{ In general: retire a project if the net present value of future cash flows is less than zero
19
Continuing Value
Long-lived Project Evaluation
{ Many large infrastructure projects, (e.g., water or gas distribution networks or electricity generators or office buildings) have lives of 50 to 100 years or more.
{ Usually require reinvestment to maintain cash flows (e.g. reparation of machines,…)
z The cash flows to use in NPV calculations are net of this maintenance cost
9
Optimal Replacement Decisions
{ The net cash flows associated with the running of the vehicle and salvage values are shown in the table on the next slide
NOPLAT = NI + D
Gross Cash Flow
NOPLAT ⋅ NOPLAT = (NI + D) NOPLAT
NOPLAT
NOPLAT
Less
Replacement Capex (Proxied by Depreciation)
NOPLAT = NI NOPLAT + D NOPLAT
z Annual equivalent is computed as the annual annuity payment over the life of the project with the same NPV as the project
7
A Decision Tree for Evaluation of Competing Projects that are on a continuing cycle
13
Year 5 Replacement Option
14
Optimal Replacement Decisions
Year
3
4
5
NPV -15,269.08 -19,657.28 -25,470.95
AE
-5,712.31 -5,672.92 -6,046.71
15
Project review
{ Replacement decision { Retirement decision
22
Net Cash Flows
{ Net cash flows will consist of:
After tax cash flows - replacement capex
NOPLAT - new investment
Net Cash Flows
23
Dividends and Reinvestment
16
Retirement decision
{ Mortlake Ltd owns machine 6 years old
{ Estimated remaining physical life no more than 2 years
{ Required return 10%
{ When should the machine be retired?
2
3
A
A
A ...
PV = A + A + A + ... (1+ r )1 (1+ r )2 (1+ r )3
PV = A r
26
Recall: Perpetuities
{ If the payment amount grows
at a constant rate of g:
0
1
2
3
A
A(1+g) A(1+g)2 …
{ In practice, we estimate this by assuming cash flows have a certain growth rate
25
Recall: Perpetuity:
an annuity with an infinite number of payments
0
1
D = NOPLAT ⋅ DR = NOPLAT (1 − NIR )
24
Continuing Value
{ Accurate forecasts are only possible for a relatively short forecast horizon
{ Value after that forecast horizon is called continuing value and must be estimated
End of year
6
7
8
Net cash Residual
flow
value
0
12000
8000 6000
5000 0
17
Solution
{ Run one more year: forego 12000 to obtain 8000 plus 6000
z NPV=-12000+(8000+6000)/1.1 =727
z In general, we proxy for this cost using depreciation
{ Growth occurs through new investment
z If there is no new investment, cash flows are the same every year
3
Comparing Projects with Different Lives
{ You are considering two vehicles:
z Vehicle A has an NPV of $3,500 over a 5-year period
z Vehicle B has an NPV of $5,000 over a 7-year period
11
Year 3 Replacement Option
NPV3
=
−20,000
+
− 1,500 1.06
+
− 3,000
(1.06)2
+
10,500
(1.06)3
= −15,269.08
AE3
=
− 15,269.08
1 − (1.06)−3
=
−5,712.31
.06
12
Year 4 Replacement Option
{ Therefore, recommend Vehicle B.
6
Comparing Projects with Different Lives
{ Principle: When evaluating competing machines that will be replaced indefinitely, choose the machine with the highest positive annual equivalent
Salvage
-
17,000 15,500 14,000 13,000 10,000
{ Principle: when determining the optimal replacement policy for a piece of equipment, choose the option with the lowest annual equivalent cost
{ What are you going to recommend to the trustees?
z Assume a required rate of return of 6% p.a.
4
Comparing Projects with Different Lives
{ Must compare projects over the same time frame
{ Company valuation involves an infinite-horizon cash flow stream
{ Require the calculation of continuing value (CV)
21
Characteristics of long lived projects
{ Run two more years: forego 12000 to obtain cash flows of 8000 and 5000
z NPV=-12000+(8000/1.1)+(5000/1.12) =-595
18
Decision
{ Retire machine at the end of the 7th year
Yes Are projects independent?
No
Select project with highest NPV annual
No
equivalent
Select all projects with + NPV