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Cost and Time Value of $$

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Cost and Time Value of $$

Prof. Eric Suuberg

ENGINEERING 90

- 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.

- What are our tools?
- Material presented by Prof. Crawford
- Discounting / Time Value of Money
- Tax Savings through Depreciation

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

- Would you be interested in investing in a company that has $1 million in annual sales?

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

Profit = Sales (revenues) - expenses (costs)

- Basis for taxation - What goes into the calculation is of great interest to Uncle Sam

- Profit = $1,000,000/yr - $900,000/yr= $100,000/yr
- Is this a good business?

- $1 million?
$2 million?

- How do you decide?
- This is one of the questions that we will answer in this part of the course.

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

Information Required

- Investment (Capital assets, working capital)
- Lifetime and Salvage Values
- Operating Costs
- Fixed
- Variable

- Interest Rate
- Tax Rate
- Depreciation Method
- Revenues

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

- Costs of process equipment may represent only 25% of actual investment!
- Costs of process equipment scale according to the “six-tenths rule”
- C2/C1 = (Q2/Q1)0.6

- See, for example:
- “Cost and Optimization Engineering” by F.C. Jelen and J.H. Black, McGraw-Hill, 1983.

- Working Capital
- Raw materials and supplies inventory
- Finished goods in stock and Work in Progress
- Accounts Receivable, Taxes payable

- Operating Costs
- Labor and Raw Materials
- Utilities and Maintenance
- Royalties

- Fixed Costs
- Insurance, rent, debt service, some taxes

- $1 today is more valuable than the promise of $1 tomorrow
- Has nothing to do with inflation

- “Discounting” is the term used to describe the process of correcting for the reduced value of future payments
- Discount rate is the return that can be earned on capital invested today

- P = Principal
- i = Annual Interest Rate
- S = Future value of investment
Compound Interest Law

S1 = P (1+i) at the end of one year

S2 = S1(1+i) = P(1+i)2 at end of year 2

Sn = P (1+i)n at end of year n

P = Sn / (1 + i)n

= Sn (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

- 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%?
- PV = 10(1.20)-15
- = $.65
- Not enough to buy a soda these days

- Not all dollars of profit are the same
- Those that come earlier are “worth” more

- Do you buy the better made equipment with the higher price tag? or the low first cost equipment that has high maintenance?

What are we doing here?

- Comparing one project to another
- Deciding to buy the expensive computer that has free maintenance versus the cheap one that makes you pay for service

vs.

- 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

- Basic Rule: All comparisons must be performed on an equal time period basis

- Repeatability Assumption (to get to same time basis)
- Annuity Comparison
- Co-termination assumption

x

x

x

x

x

x

x

x

1

(m-1)

2

m

3

4

5

6

0

x = annuity

- Uniform periodic annual payments (annuities)
- Projects frequently generate recurring income or cost streams on an annual basis

Link to summary of useful formulae

- 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?
- What set of annual revenues for N years will entice me to invest $P, given the same alternative as above?

- 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?
- What annual interest rate (bank, etc.) would be required to make an investment returning $S in N years on a present investment of $P?

- Process to be operated for 4 years and then junked
- Do you buy a new low-maintenance machine now or not???

DATA (neglect tax effects)

Options Stick w/old Buy newPurchase Price ($)04000Operating Cost ($/yr) 2000500Lifetime (yrs)44

0

3

4

2

1

$2000

$2000

$2000

$2000

OLD

NEW

0

3

4

2

1

$4000

$500

$500

$500

$500

- If management demands i = 20 %/yrPold=$5180, Pnew=$5295 old is better choice

- If management demands i = 10 %/yrPold=$6340, Pnew=$5585 new is better choice

- 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.

- Simple Example: Choose between 2 pieces of equipment, one of which is better built and has a longer lifetime
- N is not the same for both
- Not a fair comparison with N=2 unless process is to be shut down and both options have no residual value

20 year life

Well Built

Poorly Built

2 year life

Poorly Built

2 year life

- Option 1 - Repeatability

Well Built

20 year life

(Buy 1)

(Buy 10)

Alternative 1

Alternative 2

Purchase Price ($)

10,000

20,000

Annual Op. Cost ($/yr)

1500

1000

1000

Salvage Value ($)

500

3

Service Life (yrs)

2

- Convert the investment and maintenance for both options into a single annual payment

i = 0.15 / yr

Now

=

1000

=

0

0

3

2

1

2

1

3

20,000

1000

1000

1000

9472

9472

9472

In this case, choose alternative 1 because yearly cost is lower.