ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.15 : Based on Time Discounting of Money, let y = x / (1 + r) ^ n, where ^ = power, r = interest rate for the period in decimal value, n = number of periods, y = current value, x = future value. According to Present Value Addition Rule, the present value of a set of future payments is equal to sum for present value of each of the payments. In an engineering business, 3 future payments are to be made : $100 after 1 year; $500 after 2 years; $200 after 3 years. (a) Form an equation of y as a function of r. (b) Find the value of y in 5 decimal places if r = 8 %.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.16 : An engineer would like to invest his money in a home, business and bond. The implicit interest payment frequency is monthly for home loans, quarterly for business loans; semi annually for bonds. A generalized mathematical formula to calculate I = interest rate equivalence is I = (1 + i / N) ^ N - 1 where i = annual interest rate, N = number of payment per year. (a) Calculate the value of N for : (i) home loans; (ii) business loans; (iii) bonds. (b) For i = 0.08, find the value of I for : (i) home loans; (ii) business loans; (iii) bonds.

Sir i'm working in an organisation last since 4 years and kept an own company, i want to whether it is must i should get electrical license for HT works for above 33KV works. what will be the cost. Please suggest.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.18 : An investor is planning to purchase a small office for biochemical engineering consultancy on loan. In the calculation of the discount of payment in arrears, the following formula is used : y = 1 / (1 + r) ^ 1 + 1 / (1 + r) ^ 2 + 1 / (1 + r) ^ 3 + ... + 1 / (1 + r) ^ n where y = present value, r = interest rate of discount, n = number of payment, ^ = power used in certain computer languages for mathematics. (a) What is the meaning of : arrears? (b) Find a mathematical equation of y (1 + r). (c) Calculate, in less than 3 terms, y as a function of r and n.

Hard shifting of gears would result when a) There is rapid engagement of gears b) There is less heat dissipation c) The clutch inertia is very high d) There is no clutch free pedal play

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The main disadvantage of sliding mesh type of gear box is a) High noise level b) Low mechanical efficiency c) High skill required by the driver d) all of the above

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Double declutching is used for the engagement of gears in a) Sliding mesh type of gear box b) Constant mesh type of gear box c) Synchromesh type of gear box d) None of the above

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High speed gear is obtained in the case of sliding mesh gear box when a) Motion is transferred through the clutch gear to lay shaft gear b) Main shaft is coupled to the clutch shaft c) Smallest gear on the lay shaft is engaged d) Largest gear on the lay shaft is engaged

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In case of a constant mesh gear box, when the left dog clutch is slid to the left by means of the selector mechanism, a) Direct gear is obtained b) First gear is obtained c) Second gear is obtained d) Reverse gear is obtained

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In a constant mesh gear box, when the right dog clutch is engaged with the gear on the right side of the main shaft, a) Direct gear is obtained b) First gear is obtained c) Second gear is obtained d) Reverse gear is obtained

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Double declutching is required in constant mesh gear box a) To equalize the speed of main shaft gear and the sliding dog b) To provide direct gear c) To equalize the speed of countershaft gear and the sliding dog d) None of the above

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The number of spring loaded balls provided circumferentially in a synchromesh gear box is a) 3 b) 4 c) 5 d) 6

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For the second gear in a synchromesh system, the drive transfer will be through a) U1, U2, C, F2, and splines b) U1, U2, C, F1, and splines c) U1, U3, D, F3, and splines d) U1, U4, U5, E, F2, and splines

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Interlocking mechanism is provided in the gear box cover a) To prevent engagement of two gears simultaneously b) To prevent engagement of reverse gear c) To facilitate the gear selection d) To avoid the necessity of having neutral gear

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.19 : In the purchase of a unit of engineering office, a loan has been made to a bank with the following details : Term N = 30 years; interest rate R = 8.07 % / year; security : primary residence; present value pv = $450000; salary = $75000 / year or $56000 / year after tax. (a) Let the discounted present value PV = [ 1 - 1 / (1 + r) ^ n ] / r for arrears, where r = interest rate of discount, n = number of payment, ^ = symbol for power. If the loan repayment was made monthly : (i) calculate the value of r where r = R / k and R is in decimal value; (ii) find the value of n where n = kN; (iii) estimate the value of k where k = number of repayment per year; (iv) calculate the value of PV based on the formula of discounted present value. (b) Calculate the monthly repayment of the loan, MR based on the following formula : pv = PV x MR. (c) Find the percentage of salary remains after paying the loan every month.