Engineering Economics. ME 416/516. Topics. 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. Motivation.
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
First Costs or Initial Costs are the costs necessary to implement a project, including:
Some costs of starting projects may be offset by immediate savings:
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:
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?
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 = $104,000.
Savings/yr = $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%?
[Savings/yr - (First Cost/Lifetime)]/First Cost
=[$55,000/yr - ($104,000/4 yr)]/$104,000/yr
ROI = 0.279 = 27.9% (per year)
F = P(1 + i)norP = F/(1 + i)n
F/(1 + i)n = $1000/(1 +0.06)8 = $627.41
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
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.
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.
and federal, are a
fact of life
(-1) * First Cost * Deprec. Rate * Tax Rate
due to inflation in money’s value.
than others (like health care and education), the average rate of inflation is reflected by the Consumer Price Index (CPI).
CPI-1970/CPI-1991 = 38.8/136.2 = 0.285
$now = $future/(1 + I)n
so $15,000 * (1 + 0.045)10 = $23,300
$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%
(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: