 30.
3. Since ui + v1 = ci1 for occupied cells in column 2,
then u2 + 24 = 30 or u2 = 6.
4. Since u2 + vj = c2j for occupied cells in row 2, then
6 + v3 = 42 or v3 = 36, and 6 + v4 = 0 or v4 = -6.
Slide 28 of 82
Example 1: TP

Iteration 3
• MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(1,4)
(2,2)
Reduced Cost
40 - 0 - 36 = 4
0 - 0 - (-6) = 6
40 - 6 - 30 = 4
Slide 29 of 82
Example 1: TP

Iteration 3 Tableau
Since all the reduced costs are non-negative, this is
the optimal tableau.
Plant 1
Plant 2
vj
Northwood Westwood
Eastwood
Dummy
45
30
+4
40
+6
0
+4
40
10
42
20
0
5
24
20
30
24
30
36
ui
0
6
-6
Slide 30 of 82
Example 1: TP

Optimal Solution
From
Plant 1
Plant 1
Plant 2
Plant 2
To
Amount Cost
Northwood
5
120
Westwood
45
1,350
Northwood
20
600
Eastwood
10
420
Total Cost = $2,490
Slide 31 of 82
Example 2: TP
Des Moines
(100 units
capacity)
Albuquerque
(300 units
Req.)
Cleveland
Boston (200
(200 units
units req.)
Req.)
Evansville (300
units capacity)
Fort Lauderdale
(300 units capacity)
Slide 32 of 82
Example 2: TP
To (Destination)
From
Albuquerque
(Sources)
Des Moines
$5
Boston
Cleveland
$4
$3
Evansville
$8
$4
$3
Fort
Lauderdale
$9
$7
$5
Slide 33 of 82
Example 2: TP
Factory
To
From
Albuquerque
Boston
Cleveland
capacity
Des
Moines
Evansville
5
4
3
100
8
4
3
300
Fort
Lauderdale
Warehouse
requirement
9
7
5
300
300
200
200
700
Slide 34 of 82
Example 2: TP

What is the initial solution?
• Since we want to minimize the cost, it is reasonable
to “be greedy” and start at the cheapest cell ($3/unit)
while shipping as much as possible. This method of
getting the initial feasible solution is called the leastcost rule
• Any ties are broken arbitrarily.
(Des Moines to Cleveland - $3)
(Evansville to Cleveland - $3)
Let’s start ad cell 1-3 (Des Moines to Cleveland)
• In making the flow assignments make sure not to
exceed the capacities and demands on the margins!
Slide 35 of 82
Example 2: TP
Factory
To
From
Albuquerque
Boston
Cleveland
capacity
Des
Moines
Evansville
5
8
4
3
300
Fort
Lauderdale
Warehouse
requirement
9
7
5
300
300
4
200
Start
+100
200
3
100
0
700
100
Slide 36 of 82
Example 2: TP
Factory
To
From
Albuquerque
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Boston
5
Cleveland
4
capacity
3
+100
300
8
4
9
7
200
+100
200
100
100
0
3
300
5
300
0
200
700
Slide 37 of 82
Example 2: TP
Factory
To
From
Albuquerque
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Boston
5
Cleveland
4
capacity
3
+100
8
4
+200
9
300
+100
7
200
0
200
100
100
0
3
300 0
5
300
0
200
700
Slide 38 of 82
Example 2: TP
Factory
To
From
Albuquerque
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Boston
5
Cleveland
4
capacity
3
+100
8
9
+300
300
4
+200
0
+100
7
200
0
200
100
100
0
3
300 0
5
300
0
200
0
700
Slide 39 of 82
Example 2: TP



The initial feasible solution via the least-cost rule is:
• Ship 100 units from Des Moines to Cleveland
• Ship 100 units from Evansville to Cleveland
• Ship 200 units from Evansville to Boston
• Ship 300 units from Fort Lauderdale to Albuquerque
The total cost of this solution is:
• 100(3) + 100(3) + 200(4) + 300(9) = $4,100
Is the current solution optimal?
• Perhaps, but we are not sure. Hence, some more
work is needed.
Slide 40 of 82
Example 2: TP

The cells that have a number in them are referred to as
basic cells (or basic variables)
• In general, all basic variables are strictly positive
• When at least one basic variable is equal to zero, the
corresponding solution is called degenerate
• When a solution is degenerate, exercise caution in
using sensitivity reports
Slide 41 of 82
Example 2: TP


How many basic variables (cells) should there be in the
transportation tableau?
• Answer: (R+C-1), where R = number of rows and C
= number of columns. Here, (R+C-1) = 3+3-1 = 5
What are they (there are 4 of them)?
• Ship 100 units from Des Moines to Cleveland
• Ship 100 units from Evansville to Cleveland
• Ship 200 units from Evansville to Boston
• Ship 300 units from Fort Lauderdale to Albuquerque
Slide 42 of 82
Example 2: TP


Is this solution degenerate?
• Answer: Yes, because (R+C-1) = 3+3-1 = 5 > 4
• Hence, we can arbitrarily make (5-4) = 1 variable
which is at level zero basic.
• Let us make Des Moines to Albuquerque a basic
variable at level zero
How do we determine if the current solution is
optimal?
• Let us price currently nonbasic variables (cells),
which are all at level zero using the stepping stone
algorithm
Slide 43 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
0
300
3
100
3
300
5
300
+100
8
+300
capacity
+200
9
4
+100
7
200
200
700
Slide 44 of 82
Example 2: TP

Evansville – Albuquerque:
• 8–3+3–5=3>0
Slide 45 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
0
300
3
100
3
300
5
300
+100
8
+300
capacity
+200
9
4
+100
7
200
200
700
Slide 46 of 82
Example 2: TP

Des Moines - Boston:
• 4–3+3-4=0
Slide 47 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
0
300
3
100
3
300
5
300
+100
8
+300
capacity
+200
9
4
+100
7
200
200
700
Slide 48 of 82
Example 2: TP

Fort Lauderdale - Boston:
• 7 – 9 + 5 - 3 + 3 – 4 = -1 < 0
Slide 49 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
0
300
3
100
3
300
5
300
+100
8
+300
capacity
+200
9
4
+100
7
200
200
700
Slide 50 of 82
Example 2: TP

Fort Lauderdale - Cleveland:
• 5 –29 + 5 - 3 = -2 < 0
Slide 51 of 82
Example 1: TP



The objective function improvement potentials for the
nonbasic variables are:
• Evansville – Albuquerque: 3 > 0
• Des Moines - Boston: 0 = 0
• Fort Lauderdale - Boston: -1 < 0
• Fort Lauderdale - Cleveland: -2 < 0
The largest improvement (here: reduction in cost) in the
objective function can be achieved by increasing
shipments from Fort Lauderdale to Cleveland
What i2 the limit on the increase?
Slide 52 of 82
Example 2 TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
0 +K
300
3
100
3
300
5
300
+100 -K
8
+300 -K
capacity
+200
9
4
7
200
+100
+K
200
700
Slide 53 of 82
Example 2: TP



If we increase shipments from Fort Lauderdale to
Cleveland by K, we must reduce shipments in the basic
cells in the same row and column so that the totals on
the margins remain the same
Clearly, because negative quantities cannot be shipped,
we must impose that:
• K >= 0 (no problem)
• 300 – K >= 0 (K <= 300)
• 0 + K >= 0 (no problem)
• 100 – K >=0 (K <=100)
Hence, let K = 100
The new basis becomes:
Slide 54 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
capacity
3
100
3
300
5
300
+100
8
+200
300
+200
9
4
7
200
+100
+100
200
700
Slide 55 of 82
Example 2: TP



Observe that the number of basic variables (cells) is
equal to (R+C-1) = 5. Hence, the current basic solution
is feasible and not degenerate
Also observe that Des Moines – Cleveland left the basis
while Fort Lauderdale – Cleveland entered the basis
(one for one interchange)
Is the current solution optimal?
• Let us price the non-basic variables (cells):
Slide 56 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
capacity
3
100
3
300
5
300
+100
8
+200
300
+200
9
4
7
200
+100
+100
200
700
Slide 57 of 82
Example 2: TP

Evansville - Albuquerque:
• 8–3+5–9=1>0
Slide 58 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
capacity
3
100
3
300
5
300
+100
8
+200
300
+200
9
4
7
200
+100
+100
200
700
Slide 59 of 82
Example 2: TP

Des Moines - Boston:
• 4–5+9–5+3–4=2>0
Slide 60 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
capacity
3
100
3
300
5
300
+100
8
+200
300
+200
9
4
7
200
+100
+100
200
700
Slide 61 of 82
Example 2: TP

Fort Lauderdale - Boston:
• 7–5+3-4=1>0
Slide 62 of 82
Example 2: TP
Factory
To
From
Des
Moines
Evansville
Fort
Lauderdale
Warehouse
requirement
Albuquerque
Boston
5
Cleveland
4
capacity
3
100
3
300
5
300
+100
8
+200
300
+200
9
4
7
200
+100
+100
200
700
Slide 63 of 82
Example 2: TP

Des Moines - Cleveland:
• 3–5+9-5=2>0
Slide 64 of 82
Example 2: TP



The objective function improvement potentials for the
nonbasic variables are:
• Evansville - Albuquerque: 1 > 0
• Des Moines - Boston: 2 > 0
• Fort Lauderdale - Boston: 1 > 0
• Des Moines - Cleveland: 2 > 0
Since all improvement potentials are positive, it is not
possible to achieve further improvement (here:
reduction in cost) in the objective function
Hence, the current solution is optimal
Slide 65 of 82
Example 2: TP


The final optimal (and not degenerate) solution is:
• Ship 100 units from Des Moines to Albuquerque
• Ship 200 units from Fort Lauderdale to Albuquerque
• Ship 200 units from Evansville to Boston
• Ship 100 units from Evansville to Cleveland
• Ship 100 units from Fort Lauderdale to Cleveland
The total cost of this solution is:
• 100(5) + 200(9) + 200(4) + 100(3) + 100(5)= $3,900
Slide 66 of 82
Example 2: TP via LP

In our example the LP formulation is:
Min z = 5 x11 + 4x12 + 3x13 + 8 x21 + 4x22 + 3x23 + 9 x31 +
7x32 + 5x33
s.t.
x11 + x12 + x13 <= 100
(row 1)
x21 + x22 + x23 <= 300
(row 2)
x31 + x32 + x33 <= 300
(row 3)
x11 + x21 + x31 >= 300
(column 1)
x12 + x22 + x32 >= 200
(column 2)
x31 + x32 + x33 >= 200
(column 3)
xij >= 0 for i = 1, 2, 3 and j = 1, 2, 3 (nonnegativity)
Slide 67 of 82
Example 2: TP via LP

The solver formulation and solution is:
Example 1
Shipping costs:
Shipping quantities:
Total shipping cost:
s.t.
Des Moines supply:
Evansville supply:
Fort Lauderdale supply:
Albuquerque demand:
Boston demand:
Cleveland demand:
nonnegativity:
5
100
3900
4
0
3
8
0 -3.6E-15
1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
0
0
0
0
4
200
3
100
9
200
7
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
1
0
5
100
decision cells
objective function
0
100 <=
0
300 <=
1
300 <=
0
300 >=
0
200 >=
1
200 >=
0
100 >=
0
0 >=
0
0 >=
0 -3.6E-15 >=
0
200 >=
0
100 >=
0
200 >=
0
0 >=
1
100 >=
100
300
300
300
200
200
0
0
0
0
0
0
0
0
0
Slide 68 of 82
Example 2: TP via LP

The solver answer report is:
Microsoft Excel 9.0 Answ er Report
Worksheet: [Example 1 - TP via LP.xls]Sheet1
Report Created: 2/9/00 12:12:55 PM
Target Cell (Min)
Cell
$B$5
WBMIN
Adjustable Cells
Cell
$B$4
Shipping
$C$4
Shipping
$D$4
Shipping
$E$4
Shipping
$F$4
Shipping
$G$4
Shipping
$H$4
Shipping
$I$4
Shipping
$J$4
Shipping
Name
Name
quantities:
quantities:
quantities:
quantities: Example 1
quantities:
quantities:
quantities:
quantities:
quantities:
Constraints
Cell
Name
$K$10 Fort Lauderdale supply:
$K$8
Des Moines supply:
$K$12 Boston demand:
$K$9
Evansville supply:
$K$11 Albuquerque demand:
$K$13 Cleveland demand:
$K$14 nonnegativity:
$K$15
$K$16
$K$17
$K$18
$K$20
$K$21
$K$22
$K$19
Original Value
3900
Final Value
3900
Original Value
100
0
0
-3.55271E-15
200
100
200
0
100
Final Value
100
0
0
-3.55271E-15
200
100
200
0
100
Cell Value
300
100
200
300
300
200
100
0
0
-3.55271E-15
200
200
0
100
100
Formula
$K$10<=$M$10
$K$8<=$M$8
$K$12>=$M$12
$K$9<=$M$9
$K$11>=$M$11
$K$13>=$M$13
$K$14>=$M$14
$K$15>=$M$15
$K$16>=$M$16
$K$17>=$M$17
$K$18>=$M$18
$K$20>=$M$20
$K$21>=$M$21
$K$22>=$M$22
$K$19>=$M$19
Status
Slack
Binding
0
Binding
0
Binding
0
Binding
0
Binding
0
Binding
0
Not Binding
100
Binding
0
Binding
0
Binding
0
Not Binding
200
Not Binding
200
Binding
0
Not Binding
100
Not Binding
100
Slide 69 of 82
Example 2: TP via LP

The solver sensitivity report is:
Microsoft Excel 9.0 Sensitivity Report
Worksheet: [Example 1 - TP via LP.xls]Sheet1
Report Created: 2/9/00 12:12:55 PM
Adjustable Cells
Cell
$B$4
$C$4
$D$4
$E$4
$F$4
$G$4
$H$4
$I$4
$J$4
Shipping
Shipping
Shipping
Shipping
Shipping
Shipping
Shipping
Shipping
Shipping
Name
quantities:
quantities:
quantities:
quantities: Example 1
quantities:
quantities:
quantities:
quantities:
quantities:
Final
Value
Reduced
Gradient
100
0
0
-3.55271E-15
200
100
200
0
100
0
0
0
0
0
0
0
0
0
Constraints
Cell
$K$10
$K$8
$K$12
$K$9
$K$11
$K$13
$K$14
$K$15
$K$16
$K$17
$K$18
$K$20
$K$21
$K$22
$K$19
Name
Fort Lauderdale supply:
Des Moines supply:
Boston demand:
Evansville supply:
Albuquerque demand:
Cleveland demand:
nonnegativity:
Final
Value
300
100
200
300
300
200
100
0
0
-3.55271E-15
200
200
0
100
100
Lagrange
Multiplier
0
-4
6
-2
9
5
0
1.999998689
2.00000006
0.999999642
0
0
0.999998927
0
0
Slide 70 of 82
Example 2: TP via LP

The solver limits report is:
Microsoft Excel 9.0 Limits Report
Worksheet: [Example 1 - TP via LP.xls]Sheet1
Report Created: 2/9/00 12:12:55 PM
Cell
$B$5 WBMIN
Cell
$B$4
$C$4
$D$4
$E$4
$F$4
$G$4
$H$4
$I$4
$J$4
Target
Name
Adjustable
Name
Shipping quantities:
Shipping quantities:
Shipping quantities:
Shipping quantities: Example 1
Shipping quantities:
Shipping quantities:
Shipping quantities:
Shipping quantities:
Shipping quantities:
Value
3900
Value
100
0
0
-3.55271E-15
200
100
200
0
100
Lower
Limit
100
0
0
-3.55271E-15
200
100
200
0
100
Target
Result
3900
3900
3900
3900
3900
3900
3900
3900
3900
Upper
Limit
100
0
0
-3.55271E-15
200
100
200
0
100
Target
Result
3900
3900
3900
3900
3900
3900
3900
3900
3900
Slide 71 of 82
Example 3: Tropicsun (TP via LP)
Tropicsun is a grower of oranges with locations in the cities of Mt.
Dora, Eustis and Clermont. Tropicsun currently has 275,000
bushels of citrus at the grove in Mt. Dora; 400,000 bushels in Eustis;
and 300,000 bushels in Clermont. The citrus processing plants are
located in Ocala (capacity of 200,000 bushels), Orlando (600,000
bushels) and Leesburg (225,000 bushels). Tropicsun contracts with
a trucking company to transport its fruit, which charges a flat rate
for every mile that each bushel of fruit must be transported.
Observe that the problem is not balanced, since the total of the fruit
supplies (275,000 + 400,000 + 300,000 = 975,000) is not equal to the
total of the capacities (200,000 + 600,000 + 225,000 = 1,025,000). The
difference of 50,000 bushels will be assigned to a dummy source,
which represents unutilized capacity. The distances (in miles)
between the groves and processing plants are as follows:
Slide 72 of 82
Example 3: Tropicsun (TP via LP)
Distances (in miles) between groves and plants
Grove
Mt. Dora
Eustis
Clermont
Ocala
21
35
55
Orlando
50
30
20
Leesburg
40
22
25
Slide 73 of 82
Example 3: TP via LP :
Tropicsun
Supply
Groves
Distances (in miles)
Capacity
21
Mt. Dora
275,000
Processing
Plants
1
50
Ocala
4
200,000
40
35
400,000
30
Eustis
2
Orlando
600,000
5
22
55
20
Clermont
300,000
3
25
Leesburg
6
225,000
Slide 74 of 82
Example 3: Defining the Decision Variables
Xij = # of bushels shipped from node i to node j
Specifically, the nine decision variables are:
X14 = # of bushels shipped from Mt. Dora (node 1) to Ocala (node 4)
X15 = # of bushels shipped from Mt. Dora (node 1) to Orlando (node 5)
X16 = # of bushels shipped from Mt. Dora (node 1) to Leesburg (node 6)
X24 = # of bushels shipped from Eustis (node 2) to Ocala (node 4)
X25 = # of bushels shipped from Eustis (node 2) to Orlando (node 5)
X26 = # of bushels shipped from Eustis (node 2) to Leesburg (node 6)
X34 = # of bushels shipped from Clermont (node 3) to Ocala (node 4)
X35 = # of bushels shipped from Clermont (node 3) to Orlando (node 5)
X36 = # of bushels shipped from Clermont (node 3) to Leesburg (node 6)
Slide 75 of 82
Example 3: Defining the Objective Function
Minimize the total number of bushel-miles.
MIN: 21X14 + 50X15 + 40X16 +
35X24 + 30X25 + 22X26 +
55X34 + 20X35 + 25X36
Slide 76 of 82
Example 3: Defining the Constraints

Capacity constraints
} Ocala
} Orlando
} Leesburg

X14 + X24 + X34 <= 200,000
X15 + X25 + X35 <= 600,000
X16 + X26 + X36 <= 225,000
Supply constraints
X14 + X15 + X16 = 275,000
X24 + X25 + X26 = 400,000
X34 + X35 + X36 = 300,000
} Mt. Dora
} Eustis
} Clermont

Nonnegativity conditions
Xij >= 0 for all i and j
Slide 77 of 82
Example 3: Solver Formulation
Tropicsun
Grove
Mt. Dora
Eustis
Clermont
Distances From
Groves to Plant at:
Ocala
Orlando
Leesburg
21
50
40
35
30
22
55
20
25
Grove
Mt. Dora
Eustis
Clermont
Received
Capacity
Bushels Shipped From
Groves to Plant at:
Ocala
Orlando
Leesburg
0
0
0
0
0
0
0
0
0
0
0
0
200,000
600,000
225,000
Total Distance (in bushel-miles):
Bushels
Shipped
0
0
0
Bushels
Available
275,000
400,000
300,000
0
Slide 78 of 82
Example 3: Solver Solution
Tropicsun
Grove
Mt. Dora
Eustis
Clermont
Distances From
Groves to Plant at:
Ocala
Orlando
Leesburg
21
50
40
35
30
22
55
20
25
Grove
Mt. Dora
Eustis
Clermont
Received
Capacity
Bushels Shipped From
Groves to Plant at:
Ocala
Orlando
Leesburg
200,000
0
75,000
0
250,000
150,000
0
300,000
0
200,000
550,000
225,000
200,000
600,000
225,000
Total Distance (in bushel-miles):
Bushels
Shipped
275,000
400,000
300,000
Bushels
Available
275,000
400,000
300,000
24,000,000
Slide 79 of 82
Example 4: The Sentry Lock Corporation
Slide 80 of 82
Example 4: Solver Formulation
Sentry Lock Corp.
Plants
Macon
Louisville
Detroit
Phoenix
Plants
Macon
Louisville
Detroit
Phoenix
Total Shipped
Demand
Min Shipped
Total Cost
Tacoma
$2.50
$1.85
$2.30
$1.90
Tacoma
0
600
400
5800
6800
8500
6800
$3,011,360
San Diego
$2.75
$1.90
$2.25
$0.90
Unit Shipping Costs to
Distribution Center in:
Dallas
Denver St. Louis
$1.75
$2.00
$2.10
$1.50
$1.60
$1.00
$1.85
$1.25
$1.50
$1.60
$1.75
$2.00
Tampa
$1.80
$1.90
$2.25
$2.50
Unit
Production
Baltimore
Cost
$1.65
$35.50
$1.85
$37.50
$2.00
$39.00
$2.65
$36.25
San Diego
0
0
0
14200
14200
14500
11600
Quantity Shipped to
Distribution Center in:
Dallas
Denver St. Louis
0
0
0
0
0
14400
10800
12600
0
0
0
0
10800
12600
14400
13500
12600
18000
10800
10080
14400
Tampa
12000
0
0
0
12000
15000
12000
Total
Baltimore Produced Capacity
6000
18000
18000
0
15000
15000
1200
25000
25000
0
20000
20000
7200
9000
7200
Minimize:
B23
By chaning: B15:H18
Subject to: I15:I18=J15:J18
B19:H19<=B20:H20
B19:H19>=B21:H21
B15:H18>=0
Slide 81 of 82
The End of Chapter 2B
Slide 82 of 82