Why Unit Cost Problems ©Dr. B. C. Paul 2001

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 $561 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 $150,000,000.  After another 20 years the road would have to be resurfaced again for $150,000,000  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 $625,000,000

Another Bid  Ryan Buddies Inc. has brought IDOT another proposal to build an black-top highway instead.  The highway would cost only $400,000,000 and could be built over two years (two payments of $200,000,000 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 $10,000,000 in chuck hole repairs  The next year (7) the chuck holes will cost $20,000,000 to repair.

Cuddling Black Top  In year 8 the chuck holes would cost $30,000,000 to repair  In year 9 the chuck holes would cost $40,000,000 to repair  At the end of 10 years service the road would have to be completely resurfaced for $120,000,000  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 $455,000,000.

IDOTS Dilemma  Which type of highway should they build?  What Kind of Problem does this look like? All Cost Alternatives

The Classic Problem  We have an All Cost Alternatives Problem with an Unequal Lives Landmine  50 year concrete highway  20 year asphalt highway  Standard Solution One  Turn the long lived asset in for salvage  How does one role up and resell a highway with 30 years of life left?

Standard Solution Two  Replace the low cost alternative to match the life of the long lived alternative  To get a common life for 20 and 50 years we need a 100 year road plan  Is that realistic?

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 $200,000 $10,000,000 $20,000,000 $30,000,000 $40,000,000 $120,000,000 $200,000 $10,000,000 $20,000,000 $30,000,000 $40,000,000 2 Interest Rate 4.5% for tax free bonds -$620,651,717 $200,000,000

Convert to Annual Cost -$ 620,651,717 Stretch this money into equal annual Payments Over the Life of the Road * A/P 4.5,20 = $47,713,311/year

Lets Do This With Class Assistant on the Concrete Highway! -$187,000,000 each $200,000 $150,000,000

Go to the Cash Flow Analyzer Section Set in our 4.5% Interest rate Set it for annual compounding

Enter the Cash Flow

Set Up the Total Life Cycle Cost Boxes Tell it what year the road goes Into service. Tell it how long the road lasts.

Read Off the Answer The number by the Total Life Cycle Cost Heading is the Total Life Cycle Cost. - in this case $35,501,984 For interest the PV Of the cash flow at The start of the project Is also given.

Do the Same to the Concrete Road  Annual Cost of Concrete Road  -$35,501,984/year  Compare this to the Black Top Road  -$47,713,311/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) =  15.74%  If inflation is not included = Real Rate  If inflation is included = Nominal Rate