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 AlternativesYear Concrete Black Top
0 28333333 250000001 28333333 250000002 283333333456 2000007 200000 10000008 20000009 3000000
10 400000011 2500000012 20000013141516 20000017 200000 100000018 200000019 300000020 4000000
21 5500000022 300000002324252627 2000002829303132 2000003334353637 200000383940
4142 300000004344454647 2000004849505152 95000000
Lets Pick Black TopYear Concrete Black Top BT-Concrete
0 -28333333 -25000000 33333331 -28333333 -25000000 33333332 -28333333 283333333 04 05 06 -200000 -2000007 -200000 -1000000 -8000008 -2000000 -20000009 -3000000 -3000000
10 -4000000 -400000011 -25000000 -2500000012 -200000 20000013 014 015 016 -200000 -20000017 -200000 -1000000 -80000018 -2000000 -200000019 -3000000 -300000020 -4000000 -400000021 -55000000 -5500000022 -30000000 3000000023 024 025 026 0
27 -200000 20000028 029 030 031 032 -200000 20000033 034 035 036 037 -200000 20000038 039 040 041 042 -30000000 3000000043 044 045 046 047 -200000 20000048 049 050 051 052 -95000000 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 thestart of the road life.
0 1 5 6 7 8 9 10 15 16 17 18 19
$25,000,000
$200,000$1,000,000
$2,000,000$3,000,000
$4,000,000
$25,000,000
$200,000$1,000,000
$2,000,000$3,000,000
$4,000,000
2
Interest Rate4.5% for taxfree bonds
-$82,615,801
Convert to Annual Cost
-$82,615,801
Stretch this money into equal annual Payments Over the Life of the Road
* A/P4.5,20 =0.07688
-$6,351,184/year
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
• 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
• 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• 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