Financial Engineering: A Comprehensive Guide to Key Concepts and Applications
Periods: 1 year = 365 days = 52 weeks = 12 months = 4 quarters = 2 semi annuals
Nominal Interest: iperiod = inominal / (periods/year) | Converting between periods : 12% yearly to monthly : 5% = (1+12)12/1-1
Book value straight line: n = Time since purchase, P=value at purchase, S = Salvage value, N = periods until salvage value, (declining balance) d = Depreciation Rate
Bond Value = A( P|A,i,N) + F(P|F,i,N) | MARR conditions = If PW or AW > 0 is acceptable, if PW or AW = 0 marginal (equal to MARR), if PW or AW
IRR is when PW,AW and FW =0 then you rearrange for i. If not mathematically, then use trail and error or linear interpolation (formula given)
Equity Ratio: Total Equity / (Total liabilities – Total equity) [The denominator is also equal to total assets] | Return on Assets: Net post tax income / Total Assets
Return on Equity: Net post tax income / Total Equity | Optimal EAC is at the minimum EAC that is the projects life | EACcapital = (P – BV(n))(A|P,i,N) + BV
Ch.2: You purchased a saw cutter five years ago for $24,000. You project a $7000 salvage value in two years’ time. Depreciation follows the declining balance method with a depreciation rate of 25%. Find the depreciation value for this year. The answer is within $1 of which of the following?
D5 = 24,000(0.75) ^4 (0.25) = 1898.44
Ch.2: Find the effective monthly interest rate when the nominal rate is 7% with weekly compounding. Be correct to 4 decimal places. The answer is within 0.0001% of which of the following?
im = (1 + 0.07/52) ^52/12 – 1 = 0.0058464 or 0.5846%
Ch.2:How much more would you earn over 30 years on a $100,000 investment if your earning rate is 6% compounded weekly instead of 6% compounded monthly? The answer is within $5 of which of the following?
F = 100,000(1 + 0.06/52) ^52*30 – 100,000(1 + 0.06/12) ^12*30 = 2079.80
Ch.2: You deposit $2000 into an account paying 9% compounded monthly. After one year you withdraw $1000. How much do you have in the account at the end of the 3-year period.
im = 0.09/12 = 0.0075 or 0.75%
F = 2000(1.0075) ^12*3 – 1000(1.0075)^12*2 = 1420.88
= 2000(1.3087) – 1000(1.1964) = 1421
Ch. 3: You will need to buy a replacement computer, costing $3,000, in five years. If you have a bank account which earns 8% annual interest rate, how much must you put in the bank every year in order to have enough money for the replacement, assuming you make your first deposit in one year’s time?
A = 3000(A/F,8%,5) ans.= 511
Ch. 4: A municipality decided to build a new bridge to accommodate the city’s growing traffic across the river. It has to pay $200,000 to construct the bridge plus $15,000 per year for its maintenance. Assuming that the bridge will have an infinite service life, calculate the present worth of the project at 10% interest rate.
P = -200,000 – 15,000/0.10 ans. = -350,000
Ch. 4: I can invest for a pension in either the Senex or the Geriatrix pension plan. Senex requires me to invest $1,500 a year for the next 15 years (beginning one-year from now), whereas Geriatrix requires an immediate deposit of $5,000 and 15 subsequent annual investment of $1,200 a year (also starting one-year from now). If my MARR is 15%, how much greater is the equivalent uniform annual cost to me of the series of payments I would make to Geriatrix compared to the series of payments I would make to Senex?
A^Geriatrix – A^Senex = 5000(A/P,15%,15) +1200 – 1500 | ans.= 555
Ch. 5: he following table summarizes information for five projects: The data can be interpreted in the following way: The IRR on the incremental investment between project 5 and
project 4 is 16%. If the projects are mutually exclusive, which projects should be undertaken if the MARR is 15%?
Project | First Cost (in $) | IRR on Overall Investment | IRR on Increments of Investment Compared with Projects (%)
1 | 100,000 | 19% | 1 | 2 | 3 | 4
2 | 175,000 | 15% | 9%
3 | 200,000 | 18% | 17% | 23%
4 | 250,000 | 16% | 12% | 17% | 13%
5 | 300,000 | 17% | 14% | 11% | 17% | 16%
Options B and C can be excluded since the projects are mutually exclusive.
IRR^ 4-3 = 13% IRR^ 5-3 = 17% > MARR therefore 3 is rejected. | ans. = 5 only
Ch. 6: In a balance sheet, Current Assets are $4,000; Long-term Assets are $6,000; Total Owners’ Equity is $6,000; Current Liabilities are $3,000. What are Long-term Liabilities?
Current Assets + LT Assets = Current Liabilities + LT Liabilities + Owner’s Equity | ans. = 1000
Ch. 7: Currently a firm replaces its equipment every year. It has calculated the equivalent annual cost of replacement as follows: How much can the firm save by making a replacement decision based on the economic life criterion?
Replacement Period | Salvage Value | EAC Capital Costs | Annual Repair Costs | EAC Repair Costs
1 | $1,420 | $1,287 | − | −
2 | $1,102 | $1,082 | $400 | $189
3 | $910 | $976 | $600 | $290
4 | $795 | $812 | $800 | $437
5 | $580 | $673 | $1,200 | $592
Create the column EAC ^Total . Calculate EAC^ Total = EAC ^Capital + EAC ^Repair.
Find economic life: EAC* = 1249 at N* = 4.
EACN=1 – EAC* = 1287 – 1249 = 38
Ch. 7: An asset has an initial cost of $10,000. Its maintenance costs are $300 in the first year, increasing by 20% per year thereafter. Its salvage value declines by straight-line depreciation over ten years (assume it will be retired with a salvage value of zero). If your MARR is 10%, what is its economic life?
Find N*.
EAC 2 = 10k(A/P,10%,2) – 8000(A/F,10%,2) + 300[(P/F,10%,1) + 1.2(P/F,10%,2)] (A/P,10%,2)
EAC 3 = 10k(A/P,10%,3) – 7000(A/F,10%,3) + 300[(P/F,10%,1) + 1.2(P/F,10%,2) +1.2^2(P/F,10%,3)] (A/P,10%,3)
EAC 4 = 10k(A/P.10%,4) – 6000(A/F,10%,4) + 300[(P/F,10%,1) + 1.2(P/F,10%,2) + (1.2^2) (P/F,10%,3) + (1.2 ^3)
(P/F,10%,4)] (A/P,10%,4) [ To get answer to this till 6th EAC then it should be the lowest number and therefore the projects life] ans. = 6 years
Ch. 9: A bank offers me 10% annual interest on an investment, but the inflation rate is 5%. If I invest $500 in the bank, what will the value of my investment be in five years, in real dollars?
R5 = 500(1.10) ^5 /(1.05)^ 5
Ch. 9: Jason inherited $6,000 from his grandfather. He decided to invest the money in a two-year project that promised a return of $2,000 by the end of this year and $4,500 at the end of the next
year. Jason read an analytical report that inflation is going to be 3.0% over the next 5 years. He decided that he would only invest in this project if it could give him at least 10% real rate of return. What is the present worth of his investment under this rate of return?
MARRc = 0.10 + 0.03 + 0.10*0.03 = 0.133
P = -6000 + 2000/1.133 + 4500/1.133 ^2 = -729
Ch. 10: A city council is planning to build a new bridge across a creek. This will require an immediate expenditure of $50,000 but will save citizens an estimated $5,000 in travel costs every year
over the next twenty years. Alternatively, they could renovate an existing bridge. This would only cost $20,000, but the saving in travel costs would only be $2,000 every year. If the council’s social discount rate is 5%, what is the BCR for upgrading from the renovated bridge to the new bridge?
(BX – BY ) / (C X – C Y ) = (5k – 2k) / [(50k – 20k)(A/P,5%,20)] = 1.25
RANDOM QUESTIONS FROM 2023 EXAM
Q. 1: Suppose you deposit $1000 each year for 5 years into an account earning 6% per year. What is the balance of the account immediately after the final deposit? The answer is within $2 of which of the following?
F = 1000(F/A,6%,5) = 1000(5.6371) = 5637.10
Q. 2: You deposited $10,000 into an account paying 6% interest with the intention of withdrawing funds each year for 12 years. You would like the account to have a $3000 balance at the end of the 12-year period. How much can you withdraw each year? The answer is within $3 of which of the following.
10,000(F/P,6%,12) = X(F/A,6%,12) + 3000
X = 1014.94
Q. 3: You would like to begin an annual deposit plan in order to fund your retirement. You plan to retire in 20 years with enough funds in your account to be able to withdraw $30,000 per year (at the end of each year) indefinitely retirement. You will start with an initial one-time deposit of $100,000, and then add to it with the annual deposits. You plan to stop depositing at the time of retirement. If the interest rate is 6%, what is the annual deposit necessary to fund your retirement plan? The answer is within $10 of which of the following?
V20 = 30,000/0.06 = 100,000(F/P,6%,20) + X(F/A,6%,20)
X = 4874
Q. 4: You purchase a home for $1.4 million, pay 20% as a down-payment and borrow the rest. Your monthly mortgage payments are based on an interest rate of 6% compounded monthly for the current 3-year term. The mortgage is to be amortized over 20 years. Find the balance owing at the end of the term. Use the method that finds the present value of remaining payments. The answer is within $500 of which of the following?
A = 1,400,000(0.8)(A/P,0.5%,20*12=240) = 1,120,000(.0072) = 8064
B36 = 8064(P/A,0.5%,240-36=204) = 8064(127.6975) = 1,029,753
Q. 5: A project returns $100 at the end of the first year, and each year thereafter the return grows by an additional 8%. The project has a 20-year lifespan so in the end there are 20 payments. All the funds are deposited into an account earning 6%. What is the balance of the account at the end of the 20 years, immediately after the 20 th payment? The answer is within $10 of which of the following?
i0 = 1.06/1.08 – 1 = -0.0185 or -1.85%
F = 100(P/A,8%,6%, 20)(F/P,6%,20) = 100(22.6607)(3.2071) = 7267.51
Q. 6: How much is accumulated over 50 years in a fund that pays 6% compounded yearly, if $1000 is deposited at the end of every fifth year? The answer is within $100 of which of the following?
A = 1000(A/F,6%,5) = 1000(0.1774) = 177.4 | F = 177.4(F/A,6%,50) = 177.4(290.34) = 51,506.32
Q. 7: Eric is opening an online fantasy football club. Eric invested $10,000 in developing the app, and he intends to charge annual membership fees of $10 for each member, and he intends to have no more than 100 registered users. He expects to have the maximum number of 100 registered users on his launch date and maintain that number indefinitely. The membership fees are collected at the beginning of the membership year, not at the end. What is his discounted payback period if his MARR = 6%?
10,000
Q. 8: You have a choice of one of two projects. The first project has a first cost of $1000 with a promised return after 2 years of $1400. The second project has a first cost of $1500 and promised return of $2000 after 2 years. Your MARR = 6%. Find the incremental IRR of the projects. The incremental IRR is within 0.25% of which of the following?
PW = 0 = -500 + 600(P/F,i*,2) for i* = 9.55%
Q. 9: You decide to earn extra money during your Engineering program by running an uber service. You purchase a car for $60,000, which you expect to sell for $18,000 at the end of the 6-year degree. You expect annual costs for insurance, maintenance and other expenses to total $7000 per year. You project to have approximately 20 customers per shift, who you will drive 8km on average. You plan on driving 80 days each year. What is the levelized cost per km of driving? MARR = 6%. The answer is within 5 cents of which of the following?
-60,000(A/P,6%,6) + 18,000(A/F,6%,6) – 7000 + 40(80)8*C = 0 | C = $1.30
[For Retained Earnings] Q. 10: Building less Dep: 224k – 20(224k-44k)/30 = 104,000 | Equipment less Dep: 460k(0.80) 20 = 5303 | Current Assets = 344k + 2860k + 2002k +162k = 5,368k
LT Assets Assets = 104k + 5303 + 525k = 634,303 | Total Assets = 6,002,303 = Total L+OE | Current Liabilities = 949,303 + 52k + 30k = 1,031,303 | LT Liabilities = 1,220k _+ 323k = 1,543,000
Total Liabilities = 2,574,303 | Owner’s Equity = Total L+OE – Total Liabilities = 3,428,000 | Retained Earnings = Owner’s Equity – Common Shares = 1,548,000
Q. 11: The current ratio for a firm is 1.23 currently. The firm has $12,000 in cash and $140,000 in accounts receivable. The owner’s equity for the firm is $1.6 million. Which of the following is the firm’s acid test ratio? [Acid Test Ratio must be lower than current ratio]
Q. 12: A house in Toronto cost $1 million three years ago. Houses have been increasing in value by 12% per year on average. The average annual inflation rate over the past 3 years has been 4.5%. What is the real dollar cost of houses? MARRR = 6%. The answer is within $1000 of which of the following?
R 3 = 1m(1.12)^3 /(1.045)^3 = 1,231,134
Q. 13: Consider a project with a first cost of $100,000 that earns an actual dollar return of $70,000 each year for 2 years. The inflation rate is 12% in the first year and 8% in the second year. MARRR = 6%. The IRR R is within 0.25% of which of the following?
R 1 = 70,000/(1.12) = 62,500; R 2 = 70,000/(1.12*1.08) = 57,870.37
PW = 0 = -100,000 + 62,500(P/F,i*’,1) + 57,870.37(P/F,i*’,2)
PW(i*’=14%) = -646 and PW(i*’=13%) = 631; i*’ = 13.5%