Lecture # 6

TUTORIAL on numerical questions

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EquivalenceYear Plan 1 Plan 2

0 - $5,0001 $1,000

2 $1,000

3 $1,000

4 $1,000

5 $1,000

Total $5,000 -• Present value of Plan 1

– P = $1,000(3.791) = $3,791

• Present value of Plan 2– P = $5,000

• Alternative 2 is better than alternative 1 since alternative 2 has a greater present value

Interest rate = 10%

Factors • F / P = Single payment future worth factor

• P / F = Single payment present worth factor

• F / A = Equal payment series future worth factor

• A / F = Equal payment series sinking fund factor

• P / A = Equal payment series present worth factor

• A / P = Equal payment series capital recovery factor

• A / G = Arithmetic gradient series factor

• F / G = Arithmetic gradient future worth factor

• P / G = Arithmetic gradient present worth factor

• Geometric Gradient factor

Factors - Formulae

• i = annual interest rate• n = interest period• P = present principle amount• A = Equal annual payments• F = Future amount• G = Annual change or gradient

F / P = Single payment future worth factor

P / F = Single payment present worth factor

F = P (1 + i)n

1 / (1 + i)n = P/F

F / P = (1 + i)n

F / A = Equal payment series future worth factor

What will be the future worth of an amount of $ 100 deposited at the end of each next five years and earning 12 % per annum?

Answer = $ 635

A / F = Equal payment series sinking fund factor

It is desired to accumulate $ 635 by making a series of five equal annual payments at 12 % interest annually, what will be the required amount of each payment?

Answer = $ 100

P / A = Equal payment series present worth factor

i = 12%, n = 5,

A = $1000

Answer = $ 3604.579

A / P = Equal payment series capital recovery factor

A car has useful life of 5 years. The maintenance cost occurs at the end of each year. The owner wants to set up an account which earns 12 % annually on an amount of $ 3604 to cater for this maintenance cost. What is the maintenance cost per annum?

P = $ 3604i = 12%n = 5 Answer = $ 1000

We have to apply A/P and A/G Factors

A = ? , i = 10 %, n = 6A = A1 + A2A = $897.067

Sarah and her husband decide they will buy $1,000 worth of utility stocks beginning one year from now. Since they expect their salaries to increase, they will increase their purchases by $200 per year for the next nine years. What would the present worth of all the stocks be if they yield a uniform dividend rate of 10% throughout the investment period and the price/share remains constant?

Solution• How should we go about this questions• Question involves P/A Factor and P/G Factor• A = $ 1000, i = 10%, n = 9, G = $200• PW of the base amount ($1,000) is: = $5767.72• PW of the gradient is: = $3053.50• Total PW = 5767.72 + 3053.50 = $8821.22

Quiz - 1• The essential pre-requisite of successful engineering application is Economic

Feasibility.

• In evaluation of most ventures, Economic Efficiency must take precedence over physical efficiency.

• Engineers are confronted with two interconnected environments the Physical and Economical.

• Value is an appraisal of utility in terms of medium of exchange.

• Exchange is possible when the object is not valued equally by parties.

• Marginal Cost is an increment of output whose cost is barely covered by the monetary return derived from it.

• Interest is the rent for loaned money.

• The costs and benefits of engineering projects over time are summarized on a Cash Flow Diagram.

• Generally, money grows (compounds) into larger future sums and is Smaller / Discounted in the past.

• Sink cost is generally disregarded in economics.

• Life Cycle Cost considers all types of costs over the life of a product.

• Net efficiency is Physical Efficiency times economical efficiency.

Question # 2: What are the principles of Engineering

Economy?

• Analyze the idea

• Develop the alternatives

• Focus on differences in the

alternatives

• Use a consistent view point

• Use a common unit of

measurement

• Consider all relevant criteria

• Make uncertain explicit

• Revisit your decision

Question # 3: What will be the future worth of an amount of $ 100 deposited at the end of each next five years and earning 12 % per annum?

FV = PMT[(1+i)n - 1]/iPMT or A = $ 100 n = 5i = 12%Answer = $ 635

Geometric Gradient

Example

• Airplane ticket price will increase 8% in each of the next four

years. The cost at the end of the first year will be $180. How

much should be put away now to cover a students travel home

at the end of each year for the next four years? Assume

interest rate as 5%.

67.715$03.0

11928.0180

08.05.

)05.1()08.1(1180

)1()1(1

44

gi

igAP

nn

Example • A graduating TE is going to make $35,000/yr with Granite Communications. A total of 10% of

the TE salary will be placed in the mutual fund of their choice. The TE can count on a 3%

salary increase with the standard of living increases for the next 30 years of employment. If

the TE is aggressive and places their retirement in a stock index fund that will average 12%

over the course of their career, what can the TE expect at retirement?

A = 35,000 x 0.1 = 3,500i = 12%g = 3%n = 30

714,070,1$92.305350009.0

)03.1()12.1(3500

)1()1(

3030

gi

giAF

nn