Example 5.1
Transportation Models
Background Information

Midwest Electric has three electric power plants that
supply needs of four cities.

Each power plant can supply the amounts shown in
the table below.
Plant Supplies in Midwest Electric
Example
Plant 1
35
Plant 2
50
Plant 3
40
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Background Information -continued

The peak power demand (in millions of kwh) at each
city is given in this table.
City Requirements for MidWest
Electric Example
City 1
45
City2
20
City 3
30
City 4
30
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Background Information -continued

Finally, the cost (in dollars) of sending a million kwh
from each plant to each city is given in the following
table.
Plant Supplies in Midwest Electric Example
City 1
City 2
City 3
City 4
Plant 1
8
6
10
9
Plant 2
9
12
13
7
Plant 3
14
9
16
5
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Background Information -continued

Midwest Electric wants to find the lowest cost method
for meeting the demand of the four cities.
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Solution

To set up a spreadsheet model for Midwest Electric’s
problem, we need to keep track of the following:
– The power shipped (in millions of kwh) from each plant to
each city
– The total power shipped out of each plant
– The total power received by each city
– The total shipping cost incurred.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
TRANSPORT1.XLS

This file contains the spreadsheet model. The
spreadsheet is shown here.
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Developing the Model

To develop this model, proceed as follows.
– Inputs. Enter the unit shipping costs for each plant to each
city in the UnitsCosts range, the plant capacities in the
Capacities, and the cities’ demands in the Demands range.
– Amounts shipped. Enter any trial values for the shipments
from each plant to each city in the Shipped range. These are
the changing cells.
– Amounts shipped out of plants. To endure that a plant
does not ship more than its available supply, we need to
calculate the amount shipped out of each plant. In cell G13
calculate the amount shipped out of plant 1 with the formula
=SUM(C13:F13) and copy this formula to the range
G14:G15 for the other plants.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Developing the Model -continued
– Amounts received by cities. To ensure that each city
receives the needed power, we keep track of the power
received by each city. Calculate the power received by each
city. Calculate the power received by city 1 in cell C16 with
the formula =SUM(C13:C15) and copy this to the range
D16:F16 for the other cities.
– Total shipping cost. Calculate the total cost of shipping
power from the plants to the cities in the TotalCost cell with
the formula =SUMPRODUCT(UnitCosts,Shipped). This
formula simply sums all products of unit shipping costs and
amounts shipped.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Developing the Model -continued

Using Solver: Now invoke Solver with the following
specifications.
– Objective. Select the TotalCost as the objective to minimize.
– Changing cells. Select the Shipped range as the changing
cells. These cells correspond to the amounts shipped from
each plant to each city.
– Supply constraints. Add the constraints
ShippedOut<=Capacities. These constraints (called supply
constraints) ensure that no plant ships an amount of power
exceeding its capacity.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Developing the Model -continued
– Demand constraints. Add the constraint
ShippedIn>=Demands. These constraints (called demand
constraints) ensure that each city received enough power.
– Specify nonnegativity and optimize. Under SolverOptions,
check the nonnegativity box, and use the LP algorithm to
obtain the optimal solution shown.

The Solver dialog should appear as shown on the
next slide.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Developing the Model -continued

The Solver solution is illustrated graphically on the
next slide.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Solution
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Solution -- continued

A minimum cost of $1020 is incurred by using the
shipments shown.

Except for the six routes shown, no other routes are
used.

Note that all available capacity is used.

The reason is that total demand and total capacity
are both equal to 125, so that the entire capacity is
required to meet demand.

When total demand equals total supply, we call this a
balanced model.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Solution -- continued

If total capacity is greater than total demand, then
some of the capacity is less than total demand, we
need to change the model, for in this case there is no
way to meet total demand – Solver will report “no
feasible solution.”

We need to drop the “greater than or equal to”
demand constraints and probably include unit penalty
costs for not meeting demand at the various cities.

A formulation along these lines appears in the Figure
on the next slide, with the completed Solver dialog
box shown on the slide after that.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Solution -- continued

Here we treat the unmet demand in row 21 as an
extra set of changing cells, we add a constraint that
requires the sums in row 22 to equal demand, and
we account for the penalty cost of unmet demand in
the total cost.

The optimal solution trades off shipping costs with the
penalties from unmet demand.

In this case, the least-cost solution meets all demand
except at city 3.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Sensitivity Analysis

There are many sensitivity analyses we could
perform on the basic transportation model.

We could vary one (or two) shipping costs, or we
could vary capacities or demands.

One interesting analysis is to keep shipping costs
and demands constant and allow all of the capacities
to increase by a certain percentage.

This percentage becomes the input to SolverTable.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Sensitivity Analysis -- continued

Then we keep track of the total cost and any
particular amounts shipped of interest.

The key is to modify the model slightly before running
SolverTable.

The appropriate modifications appear in the model on
the next slide.

Now we store the original capacities in column K, we
enter a percent increase in the PCtIncrease cell, and
we enter formulas in the Capacities range.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Sensitivity Analysis -- continued

Then we run the SolverTable with the PctIncrease
cell as the single input cell, allowing it to vary from
0% to 50% in increments of 10%, and we keep track
of total cost, as well as the shipments out of plant 1.

The results are possibly surprising. Because total
demand is not changing, the extra capacity does not
imply that we will ship more units total – there is no
incentive to send more than the demands require.

However, the increases capacity gives us more
flexibility to use lower-cost shipping routes.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Sensitivity Analysis -- continued

As we see, the total cost steadily decreases as more
capacity is available, and we tend to take more
advantage of the routes out of plant 1.

This sensitivity analysis demonstrates that even
though transportation models are among the simplest
of all LP models to formulate, their optimal solutions
can have somewhat unintuitive properties.
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An Alternative Formulation

The transportation model is a very natural one. If we
consider the graphical representation we note that
“flows” go from left to right, from suppliers to
demanders.

Therefore, the rectangular range of shipments allows
us to calculate shipments out of plants as row sums
and shipments into cities as column sums.

In anticipation of later models in this chapter,
however, where the graphical network can be more
complex, we present an alternative formulation of the
transportation model.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
TRANSPORT2.XLS

This file contains the model.

First, it is useful to introduce some standard
terminology.

When we represent a network model graphically, we
generally connect circles with arrows.

The circles are called nodes. They generally
represent cities, warehouses, manufacturing plants,
or other locations.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Network Models

The arrows are called arcs. They generally represent
routes, such as roads, train tracks, or rivers.

The numbers on the arcs represent flows, the
number of units sent along the arcs.

Sometimes arcs have capacities, the upper limit of
flows on these arcs. They are normally shown along
the arc with the flows. They must be noted as such.

The direction of the arrows indicates which way the
flows are allowed to travel. An arc point into a node is
call an inflow, whereas an arrow pointed out of a
node is called an outflow.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
Network Models -- continued

In the basic transportation model, all outflows
originate from suppliers, and all inflows go toward
demanders. However, general networks can have
both inflows and outflows corresponding to any given
node.

The typical network model has one changing cell per
arc. It indicates how much to send along that arc.

Therefore it is often useful to model network
problems by listing all of the arcs and their
corresponding flows in one long list.
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Network Models -- continued

Specifically, for each node in the network there will be
a flow balance constraint. These flow balance
constraints for the basic transportation model are
simply the supply and demand constraints we have
already discussed.

The alternative formulation of the Midwest Electric
model appears on the next slide.

In the range A12:B23, we manually enter the plant
and city indexes.
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5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

Each of these corresponds to a given name – that is
an arc is the network.

In column C we enter the unit shipping costs.

If they have already been entered in a rectangular
range, as in the CostMatrix range, we can easily
“transfer” them to the appropriate cells in the
UnitCosts range by entering the formula
=VLOOKUP(A12,CostMatrix,B12+1) in cell C12 and
copying it down.
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An Alternative Formulation -continued

Then we enter a column of changing cells for the
flows in column D.

The flow balance constraints are conceptually
straightforward.

Each cell in the Outflows and Inflows ranges contains
the appropriate sum of changing cells.

Is there an easy way to take advantage of copying
when entering these formulas?
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

Fortunately, the answer is “yes”. We use Excel’s built
in SUMIF function, in the form
=SUMIF(Range,Criteria,SumRange).

For example, the formula in cell G13 is
=SUMIF(Origins,F13,Flows). This compares the
plant number in cell F13 to the Options range in
column A and sums all flows where they are equal –
that is, it sums all flows out of plant 1.

By copying this down, we obtain the flows out of the
other plants.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

For flows into cities, we enter the similar formula
=SUMIF(Dests,F19,Flows) in cell G19 to sum all
flows into city 1, and we copy it down for flows into
the other cities.

In general, the SUMIF function finds all cells in the
first argument that satisfy the criteria in the second
argument and then sums the corresponding cells in
the third argument – a very hand function.

This use of the SUMIF function, along with the list of
origins, destinations, unit costs, and flows in columns
A-D, is the key to the network formulation.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

From there on, the model is straightforward.

We calculate the total cost as the SUMPRODUCT of
UnitCosts and Flows, and we set up the Solver dialog
box exactly as before.

To a certain extent this makes all network models
alike.

There is an additional benefit from this model
formulation.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

Suppose that, for whatever reason, flows from certain
plants to certain cities are not allowed.

It is easy to disallow such routes in the original
formulation. The usual trick then is to allow the
“disallowed” routes but impose extremely large unit
shipping costs on them.

This works, but it is wasteful because it adds
changing cells that do not really belong in the model.
However, the current formulation simply omits arcs
that are not allowed.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a
An Alternative Formulation -continued

This creates a model with exactly as many changing
cells as allowable arcs.

This additional benefit can be very valuable when the
number of potential arcs in the network is huge, even
though the vast majority of them are disallowed – and
this is exactly the situation in most large network
models.
5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10 | 5.10a