# Cost and Time Value of \$\$ Prof. Eric Suuberg ENGINEERING 90.

## Presentation on theme: "Cost and Time Value of \$\$ Prof. Eric Suuberg ENGINEERING 90."— Presentation transcript:

Cost and Time Value of \$\$ Prof. Eric Suuberg ENGINEERING 90

Cost and Time Value Lecture l What is our goal? »To gain an understanding of what is and what is not a good project to undertake from a financial point of view. l What are our tools? »Material presented by Prof. Crawford »Discounting / Time Value of Money »Tax Savings through Depreciation

So, what are we starting today? l Go through some of the “fun” math for Present Value Calculations l Do a teaching example of purchasing a machine for a manufacturing plant l Talk about costs – both the obvious kind as well as the non-obvious types l Time value of money calculations l Cost Comparisons l Depreciation l Put it all together – inc. continuous discounting and after-tax cost comparisons

Have I Got a Deal for You l Would you be interested in investing in a company that has \$1 million in annual sales?

What More Would You Like to Know? l Annual operating expenses (salaries, raw materials, etc.) l Suppose these were \$900,000/yr l Are you interested? (Come on - I’ve got to know now. There are a lot of people interested)

Profit Profit = Sales (revenues) - expenses (costs) l Basis for taxation - What goes into the calculation is of great interest to Uncle Sam

In Our Example l Profit = \$1,000,000/yr - \$900,000/yr = \$100,000/yr l Is this a good business?

What Would You be Willing to Pay Me for this Business? l \$1 million? \$2 million? l How do you decide? l This is one of the questions that we will answer in this part of the course.

Present Value Calculations l Essential element of evaluating a business opportunity l Different variants »Simple discounting »Replacement and abandonment »Venture Worth, Present Value, Discounted Cash Flow Rate of Return

What information do we need? l Investment (Capital assets, working capital) l Lifetime and Salvage Values l Operating Costs »Fixed »Variable l Interest Rate l Tax Rate l Depreciation Method l Revenues Information Required

Capital Investment - Facility l Purchased Process Equipment l Field Constructed Equipment l Wiring, Piping, Instrumentation l Construction, Installation Costs l Site Preparation, Buildings l Storage Areas l Utilities l Services (Cafeterias, Parking lots, etc.) l Contingency

Capital Investment- Manufacturing l Costs of process equipment may represent only 25% of actual investment! l Costs of process equipment scale according to the “six-tenths rule” »C 2 /C 1 = (Q 2 /Q 1 ) 0.6 l See, for example: »“Cost and Optimization Engineering” by F.C. Jelen and J.H. Black, McGraw-Hill, 1983.

Other Items l Working Capital »Raw materials and supplies inventory »Finished goods in stock and Work in Progress »Accounts Receivable, Taxes payable l Operating Costs »Labor and Raw Materials »Utilities and Maintenance »Royalties l Fixed Costs »Insurance, rent, debt service, some taxes

Time Value of Money l \$1 today is more valuable than the promise of \$1 tomorrow »Has nothing to do with inflation l “Discounting” is the term used to describe the process of correcting for the reduced value of future payments l Discount rate is the return that can be earned on capital invested today

Future Worth of an Investment l P = Principal l i = Annual Interest Rate l S = Future value of investment Compound Interest Law S 1 = P (1+i) at the end of one year S 2 = S 1 (1+i) = P(1+i) 2 at end of year 2 S n = P (1+i) n at end of year n

Present Value of a Future Amount P = S n / (1 + i) n = S n (1 + i) -n (1 + i) -n = Present Value Factor or Discount Factor The promise of \$1 million at a time 50 years in the future @ i = 15%/yr P = \$1,000,000(1+0.15) -50 = \$923

Simple Example l What is the PV of \$10.00 today if I promise to give it to you in fifteen years, given a discount rate of 20%? l PV = 10(1.20) -15 l = \$.65 l Not enough to buy a soda these days

Take Home Message l Not all dollars of profit are the same l Those that come earlier are “worth” more

Start with Simple Example from Everyday Life l Do you buy the better made equipment with the higher price tag? or the low first cost equipment that has high maintenance?

Cost Comparisons What are we doing here? l Comparing one project to another l Deciding to buy the expensive computer that has free maintenance versus the cheap one that makes you pay for service vs.

Simple Cost Comparisons l Strategy »Reduce costs (and/or revenues) to a common instant, usually the present time »Work on full year periods »approximate costs or revenues which occur over the year as single year-end amounts l Basic Rule: All comparisons must be performed on an equal time period basis

Unequal Lifetime Cost Comparisons l Repeatability Assumption (to get to same time basis) l Annuity Comparison l Co-termination assumption

First Some Useful Mathematical Machinery l Uniform periodic annual payments (annuities) l Projects frequently generate recurring income or cost streams on an annual basis 1 65 43 2 (m-1) m 0 x x xxx x xx x = annuity

Discounting a Series of Payments

Discounting a Series of Payments con’t

Capital Recovery Factor

Future Equivalents of Annuities Link to summary of useful formulae

Examples l What future payment N years from now shall I accept in return for an investment of \$P now, given I could instead invest my money elsewhere (e.g. a bank) and earn i %/yr? l What set of annual revenues for N years will entice me to invest \$P, given the same alternative as above?

l What price should I pay for an investment which returns \$X/yr for N years, if i %/yr is available to me in a bank? l What annual interest rate (bank, etc.) would be required to make an investment returning \$S in N years on a present investment of \$P? Examples

A Simple Replacement Problem l Process to be operated for 4 years and then junked l Do you buy a new low-maintenance machine now or not??? DATA (neglect tax effects) Options Stick w/old Buy new Purchase Price (\$)04000 Operating Cost (\$/yr) 2000500 Lifetime (yrs)44

Cash Flow Time Lines 1 43 2 0 \$2000 OLD 1 43 2 0 \$500 NEW \$4000

The Key Role of Interest Rates l If management demands i = 10 %/yr P old =\$6340, P new =\$5585 new is better choice l If management demands i = 20 %/yr P old =\$5180, P new =\$5295 old is better choice

Note l In a replacement problem like this you could have added revenues to the analysis, but no need to do so if they are the same for both options.

Financial Comparisons with Unequal Lifetimes l Simple Example: Choose between 2 pieces of equipment, one of which is better built and has a longer lifetime l N is not the same for both l Not a fair comparison with N=2 unless process is to be shut down and both options have no residual value Well Built Poorly Built 20 year life 2 year life

What to Do? l Option 1 - Repeatability Well Built20 year life Poorly Built2 year life (Buy 1) (Buy 10)

Option 2 - Annualized Costs l Convert the investment and maintenance for both options into a single annual payment Alternative 1Alternative 2 Purchase Price (\$) Annual Op. Cost (\$/yr) Salvage Value (\$) Service Life (yrs) 10,00020,000 15001000 500 1000 2 3 i = 0.15 / yr

Annualized Cost of Alternative 1 = Now

Annualized Cost of Alternative 2 0 1000 20,0001000 1 0 9472 1 3 2 = In this case, choose alternative 1 because yearly cost is lower. 2 3

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