Question { 1199 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.14 : For a formula of y = [ 1 - 1 / (1 + r) ^ n ] / r, let y = present value, r = interest rate / year, n = number of years to future value of 1 : (a) Find a simple mathematical relationship of y as a function of r when n = 1 million years. (b) What is the present value of $10000, if the annual discount rate is 10 %, forever?

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.14 : (a) When n = 1 million years, it could be approximated as n = infinity. Then y = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + r) ^ (infinity) ] / r = [ 1 - 1 / (infinity) ] / r = (1 - 0) / r = 1 / r. (b) For future value of $10000 and n = infinity forever, use the equation of y = 1 / r for future value of 1 and interest rate r. As r = 10 % = 0.1, then y = x / r = $10000 / 0.1 = $100000, where x = future value. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1229 }

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 %.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.15 : (a) First future payment : n = 1, x = $100. Second future payment : n = 2, x = $500. Third future payment : n = 3, x = $200. Then total y = sum of x / (1 + r) ^ n = 100 / (1 + r) + 500 / (1 + r) ^ 2 + 200 / (1 + r) ^ 3. (b) When r = 8 % = 0.08, y = 100 / (1 + 0.08) + 500 / (1 + 0.08) ^ 2 + 200 / (1 + 0.08) ^ 3 = 680.02845 (5 decimal places). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1427 }

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.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.16 : (a) (i) There are 12 months / year, then N = 12. (ii) There are 4 quarters / year, then N = 4. (iii) There are 2 semiannuals / year, then N = 2. (b) Use I = (1 + i / N) ^ N - 1 as a reference. (i) I = (1 + 0.08 / 12) ^ 12 - 1 = 0.083. (ii) I = (1 + 0.08 / 4) ^ 4 - 1 = 0.08243216. (iii) I = (1 + 0.08 / 2) ^ 2 - 1 = 0.0816. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1284 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.17 : In the engineering calculations of interest rate caused by inflation, General Inflation Effect and Fisher Effect may be considered. Let I = inflation rate, R = nominal interest rate, r = real interest rate. According to Fisher Effect, (1 + R) = (1 + r) (1 + I). According to General Inflation Effect, r = R - I. (a) If I = 0.1 for all effects, both the values of R and r in the Fisher Effect are the same as R and r in the General Inflation Effect, find the values of R and r. (b) If R has the same value caused by both General Inflation Effect and Fisher Effect, find the possible values of R, r and I in term of R etc.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.17 : (a) For Fisher Effect, 1 + R = 1.1 (1 + r) = 1.1 + 1.1 r, 0.1 = R - 1.1 r (Equation 1). For General Inflation Effect, r = R - 0.1, 0.1 = R - r (Equation 2). Equation 2 - Equation 1 : r = 0, then R = 0.1. (b) (1 + R) = (1 + r) (1 + I) = 1 + I + r + rI, then R = (I + r) + rI (Equation 3). Let R = I + r (Equation 4). Equation 3 - Equation 4 : 0 = Ir, where (I = 0) or (r = 0) or (I = 0 and r = 0). If I = 0, r = R. If r = 0, I = R. If I = 0 and r = 0, R = 0 = 0 R = 1 R. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1292 }

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.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.18 : (a) Backward. (b) Let Equation 1 : y = 1 / (1 + r) ^ 1 + 1 / (1 + r) ^ 2 + ... + 1 / (1 + r) ^ n. Multiply Equation 1 with (1 + r) will produce Equation 2 : y (1 + r) = 1 + 1 / (1 + r) ^ 1 + ... + 1 / (1 + r) ^ (n - 1). (c) Equation 2 - Equation 1 will produce yr = 1 - 1 / (1 + r) ^ n, then y = [ 1 - 1 / (1 + r) ^ n ] / r. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1444 }

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.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.19 : Value of k has to be found first : 12 months / year, then k = 12. (a) (i) r = R / k = 0.0807 / 12 = 0.006725. (ii) n = kN = 12 x 30 = 360. (iii) k = 12. (iv) PV = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + 0.006725) ^ 360 ] / 0.006725 = 135.382. (b) MR = pv / PV = $450000 / 135.382 = $3323.928. (c) Annual repayment = MR x k = $3323.928 x 12 = $39887.136. Percentage of salary remains after paying loan = 100 [ 1 - (annual repayment) / (annual salary after tax) ] = 100 (1 - 39887.136 / 56000) = 28.773 %. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1257 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.20 : Assume an engineer buys a $1 bond in period T while the nominal interest rate is R. The inflation rate at T + 1 is anticipated to be I. (a) If the bond is redeemed in period T + 1, how much money will the buyer engineer receive, in term of R, which is not affected by inflation? (b) Find the present value, PV of the proceeds from the bond, in term of R and I. (c) If the bond is redeemed in period T + 1, calculate the real growth or real value of the money that the buyer engineer will receive, in term of r = real interest rate, which is affected by inflation. (d) From the answers in (b) and (c), find the values of x, y and z in the following Fisher equation : (1 + x) = (1 + y) / (1 + z), in term of r, R and I.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.20 : (a) Money received without inflation = $(1 + R). (b) PV = $ [ (1 + R) / (1 + I) ]. (c) Money received with inflation = $(1 + r). (d) Answer in (b) = (c) = $ [ (1 + R) / (1 + I) ] = $(1 + r); then x = r, y = R, z = I. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1293 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.21 : The cost of building a biochemical processing plant is increasing due to inflation. Let I = inflation rate, R = nominal interest rate, r = real interest rate. According to Fisher Equation, (1 + r)(1 + I) = (1 + R). According to General Inflation Equation, R = r + I. (a) By assuming that both r and I are fairly small, prove by mathematical calculations that complicated Fisher Equation could be simplified into the General Inflation Equation. (b) By using 2 first order Taylor expansions in the linear approximation, namely 1 / (1 + x) ≈ (1 - x), (1 + x)(1 + y) ≈ 1 + x + y, show by mathematical calculations that (1 + r) = (1 + R) / (1 + I) could be approximated by r ≈ R - I.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.21 : (a) Let (1 + R) = (1 + r)(1 + I) = 1 + r + I + rI, R = r + I + rI. When r ≈ I ≈ fairly small, R ≈ r + I (proven). (b) Let 1 / (1 + I) ≈ (1 - I), then (1 + r) ≈ (1 + R)(1 - I) ≈ 1 + R - I, then r ≈ R - I (shown). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1272 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.22 : An engineering company intends to produce a small piece of biochemical instrument for sales. Let A = overall fixed cost of production, B = variable cost of production per unit, C = selling price per unit, D = quantity of unit produced. Breakeven Analysis is used where revenue = cost. (a) Explain the role of Breakeven Analysis by using A, B, C and D. (b) Find the value of Contribution Margin in term of A, B, C and / or D. (c) At the breakeven point where revenue = cost, derive an equation of D as a function of A, B and C.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.22 : (a) Breakeven Analysis examines the effect of volume (D) on revenue (DC), costs (DB + A) and profit [ DC - (DB + A) ]. (b) Contribution Margin = C - B. (c) When revenue = cost, then DC = DB + A, D (C - B) = A, D = A / (C - B). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1295 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.23 : An engineering company would like to produce 10000 units of control instruments. Let selling price per unit = $10, variable cost per unit = $2, overall fixed cost = $30000. (a) Calculate the income obtained when all units are sold out successfully. (b) Find the overall cost of production. (c) Calculate the percentage of gross profit obtained based on the answers in (a) and (b). (d) Find the minimum units that need to be sold out successfully in order to prevent losses. (e) How many minimum units of instruments that need to be produced in order to prevent losses, if all units produced are sold out successfully?

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.23 : (a) Income = 10000 units x $10 / unit = $100000. (b) Overall cost of production = overall variable cost + overall fixed cost = 10000 units x $2 / unit + $30000 = $50000. (c) Percentage of gross profit = 100 x [ answer (a) - answer (b) ] / [ answer (a) ] = 100 x ($100000 - $50000) / ($100000) = 50 %. (d) Let N units x $10 / unit = $50000 = answer (b) where N = units successfully sold = $50000 / ($10 / unit) = 5000 units. (e) Let contribution margin = (selling price - variable cost) / unit = $ (10 - 2) = $8. M x (contribution margin) = overall fixed cost = $30000. M = ($30000) / ($8) = 3750 units need to be produced at minimum. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1363 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.24 : (a) The net present value of an engineering project is given by V. Let Q = summation of [ T / (1 + I) ^ N ] from N = A to N = B, where P = project cost, T = tax inflow, N = duration. If V = P + Q where P = -$1000, T = $275, I = 0.12 (means 12 % , capital cost of project per year), A = 1 year, B = 5 years, calculate the value of V. (b) The present value tax shields are W = $ (11.9725, 9.1486, 6.5579, 4.1811, 2.0005) in 5 beginning balances of an engineering project. Find the adjusted present value of the engineering project X, when X = V + (summation of the values of W).

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.24 : (a) Let ^ to be mathematical symbol for power. From N = A = 1 year to N = B = 5 years, Q = summation of [ T / (1 + I) ^ N ] = T { summation of [ (1 + I) ^ (-N) ] } = $275 { summation of [ (1.12) ^ (-N) ] } = $275 [ (1.12) ^ (-1) + (1.12) ^ (-2) + (1.12) ^ (-3) + (1.12) ^ (-4) + (1.12) ^ (-5) ] = $991.31. Then V = P + Q = -$1000 + $991.31 = -$8.6865. (b) X = V + (summation of the values of W) = -$8.6865 + $ (11.9725 + 9.1486 + 6.5579 + 4.1811 + 2.0005) = $25.1741. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1123 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.25 : (a) If an engineering company has a policy to maintain A % debt ratio, it will limit its total debt B, to A % of all company assets with total assets C. Find the value of A as a function of B and C. (b) An engineering consultancy that designs biochemical processing plants has a beginning balance in a bank of value D. The interest paid by the bank to the consultancy for the saving is E. The consultancy withdraws F amount of money from the bank to cover the project cost. (i) Calculate the end balance in a bank of the consultancy by using the symbols of D, E and F etc. (ii) Find the value of the principal due to E and F.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.25 : (a) Debt ratio (A) = 100 [ total debt (B) ] / [ total assets (C) ], based on the definition. (b) (i) End balance = D + E - F. (ii) Principal = F - E. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1235 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.26 : In a construction accounting for a biochemical processing plant, $A has been spent to date, and another $B is expected required to complete the construction project. The biochemical processing plant is expected to generate a total revenue of $C. Calculate the : (a) estimated total costs; (b) percentage completion of project; (c) total estimated profit; (d) profit to date.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.26 : (a) Costs = $A + $B. (b) Percentage = 100 ($A) / ($A + $B). (c) Total profit = $C - ($A + $B). (d) Profit to date = [ answer (b) ] [ answer (c) ] / 100 = (100 / 100) [ ($A) / ($A + $B) ] [ $C - ($A + $B) ] = ($A) ($C) / ($A + $B) - ($A). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1162 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.27 : A biochemical engineering consultancy applies construction accounting in its finance. Its project began on 1 January 2010. Total revenue generated from the project was $9000. On 1 January 2011 as the budget, $2000 had been spent, with $6000 expected. However, the project cost increased latter, causing deviation from its initial budget on 1 January 2012, where $7000 had been spent, with $1400 expected. Let (estimated total cost) = (spent cost) + (expected cost to be spent), (percentage completion) = 100 (spent cost) / (estimated total costs), (total expected profits) = (total revenue) - (estimated total costs). Calculate : (a) total expected profits on 1 January 2011 and 1 January 2012; (b) estimated total cost as and not as the budget; (c) percentage completion of the project since the project began, in the first and second years.

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.27 : On 1 January 2011 (as the budget), estimated total costs = $2000 + $6000 = $8000, total expected profits = $9000 - $8000 = $1000, percentage completion on first year = 100 ($2000 / $8000) = 25%. On 1 January 2012 (not as the budget), estimated total costs = $7000 + $1400 = $8400, total expected profits = $9000 - $8400 = $600, percentage completion on second year = 100 ($7000 / $8400) = 83.33%. Final answer : (a) Profits : $1000 (1 January 2011), $600 (1 January 2012). (b) Costs : $8000 (as the budget), $8400 (not as the budget). (c) Percentage completion : 25% (first year), 83.3% (second year). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1294 }

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.28 : In the development of a small engineering facility for biochemical processing, completed contract method is applied in its construction accounting. In a short contract period, let the (contract price) = A. Let the (costs to date) = B; (estimated cost to complete) = C. (a) Find the values of : (i) estimated total costs; (ii) total estimated profit; (iii) percent completion to date. (b) If the project contract is 100 % completed : (i) find the total gross profit recognized; (ii) what is the exact numerical value of C?

Answer

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.28 : (a)(i) Cost = B + C. (ii) Profit = A - (B + C). (iii) Percent = (100) B / (B + C). (b)(i) Profit = A - (B + C). Profit = A - B when C = 0. (ii) C = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.