1 / 19

# Lecture # 6 TUTORIAL on numerical questions

Lecture # 6 TUTORIAL on numerical questions. Remember you are welcome to contact me for queries any time Email : ejazjazz@yahoo.com Cell : 0300 - 5906405. Equivalence. Interest rate = 10%. Present value of Plan 1 P = \$1,000(3.791) = \$3,791 Present value of Plan 2 P = \$5,000

## Lecture # 6 TUTORIAL on numerical questions

E N D

### Presentation Transcript

1. Lecture # 6TUTORIAL on numerical questions

2. Remember you are welcome to contact me for queries any time • Email : ejazjazz@yahoo.com • Cell : 0300 - 5906405

3. Equivalence Interest rate = 10% • 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

4. 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

5. 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

6. F / P = Single payment future worth factor F = P (1 + i)n F / P = (1 + i)n P / F = Single payment present worth factor 1 / (1 + i)n = P/F

7. 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

8. P / A = Equal payment series present worth factor i = 12%, n = 5, A = \$1000 Answer = \$ 3604.579

9. 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 = \$ 3604 i = 12% n = 5 Answer = \$ 1000

10. A = ? , i = 10 %, n = 6 A = A1 + A2 A = \$897.067 We have to apply A/P and A/G Factors

11. 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

12. 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 Physicaland 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 FlowDiagram. • 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.

13. 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 FV = PMT[(1+i)n - 1]/i PMT or A = \$ 100 n = 5 i = 12% Answer = \$ 635 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? Question # 2: What are the principles of Engineering Economy?