DECISION SCIENCES

Dr. Endang Chumaidiyah

ENGINEERING ECONOMY•Engineering economy is concerned with the

financial analysis of engineering projects.•Solution to answers one of two questions:

1) which of the choices considered is best from a financial point of view? (i.e., which equipment offers the required service at lowest cost?), or

2) what is the expected return on investment (ROI) in using this equipment?

Interest Calculations

•The four variables in an engineering economy problem:▫p = present worth▫f= future worth▫i = interest rate▫n= number of interest periods

•Interest rate is the rent charged on the money lent for a defined period of time.

•The compound interest equation is: F=P(1+i)n

Example :If lent you $500.00 three years ago at 6%

interest.How much do you owe me now?F=P(1+i)n= 500(1.06)3=$595.51

•Engineering economy problems have traditionally been solved employing interest tables, and their use is generally instructive.

•Values for four different interest rates are tabulated in table 8-1.

• Example 2: You paid me $595.50 today for a loan made three years

ago. If the interest charged was 6 percent, how much was the original loan?F is known ($595.50) and the prior present value P is unknownP=F(PF,i,n)= 595.5 (PF,6,3) = 595.5 (0.840) = 500.22 = $500

• Cash flow diagram for this problem is shown in this figure.

• Another variable in engineering economy problems is equivalent annual amount. It represents a constant value (A).ExampleIf I put $1000 in a bank that pays 6 percent interest, what equivalent annual amount can I withdraw from this account at the end of each of the next five years?A=P(AP,I,n)=$1000(AP,6,5)=$1000(0.237)=$237

• Cash flow diagram shown in this figure.

Annual Cost Method• Equivalent annual cost (EAC) represents an equal

annual amount required to obtain the use of some service over period time, referred to as the economic life of the investment.

• Annual operating cost (AOC) refers to the assumed equal net outlays one is expected to make each year to support the investment over its economic life (lubricating oil and gasoline, maintenance cost).

• Purchasing equipment, providing net operating costs over its economic life, and obtaining salvage value at the end of its economic life.

• Interest rate i employed in an annual cost problem as the Minimum Acceptable Rate of Return (MARR).

• Example:Machines A and B perform equally well. Machine A initially costs $20.000 has estimated net annual operating expenses of $3000, and has an estimated salvage value at the end of a 10-year economic life of $4000. Machine B will cost $25.000 initially, with annual operating expenses of $2000, and a salvage value at the end of a 15-year economic life of $5000. If your minimum acceptable rate of return (MARR) is 20 percent, which machine should you purchase?.

• The typical annual cost method problem for machine service over a period of years.

EAC=P(AP,I,n)+AOC-F(AF,I,n)

•Equivalent annual cost for Machine A

•Equivalent annual cost for Machine B

Return on Investment (ROI)• To answer the question “If I make this investment,

will there be a sufficient return on the money invested?”

• Example.If I invest $1 by depositing it in a bank paying compound annual interest and one year later I have $1.06 in the account, what is my ROI?

• Employing interest table symbolism.

• PF,i,1 factor of 0.943 is the 6% table

Payback Method• The alternative that returns the initial investment

in the shortest time period is preferred.Example.

Machine A Machine BInitial investment $ 10.000 $ 10.000Net annual income 2.000 2.500

Machine A will pay back the original investment in five years, and Machine B in four years. If we employ the payback criterion, Machine B would be preferred.

• The weakness payback period criterion is in what it fails to consider, return after payback.

• Another weakness is that it does not consider the time value of money during the economic life of equipment.

After tax Analysis• The solutions to all the examples in this section

thus far are for pretax problems.• The following example compares a pretax solution

and after tax solution to the same problem.• Example.

The purchase of a minicomputer control system can reduce clerical labor required in controlling a process from $85.800 to $55.000 annually. The minicomputer control system can be purchased for $80.000 and will have an expected useful economic life of four years at which time it will have no salvage value. Assume straight-line depreciation, a 20 percent minimum attractive rate of return (MARR) before taxes, and an incremental tax rate of 40 percent. Should the minicomputer control system be purchased?.

DECISION THEORY AND UTILITY• In the real world decisions are rarely made with

certainty; most decisions involve risk.• A common criterion for valuing risky decisions is the

maximum (or minimum) expected value criterion.• For example. A contractor constructing a new building is installing

wood flooring this week and also moving in certain pieces of equipment. One major piece of equipment will not arrive until next week. According to weather information, some change in the weather is possible this coming weekend. Although there is an 80 percent chance that the weather will be ideal, the forecast calls for a 15 percent chance of high humidity and a 5 percent chance of rain.

• Install temporary roof : 0.8($1600)+0.15($7600)+0.05($22.600)= $ 3550• No roof : 0.8 ($0)+0.15($12.000)+0.05($30.000) = $ 3300

The best choice using the expected value approach is to construct no roof and wait to see what happens

• Value is specific to individuals. It depends on a person’s relative needs for security, pleasure, peer approval, aspirations, etc.

COST CONTROL ENGINEERING• Inventory, production, and quality control are all

concerned with developing procedures that ultimately affect overall cost control.

• Some control techniques can only be classified as cost control technique.

• If potential purchaser offers to purchase 1000 units of Product A, which could be produced next month for a unit price of $41, the order would not be accepted. Assume 70% normal capacity.

• The order 1000 unit of product A would bring the plant to 90 % capacity. Even though order will not produce profits in the traditional sense, it will help pay some of the overhead expenses of the plant; in fact, to the amount of 1000x$10 = $10.000. So we accept an offer.

THE END