Example 5.10
Project Scheduling Models
Background Information

Tom Lingley, an independent contractor, has agreed
to build a new room on an existing house.

He plans to begin work on Monday morning, June 1.

The main question is when will he complete his work,
given that he works only on weekdays.

The owner of the house is particularly hopeful that
the room will be ready in 15 or fewer working days –
that is, by the end of Friday, June 19.

The work proceeds in stages, labeled A through J, as
summarized in the table on the next slide.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Background Information –
continued
Activity Time Data
Description
Prepare foundation
Put up frame
Order custom windows
Erect outside walls
Do electrical wiring
Do plumbing
Put in duct work
Hand dry wall
Install windows
Paint and clean up

Index
A
B
C
D
E
F
G
H
I
J
Predecessors
None
A
None
B
D
D
D
E, F, G
B, C
H
Durations (Days)
4
4
11
3
4
3
4
3
1
2
Three of these activities, E, F, and G, will be done by
separate independent subcontractors.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Background Information –
continued

Lingley wants to know how long the project will take,
given the activity times (durations) in the table.

He also wants to know the critical activities.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Solution

The project network appears here. The activity time
for each activity is shown on its arc.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Solution – continued

We suggest that you verify rule 3 from our general
discussion – if an activity’s arc leads out of a node,
then all of this activity’s predecessors should have
arcs leading into the node.

The key to the solution is finding event times for each
node in the network, where the event time for node i
is the earliest time we could reach that point in the
network.

Denote the event time for node i by Ei. We begin by
setting E1=0. Node 1 is the start node, so its event
time is 0 – right away.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Solution – continued



Also, if node n is the finish node, then the earliest the
entire project can be completed is at time En.
Therefore, the total project time is E8 for our example.
In general, let i and j be any nodes joined by an arc
with activity time tij.
Then we must have Ej  Ei + tij. The reasoning is that
the event at node j cannot occur until at least time tij
after the event at node i.
Actually, we can do better than this. The event time Ej
is equal to the maximum of the quantities Ei +tij,
where the maximum is taken over all arcs that lead
into node j.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Solution – continued

We could use this relationship to find the event times.
In fact, we will do this when we revisit this example in
Chapter 12.

For right now, however, we will not use maximums.

Instead, we will exploit inequalities.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Developing the Model –
continued

The completed spreadsheet model appears on the
next slide.

It can be developed with the following steps.
– Enter data. The data for this model include the
predecessors and activity times (durations) in the range
C5:D14.
– Event times. We will eventually use the Solver to find the
event times that satisfy inequalities. For now, enter any
event times in the EventTimes range. The EventTime range
will be the changing cells range.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Developing the Model –
continued
– Network Information. Enter the information in the range D18:F28.
This comes directly from the network shown earlier.
– Inequalities. Now we implement inequalities in the range G18:I28.
We could enter the formulas one cell at a time, but it is more
convenient to use VLOOKUP functions. To do so, enter the formula
VLOOKUP(F18,LTable2,2) in cell G18 and copy it down. Each of
these values corresponds to Ei . Then enter the formula
=VLOOKUP(E18,LTable2,2)+VLOOKUP(D18,LTable1,3) in cell
I18 and copy it down. Each of these values corresponds to Ei . + tij.
As usual, make sure you understand exactly how these formulas
work. Lookup functions can be intimidating, but they can save a lot
of work!
– Project time. Enter the formula =B25 in cell B30. This creates a
link to the event time for the final node in cell B25 – that is, to the
time to complete the project.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Using Solver

We want to choose the event times so as to minimize
the project time and satisfy inequalities. Therefore,
set up the Solver as shown here.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Using Solver – continued

When we click on Solve and see the dialog box
indicating that the Solver has found the optimal
solution, we now request a sensitivity report, which is
shown on the next slide.

We stated in Chapter 3 that these sensitivity reports
can sometimes be misleading, but they provide
useful information in this example.

Specifically, consider the Shadow Price column of the
sensitivity report.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Using Solver – continued

Each row of this table corresponds to one of the
inequalities in the model – that is, to one of the
activities in the project.

In general, a shadow price indicates the change in
the target cell if the right side of a constraint
increases by 1.

Because our constraints include activity times on
their right sides, each shadow price indicates how
much the total project time will increase if the
corresponding activity time increases by 1.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a
Using Solver – continued

We see that some activity time increases have no
effect on the project time, whereas others have a
positive effect.

This is how we find the critical path. It includes
exactly those activities with positive shadow prices.
These are indicated by asterisks.

The activities that are on the critical path are A, B, D,
E, H, and J.

If the activity times for activities not on the critical
path could increase, at least a little, with no effect on
the total project time.
5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.6 | 5.7 | 5.8 | 5.9 | 5.10a