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Engineering Economics

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Engineering Economics

ME 416/516

- Motivation
- Types of Costs
- Two Typical Scenarios for Analysis
- “Simple” Methods
- The Time Value of Money
- The Present Value Method
- The Project Timetable
- Accounting for Taxes and Depreciation
- Accounting for Inflation

- The objective is to introduce methods of economic analysis for energy engineering decision making in a corporate, institutional or governmental setting.
- Legal, environmental, public relations, energy efficiency, safety and ethical considerations are important, but the most important consideration for engineering design decisions is economics-- “dollars and cents”.

- If factors can be valued in $, they should be included in the economic analysis.
- The analysis methods here do not require tables. The use of computer spreadsheets (like EXCEL) will be emphasized.
- A spreadsheet is a computerized accountant’s ledger- ideal for economic analysis.
- Spreadsheets give greater flexibility
than table-based methods.

There are usually two types of costs associated with an engineering project, one-time costs, which include first costs and salvage costs, and annual costs

(or benefits) that occur every

year or several

years of the

project.

First Costs or Initial Costs are the costs necessary to implement a project, including:

- Costs of new equipment
- Costs of shipping and installation
- Costs of renovations needed to install equipment
- Cost of engineering
- Cost of permits,
licenses, etc.

Some costs of starting projects may be offset by immediate savings:

- Deduct gains from sale of replaced equipment
- Deduct investment tax credits. These are government tax incentives to purchase certain equipment, and reduce first costs by reducing taxes.

We are attempting to estimate the total cost of doing a project. Cost is reduced if we can sell the equipment at end of project.

Salvage value is the money that can be obtained at the end of the project by selling equipment. Salvage value is a benefit rather than a cost.

Annual Costs and Benefits are costs and benefits of the project that accrue over two or more years of the project, including:

- Direct operating costs such as labor, supervision, janitorial, supplies, maintenance, material, electricity, fuel, etc.
- Indirect operating costs sometimes included, such as a portion of building rent, a portion of secretarial expenses, etc.

- Sometimes include tax costs or benefits. If the project increases profits, it also increases taxes, which are an additional cost. There is a tax benefit if project reduces profits.
- Depreciationof equipment for tax purposes is also a tax benefit.
- Finally, need to include savings or profits from the project.

New Project- selecting from two or more alternative solutions. The objective in this scenario is to find the lowest cost solution that accomplishes some objective.

Example- Your client, a school district is building a new school. You are designing the heating and cooling system. Should you select: (1) a gas boiler plus an air conditioner, (2) an electric air-to-air heat pump, or (3) a ground-source heat pump?

Replacement Project- a method for accomplishing the goal is already in place, but a new alternative solution can accomplish the goal more cheaply. Do the future savings from the new method justify spending money now to cover the first costs?

Example- Should 2 existing machines with an operator each be replaced with a single new machine with the same output and a single operator?

- Simple Payback Period (SPP)- The time required for savings to offset first costs.
- Simple Return on Investment (ROI)- The simple percent return the project pays over its life.
- These methods are “simple” because they do not consider the time value of money.
- Simple methods are OK for investments that are very good and pay off over short time periods.

For a replacement project:

For comparing two projects A and B, where the first cost of A is greater than B, but the annual costs of A are lower than B:

The simple rate of return on investment is:

Lifetime is the life of the project.

The Acme Threaded Products Corp. now uses 2 machinists to operate its 4 screw machines. Each screw machine costs $12,000/yr for electricity, maintenance and wasted materials. Each machinist costs $35,000/yr. Find SPP for replacing 4 old machines with 2 new machines @ $60,000 each that produce same output, cost $14,000/yr each to operate, and can be run by only one machinist? The old machines can be sold for $4000 each.

- First cost is 2 * $60,000 - 4 * $4000, or:
First cost = $104,000.

- Savings/yr is $35,000 (labor), + 4 * $12,000 - 2 * $14,000 (operation), or:
Savings/yr = $55,000/yr

- SPP = First cost/(Savings/yr)
= $104,000/($55,000/yr)

or SPP = 1.89 yr

For the previous example, which had a first cost of $104,000 and an annual savings of $55,000/yr, the expected useful life of the equipment is 4 years. Acme Threaded Products Corp. requires a minimum simple return on investment, ROI, of 10%. (This minimum simple ROI is often called the corporate hurdle rate.) Does this investment exceed the corporate hurdle rate of 10%?

ROI =

[Savings/yr - (First Cost/Lifetime)]/First Cost

=[$55,000/yr - ($104,000/4 yr)]/$104,000/yr

ROI = 0.279 = 27.9% (per year)

- “Would you prefer $100 now or a year from now?” Most would prefer “cash up front” because $1 today is worth more than $1 some time in the future. You could invest $1 today and have more than $1 in a year.
- A project with a life of several years has cash flows at various times. To consider dollar amounts at different times, we need to put all amounts on an equal basis taking the time value of money into account.

- Present Value is the value now of an amount of money Freceived n years in the future.
- Future Value is value n years in the future of an amount of money Preceived now.
- If we can earn interest rate i on investments, the relationship between P and F is:
F = P(1 + i)norP = F/(1 + i)n

- Problem- If a savings bond with a yield of 6% matures with a value of $1000 in 8 years, what does it cost now?
- Solution- This is equivalent to asking what is the present value P of F = $1000 received n = 8 years in the future at an interest rate of i = 6%, soP =:
F/(1 + i)n = $1000/(1 +0.06)8 = $627.41

- A savings bond with these terms would cost this amount today.

- Problem- If you invest $500 today at 9% interest, how much time is required for the investment to be worth $1200?
- Solution- Here P = $500, F = $1200, i = 9% and n is the unknown, so:
F = P(1 + i)n or F/P = (1 + i)n

ln F/P = n ln(1 + I)

or: n = [ln F/P]/[ln(1 + i)]

n = [ln 1200/500]/[ln(1 + 0.09)] = 10.2 yr

- Method determines the value in today’s dollars of a set of cash inflows and outflows at various times from now into the future.
- Set up a table (spreadsheet!) of all costs and savings associated with the project, making costs negative and income/savings positive.
- Sum up all cash flows from each year to get the net cash flow (NCF) for that year.

Convert each NCF from all years n to present value by multiplying NCF * 1/(1 + i)n. This process is called discounting the NCF to account for present value. Most large companies have a “discount rate” or “rate of return” i that they use for all projects.

Add up discounted cash flows for all years. This is the present value of the investment.

- Replacement projects: If present value is greater than zero, then the accumulated, discounted savings exceed costs, and the investment should be made.
- Using a spreadsheet, i can be adjusted so that P is exactly zero. The value of i for which P = 0 is the discounted rate of return. This means that the return on the proposed new investment is the same as putting the money in the bank at an interest rate of i.

- New projects: The present value will normally be negative for each alternative. The most economically attractive alternative is the one having the smallest (negative) value of P.
- The smallest P corresponds to the alternative that will cost the least- cheapest.

- One potentially confusing aspect of the present value method is in which year to place various expenses.
- The lifetime of the project must first be determined. The economic analysis covers all years of the project life.
- For analysis, the project starts on the very last day (Dec. 31) of Year 0. All first costs plus investment tax credits are considered to occur in Year 0 for most projects.

- All operating costs and savings plus any tax and depreciation effects that occur at any time in Year 1 are assumed to take place on the final day of Year 1 , that is, one year after project start. Similarly, all costs or savings occurring in any year are considered to occur at the end of that year.
- The salvage value is a positive cash flow taken at the end of the final project year.

Given: CrimsonCorp lights its factory with fluorescent lights with a power bill of $12,000/yr and a lamp replacement cost of $5200/year. CrimsonCorp is considering high pressure sodium lights that could be operated for a power cost of $5000/yr and a replacement cost of $2600/yr. The cost of removing the old fixtures and installing new fixtures is $35,000. The life of the new lamp fixtures is estimated to be at least 8 years.

- Find- (a) For a 10% minimum rate of return, should lights be replaced? (b) What is the actual rate of return on this investment?
- Solution- The First Cost is $35,000. The annual savings are the cost of operating old lights less the cost of operating new lights. So, the annual savings are ($12,000 + $5200) - ($5000 + $2600) = $9600/yr. A spreadsheet solution is used for this replacement problem.

- The actual rate of return is the value of i that causes the present value to exactly equal zero.
- This is easily determined once the Excel spreadsheet is set up by using Goal Seek...

- Taxes, local, state
and federal, are a

fact of life

- Taxes are applied to
profits, therefore:

- an expense that decreases profit also decreases taxes owed by the company.
- a cash inflow that increases profit also increases taxes owed.

- To account for taxes on the cash flow in a particular year, add to the NCF the quantity: -1 * NCF * tax rate. The negative sign shows that a “+” cash flow has a “-” tax effect.
- The federal corporate tax rate varies over time (with politics and circumstances). Consult your accountant for details!!
- The primary tax effect described here affects the net cash flow any time there is any cash flow in a year-- positive or negative.

- In addition to the primary tax effect, in some years there can be a tax effect from investment tax credits or depreciation.
- To encourage investment in some goods (like energy efficiency), the government some-times legislates an investment tax credit.
- An investment tax credit is a specified percentage of the purchase price that can be subtracted directly from tax bill in the year of the purchase (Year 0 for our purposes).

- A tax credit is an amount subtracted directly from the taxes owed. The taxpayer’s taxes are reduced by the entire amount of a credit.
- A tax deduction is an amount subtracted from income. The company benefits only by the tax rate times the deduction for a tax deduction.

- A company cannot deduct the cost of capital purchases from its income as a “cost of doing business.” The IRS regards capital purchases as investments rather than expenses.
- The company can “write off,” or depreciate capital purchases over a number of years.
- Depending on type of equipment, a company can claim a depreciation deduction of a specified amount for a specified number of years, following a “depreciation schedule.”

- The tax effect of a depreciation deduction can be calculated as:
(-1) * First Cost * Deprec. Rate * Tax Rate

- Note that First Cost is a negative quantity, so the depreciation effect is a positive cash flow.
- The most commonly used gov’t approved depreciation schedule is the Accelerated Cost Recovery System (ACRS):

- Depending on what the government decides is the lifetime of a type of capital equipment, a 3, 5, 7 or 10-year depreciation schedule is followed.
- Note: the fact that an item has been depreciated 100% for tax purposes has no relation to its actual value or salvage value.

- Given- Previously, CrimsonCorp was considering an investment with a first cost of $35,000 and annual savings of $9600/yr. Assume that CrimsonCorp is entitled to a 5% investment tax credit on the first cost, that pays a 34% tax on profits, and that the federal gov’t classifies the lighting equipment in the “5-year” property class.

- Find- (a) the present value of the investment assuming a 10% discount rate, and (b) the actual rate of return for the investment.
- Solution- Again, a computer spreadsheet solution is recommended. The 5% investment credit is the only Year 0 tax effect, there is a primary tax effect for Years 1 - 10, and the 5-year class depreciation schedule is followed in Years 1 - 6...

- Prices tend to increase over time
due to inflation in money’s value.

- Although some goods inflate faster
than others (like health care and education), the average rate of inflation is reflected by the Consumer Price Index (CPI).

- The ratio of the CPI for one year to that of another is inversely proportional to the ratio of the buying power of a dollar for the same two years.

- Given- The average CPI (based on 1982-84 = 100) for 1970 was 38.8 and the CPI for 1991 was 136.2. How much could a dollar buy in 1991 compared to 1970.
- Sol’n- The ratio of buying power is inversely proportional to the ratio of CPI:
CPI-1970/CPI-1991 = 38.8/136.2 = 0.285

- Therefore, the 1991 dollar is worth only 28.5% of the 1970 dollar for the average consumer purchase.

- There is a simple relationship between the purchasing power of a dollar now compared to the value of a dollar n years in the future assuming that the inflation rate is I:
$now = $future/(1 + I)n

- Note that i is used for interest rate, I for inflation rate.

- Given- If it costs you $15,000/year to live now, how much will it cost in 10 years if the inflation rate is 4.5%?
- Sol’n- $Future = $Now * (1 + I)n
so $15,000 * (1 + 0.045)10 = $23,300

- Given- For medical care, the 1970 CPI was 34.0 and the 1991 CPI was 177.
- Find- the average inflation rate for medical care over the 21 year period.
- Sol’n- $34 in 1970 = $177 in 1992, so
$177 = $34 * (1 + I)21

ln(177/34) = 21 ln(1 + I)

exp[(1/21) ln(177/34)] = 1 + I

I = exp[(1/21) ln(177/34)] - 1= 0.0817= 8.17%

- Note that a dollar in the future is worth less than a dollar now for two different reasons:
(1)You could invest your dollar today and have more than a dollar at the future time, so it takes more than a dollar in the future to have the present value of a dollar today (time value of money).

(2)Your dollar in the future won’t be able to buy as much as it could today because of inflation.

(1) Inflate all costs and savings using the available information, so that net cash flow before taxes is accurate for the year in question. Lacking better information, use the average inflation rate.

(2) Calculate NCFAT the usual way.

(3) Calculate the present value discount factor the same as before, using discount rate i.

PV Discount = 1/(1 + i)n

(4) Calculate an inflation discount factor using inflation rate I:

Inflation Discount = 1/(1 + I)n

(5) Calculate an overall discount factor:

Overall Disc. = Inflation Disc. * PV Disc.

(6) The Present Value for a year is the NCFAT times the overall discount factor.

Given: Acme Axles needs a heat treating oven for its account with Deutschmobile AG. An electric oven and natural-gas fired oven are considered. For both options assume a tax rate of 34%, 7-year depreciation, a 10-year life, no tax credits, salvage value of 10% of first cost, discount rate of 10%, inflation rate of 4%, electricity inflation rate of 5%, and natural gas inflation rate of 7%, All costs of the two ovens are identical except for the following:

Electric Oven- First cost is $125,000, electricity cost is $35,000/yr, maintenance cost is $6000/yr.

Gas Oven- First cost is $185,000, fuel cost is $10,000/yr, maintenance cost is $9000/yr.

Find: which option should Acme select?

Sol’n: Use spreadsheet solution method, as follows: