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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

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## Lecture # 6 TUTORIAL on numerical questions

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**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 • Alternative 2 is better than alternative 1 since alternative 2 has a greater present value**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**F = P (1 + i)n F / P = (1 + i)n P / F = Single payment present worth factor 1 / (1 + i)n = P/F**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 = $ 3604 i = 12% n = 5 Answer = $ 1000**A = ? , i = 10 %, n = 6**A = A1 + A2 A = $897.067 We have to apply A/P and A/G Factors**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 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.**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?**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%.**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

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