Transcript Engineering Econ Problems and Evaluation Techniques

Unit Cost Problems to Simplify
Analysis
• Some problems can be awkward to solve
• Example – IDOT plans to make 127 from Murphysboro to
Interstate 64 into a 4 lane
– IDOT considers building a concrete roadway
• The road will cost 85 million spread in three equal
payments over a 3 year construction period (first
payment now)
• After construction the road will have to be restriped
every 5 years for $200,000
IDOTs Concrete Baby
• After 20 years of use the surface would become pitted and
the road would have to be resurfaced at a cost of
$30,000,000.
• After another 20 years the road would have to be
resurfaced again for $30,000,000
• After 50 total years of use the highway base will begin to
break-up and the entire roadway will have to be taken up
and rebuilt from scratch for $95,000,000
Another Bid
• Ryan Buddies Inc. has brought IDOT another proposal to
build an black-top highway instead.
• The highway would cost only $50,000,000 and could be
built over two years (two payments of $25,000,000 the
first being now)
• The road would have to be striped every 5 years of use for
$200,000.
• After 6 years of use the highway will need $1,000,000 in
chuck hole repairs
• The next year (7) the chuck holes will cost $2,000,000 to
repair.
Cuddling Black Top
• In year 8 the chuck holes would cost $3,000,000 to repair
• In year 9 the chuck holes would cost $4,000,000 to repair
• At the end of 10 years service the road would have to be
completely resurfaced for $25,000,000
• After 5 years of post resurfacing use the chuck hole saga
would repeat again.
• After 20 total years of surface the roadbase would fail and
the entire road will have to be ripped up and rebuilt at a
cost of $55,000,000.
IDOTS Dilemma
• Which type of highway should they build?
• What Kind of Problem does this look like?
All Cost Alternatives
Cash Flow for Two Alternatives
Year
Concrete Black Top
0 28333333 25000000
1 28333333 25000000
2 28333333
3
4
5
6
200000
7
200000 1000000
8
2000000
9
3000000
10
4000000
11
25000000
12
200000
13
14
15
16
200000
17
200000 1000000
18
2000000
19
3000000
20
4000000
21
55000000
22 30000000
23
24
25
26
27
200000
28
29
30
31
32
200000
33
34
35
36
37
200000
38
39
40
41
42 30000000
43
44
45
46
47
200000
48
49
50
51
52 95000000
Lets Pick Black Top
Year
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
Concrete Black Top BT-Concrete
-28333333 -25000000 3333333
-28333333 -25000000 3333333
-28333333
28333333
0
0
0
-200000
-200000
-200000 -1000000
-800000
-2000000 -2000000
-3000000 -3000000
-4000000 -4000000
-25000000 -25000000
-200000
200000
0
0
0
-200000
-200000
-200000 -1000000
-800000
-2000000 -2000000
-3000000 -3000000
-4000000 -4000000
-55000000 -55000000
-30000000
30000000
0
0
0
0
27
-200000
28
29
30
31
32
-200000
33
34
35
36
37
-200000
38
39
40
41
42 -30000000
43
44
45
46
47
-200000
48
49
50
51
52 -95000000
200000
0
0
0
0
200000
0
0
0
0
200000
0
0
0
0
30000000
0
0
0
0
200000
0
0
0
0
95000000
How Do You Like My Cash
Flow?
• Like many comparisons of very different
alternatives this cash flow swings negative to
positive (wrecking IRR)
• One of the features is that my Black Top choice
saves me a lot of money after the first 20 years
– But that’s because the Black Top highway is
dead and gone - the concrete highway is still
serving
– Does that sound like valid data for decision
making?
The Unequal Lives Problem
• Very often when comparing alternatives for
doing something there is a cheap version
that has a short life and lot of maintenance
and an expensive version with low
maintenance and long life
• My first attempt to saying I’ll get something
run it till it dies and then leave the business
is usually not the way people do things.
Standard Solutions
• Try to do enough replacements of the cheap
choice to get an equal life to the long lived
item
– Getting a number with an integer number of
long life and short life choices can be a joke try to get a common life on my roadway.
• The Salvage Value Approach
– Run the analysis to the life of the short lived
alternative
– Turn the long lived item in for salvage (a
positive value)
The Salvage Value Approach
• Issue #1 - Unlikely in real life that people will
trade in well built equipment every time cheap
equipment would have worn out
– When you do things in cash flows that are vastly
different from real life you risk getting wrong answers
• Issue #2 - Most people buy equipment because it
produces value for them.
– Salvage market seldom pays the value of what a long
lived item could still produce - (When was the last time
you know of someone who had their car totaled and
thought the insurance company gave them a fair
settlement?)
More Salvage Value Issues
• Salvage Markets are highly variable and
regional for used heavy equipment
– May mean that where a plant is located makes a big
difference because of a salvage market that no one will
really use
• Some things don’t salvage well
– Especially things that don’t move well (a lot of civil
engineering structures are not very mobile)
– There may not be a salvage market (such as salvage for
a concrete highway with 30 years life left)
Truncating the Problem
• Remember for the endless life problems I
said usually the first 20 to 30 years makes
the NPV - just cut it off
• Issues with truncation for unequal life
– often the cheap choice cuts off before end of
decisive time for cash flow
• In the case of highway projects were low cost
government financing may be involved the critical
life can be longer than 30 years
Problems with Truncation
• Can be prone to manipulation
– May try to cut-off just before a big expense or
earning from one alternative or the other
• Even if not manipulating you may not know
how close to a major cost you can get
without distorting result
• Perhaps best truncation choice is to do
multiple cycles of the cheap product till
someplace close to long life if that is
beyond the NPV forming years.
The Most Clear Conclusion
• Problems comparing long and short lived
investments (especially all cost alternatives
problems) force people to make shaky
approximations and assumptions to handle
unequal lives
– If I have a big list the chances that some of the
choices will create iffy cash flows increases
• Civil Projects for Highways can be
especially prone to trigger problem patterns
The Unit Cost Solution
Pick an interest rate and discount a full life cycle of each roads costs back to the
start of the road life.
0
1
2
5 6 7 8
9 10
$200,000
$1,000,000
$2,000,000
$3,000,000
$4,000,000
$25,000,000
$25,000,000
-$82,615,801
15 16 17 18 19
$200,000
$1,000,000
$2,000,000
$3,000,000
$4,000,000
Interest Rate
4.5% for tax
free bonds
Convert to Annual Cost
Stretch this money into equal annual Payments
Over the Life of the Road
-$82,615,801 * A/P4.5,20 = -$6,351,184/year
0.07688
Do the Same to the Concrete
Road
• Annual Cost of Concrete Road
– -$5,661,386/year
• Compare this to the Black Top Road
– -$6,351,184/year
• Which Road is Most Cost Effective?
• We converted an All Cost Alternatives
Problem with a different lives problem into
a unit cost problem
Unit Cost Problems
• Problems of this type are sometimes called
“Total Life Cycle Cost” in highway
engineering
• Most professions have some type of
arrangement for unit cost
– Get all the money into an annual cost
– Divide the money by the number of units of
interest you get (in highways it’s the service
year)
– Compare the cost/unit
Why Do We Do Unit Cost
Problems?
• Its not really even its own type of problem (most
are just all cost alternatives problems)
– It’s a method of solution
• Done in most fields because it presents the answer
in one number easy for someone in the business to
understand
– If I tell you it costs 25 cents/mile to own and operate an
automobile you understand fast
• Done because it covers up nasty practical
problems with simple All Cost Alternatives
– Especially the infamous unequal lives problem
Summary of 5 Types of Problems
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Invest and Earn Problem
All Cost Alternatives Problem
Incremental Investment Problem
Competing Investments Problem
Unit Cost Problem
Basics Needed
• Identify the investor and build a cash flow
showing money in and out of his pocket
• Identify the point of decision and put the
pot at that location
– identify needed locations for any temporary
pots
The Six Magic Numbers
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P/F
F/P
P/A
A/P
F/A
A/F
There are a few other minor numbers
Interest Rates
• Interest Rates are almost always reported
annually
– can be adjusted to other compounding periods
so they can be used as i in magic numbers
– Example - Convert to Monthly Interest
• Annual Rate% / 12 (convert to months) / 100
(convert from percent to fraction)
Components of Interest
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•
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Safe Rate (about 2%)
Inflation Rate (now around 4%)
Risk Premium (depends on investment)
Motivation Premium (usually small)
Dealt with by Multiplication
– (1.02)(1.04)(1.09)(1.001) = 1.1574
– 15.74%
• If inflation is not included = Real Rate
• If inflation is included = Nominal Rate