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Engineering Economics

Engineering Economy

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It deals with the concepts and techniques of analysis useful in evaluating the worth of systems, products, and services in relation to their costs.

Engineering Economy

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It is used to answer many different questionsWhich engineering projects are worthwhile?

Has the mining or petroleum engineer shown that the mineral or oil deposits is worth developing?

Which engineering projects should have a higher priority?Has the industrial engineer shown which factory

improvement projects should be funded with the available dollars?

How should the engineering project be designed?Has civil or mechanical engineer chosen the best

thickness for insulation?

Basic Concepts

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1. Cash flow2. Interest Rate and Time value of money3. Equivalence technique

1. Cash Flow

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Engineering projects generally have economic consequences that occur over an extended period of time› For example, if an expensive piece of

machinery is installed in a plant were brought on credit, the simple process of paying for it may take several years.

› The resulting favorable consequences may last as long as the equipment performs its useful function.

Each project is described as cash receipts or disbursements (expenses) at different points in time

Categories of Cash Flows

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The expenses and receipts due to engineering projects usually fall into one of the following categories:

1. First cost: expense to build or to buy and install

2. Operations and maintenance (O&M): annual expense, such as electricity, labor, and minor repairs

3. Salvage value: receipt at project termination for sale or transfer of the equipment (can be a salvage cost) –future

4. Revenues: annual receipts due to sale of products or services

5. Overhaul: major capital expenditure that occurs during the asset’s life

Cash Flow diagrams

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Cash flow diagrams are a mean of visualizing (and simplifying) the flow of receipts and disbursements (for the acquisition and operation of items in an enterprise).

The diagram convention is as follows: 1. Horizontal Axis : The horizontal axis is marked

off in equal increments, one per period, up to the duration of the project.

2. Revenues : Revenues (or receipts) are represented by upward pointing arrows.

3. Disbursements : Disbursements (or payments) are represented by downward pointing arrows.

Cash Flow diagrams

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All disbursements and receipts (i.e. cash flows) are assumed to take place at the end of the year in which they occur. This is known as the "end-of-year" convention.

Arrow lengths are approximately proportional to the magnitude of the cash flow.

Expenses incurred before time = 0 are sunk costs, and are not relevant to the problem.

sunk costs are retrospective (past) cost that have already been incurred and cannot be recovered.

Since there are two parties to every transaction, it is important to note that cash flow directions in cash flow diagrams depend upon the point of view taken.

Typical cash flow diagram

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Figure shows cash flow diagrams for a transaction spanning five years. The transaction begins with a $1000.00 loan. For years two, three and four, the borrower pays the lender $120.00 interest (riba-haram). At year five, the borrower pays the lender $120.00 interest plus the $1000.00 principal.

An Example of Cash Flow Diagram

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A man borrowed $1,000 from a bank at 8% interest. Two end-of-year payments: at the end of the first year, he will repay half of the $1000 principal plus the interest that is due. At the end of the second year, he will repay the remaining half plus the interest for the second year.

Cash flow for this problem is:End of year Cash flow 0 +$1000 1 -$580 (-$500 - $80) 2 -$580 (-$500 - $80)

Cash Flow Diagram

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$1,000

0

1 2

$580 $580

year

2. Time value of money

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The time-value of money is the relationship between interest and time. i.e.

Time Value of Money

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Money has valueMoney can be leased or rentedThe payment is called interestIf you put $100 in a bank at 9% interest for one

time period you will receive back your original $100 plus $9

Original amount to be returned = $100Interest to be returned = $100 x .09 = $9

Interest Rate

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Interest is a rental amount charged by financial institutions for the use of money.

Also called as the rate of capital growth, it is the rate of gain received from an investment.

It is expressed on an annual basis.

For the lender (BANK), it consists, for convenience, of (1) risk of loss, (2) administrative expenses, and (3) profit or pure gain.

For the borrower, it is the cost of using a capital for immediately meeting his or her needs.

Types of Interest

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Simple Interest Simple Interest : I = Pni. P = Principal i = Interest rate n = Number of years (or periods) I = Interest

Interest is due at the end of the time period. For fractions of a time period, multiply the interest by the fraction.

Example : Suppose that $50,000 is borrowed at a simple interest rate of 8% per annum. At the end of two years the interest owed would be: I = $ 50,000 * 0.08 * 2

= $ 8,000

Compound Interest

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Compound interest arises when interest is added to the principal, so that, from that moment on, the interest that has been added also earns interest. This addition of interest to the principal is called compounding.

The effect of earning 20% annual interest on an initial $1,000 investment at various compounding frequencies

Compound Interest

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Compound interest is most commonly used in practice.

Total interest earned = In = P (1+i)n - PWhere,

P – present sum of money = 100i – interest rate = 9%n – number of periods (years) = 2

I2 = $100 x (1+.09)2 - $100 = $18.81

Future Value of a Loan With Compound Interest

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Amount of money due at the end of a loanF = P(1+i)1(1+i)2…..(1+i)n or F = P (1 + i)n

Where,F = future value and P = present value

i = 9%, P = $100 and say n= 2. Determine the value of F. F = $100 (1 + .09)2 = $118.81

Notation forCalculating a Future Value

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Formula:F=P(1+i)n is the

single payment compound amount factor.

Try this…

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Example 1 : Let the principal P = $1000, the interest rate i = 12%, and the number of periods n = 4 years. The future sum is:F = $1000 [1 + 0.12]4

= $1,573.5

Cash Flow for Single Payment Compound Amount

Notation for Calculating a Present Value

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Future Value, F=P(1+i)n

Present Value, P=(F/(1+i)n )single payment present worth factor.

Example 1 : Let the future sum F = $1000, interest rate i =

12%, and number of periods n = 4 years. The single payment present-worth factor is:

P = F = $1000 =

$635.50 [1 + i]n [ 1 + 0.12 ]4

The present worth P = $635.50.

Notation for Calculating Equal Payment Series - Sinking Fund Factor

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Given a future amount F, the equal payments compound-amount relationship is: i A = F * --------------- [ 1 + i ]n - 1

A = required end-of-year payments to accumulate a future amount F.

Example 1: Let F = 1000, i = 12%, and n = 4 years.

0.12 A = 1000 * ------------------ = 209.2 [ 1 + 0.12 ]4 - 1

Notation for Calculating Equal Payment Series - Presents Worth Factor

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This can be described as [ 1 + i ]n - 1 P = A * --------------- i * [ 1 + i ]n

Example 1: Let A = 100, i = 12%, and n = 4 years. [ 1 + 0.12 ]4 - 1 P = 100 * --------------------- = 303.7 0.12 * [ 1 + 0.12 ]4

An Example of Present Value

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Example : If you wished to have $800 in a savings account at the end of four years, and 5% interest we paid annually, how much should you put into the savings account?

n = 4, F = $800, i = 5%, P = ?

Method 1 : Manually calculated

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n = 4, F = $800, i = 5%, P = ? P=(F/(1+i)n )P = 658.16

Method 2 : Compound Interest table

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n = 4, F = $800, i = 5%, P = ?P = PV(5%,4,,800,0) = -$658.16Refer Compound Interest TableSingle Payment, Present Worth Factor = 0.8227P = F * 0.8227 = 658.16

An Example of Future Value

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Example: If $500 were deposited in a bank savings account NOW, how much would be in the account 3years, if the bank paid 6% interest compounded annually?

Given P = 500, i = 6%, n = 3,

Method 1 : Manually calculate

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Given P = 500, i = 6%, n = 3

F=P(1+i)n

F = 595.5

Method 2 : Compound Interest table

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Given P = 500, i = 6%, n = 3use F = FV(6%,3,,500)

Single paymentCompound Amount FactorFactor = 1.191F = P(1.191) = 595.5

Engineering Economic Analysis Calculation

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Generally involves compound interest formulas (factors)

Compound interest formulas (factors) can be evaluated by using one of the three methods

1. Interest factor tables2. Calculator3. Spreadsheet

Economic Analysis Methods

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Three commonly used economic analysis methods are

1. Present Worth Analysis2. Annual Worth Analysis3. Rate of Return Analysis

1. Present Worth Analysis

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Steps to do present worth analysis for selecting a single alternative (investment) from among multiple alternatives› Step 1: Select a desired value of the return on

investment (i) › Step 2: Using the compound interest formulas

bring all benefits and costs to present worth for each alternative

› Step 3: Select the alternative with the largest net present worth (Present worth of benefits – Present worth of costs)

Present Worth Analysis

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A construction enterprise is investigating the purchase of a new dump truck. Interest rate is 9%. The cash flow for the dump truck are as follows:First cost = $50,000, cash Pannual operating cost = $2000, annual income = $9,000, Asalvage value is $10,000, S (depreciation 9%

per year)life = 10 years. n

Is this investment worth undertaking?Evaluate net present worth = present worth

of benefits – present worth of costs

Present Worth Analysis

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Present worth of benefits = $9,000(PA,9%,10) = $9,000(6.418) = $57,762

Present worth of costs = $50,000 + $2,000(PA,9%,10) - $10,000(PF,9%,10)= $50,000 + $2,000(6.418) - $10,000(0.4224) = $58,612

Net present worth = $57,762 - $58,612 < 0 Do not invest

What should be the minimum annual benefit for making it a worthy of investment at 9% rate of return?

Present Worth Analysis

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Present worth of benefits = A(PA,9%,10) = A(6.418)

Present worth of costs = $50,000 + $2,000(PA,9%,10) -

$10,000(PF,9%,10)= $50,000 + $2,000(6..418) - $10,000(.4224) = $58,612A(6.418) = $58,612 A = $58,612/6.418 =

$9,312.44- if annual income more than $ 9,312.44 it

worth to buy

Cost and Benefit Estimates

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Present and future benefits (income) and costs need to be estimated to determine the attractiveness (worthiness) of a new product investment alternative

Annual costs and Income for a Product

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Annual product total cost is the sum of annual material, Labor, overhead (salaries, taxes, marketing expenses,

office costs, and related costs), annual operating costs (power, maintenance,

repairs, space costs, and related expenses), and annual first cost minus the annual salvage

value.Annual income generated through the sales of

a product number of units sold annually x unit price