Financial Calculations and Analysis
2) Compound Interest
$1000 for 3 years at 10% per year compound interest F = P(1 + i)^N = 1000(1 + 0.10)^3 = $1331 | $1000 for one year at 1% per month: $1000(1.01)^12 = $1,126.83 | for one year at 12% annual: $1000(1.12) = 1,120 | at 0.2% per week three years 1000(1.002) 52*3 = 1,365.73 |
2) Bond Valuation
$1000 state govt. bond is issued in 1865. It carries an annual interest rate of 24%. How much is the fair market value of the bond 128 years late in 1993? Compound: 1000(1.24)^128 = 908 T Simple interest: F = P + P*i*N 1000+1000*.24*128 = 31,720
2) Effective Interest Rate
The CCC charges a nominal 24% interest on overdue accounts compounded daily. What is the effective interest rate? Is = r / m; 0.24/365 = 0.0006575 (or 0.06575%); Ie = (1 + Is)^m − 1; ((1 + 0.0006575)^365) – 1 = 0.271; Effective (annual) interest rate is 27.1%
2) Nominal and Effective Interest Rates
Nominal interest rate is 12% and interest is compounded semi annually; Is = r/m = 0.12/2 = 0.06; Ie = (1 + Is)^m − 1 = (1 + 0.06)^2 − 1 = 12.36% | Monthly? Is = r/m = 0.12/12 = 0.01; Ie = (1 + Is)^m − 1 = (1 + 0.01)^12 − 1 = 12.68% |
2) Equivalent Annual Cost
$1000 for two years at 2%/month vs 6%/quarter 1000(1.02)^12*2 – 1000(1.06)^4*2 = 14.59 | 2) Purchase Price P=$380000 4 yrs ago, Sal. Value S=$30000 2 in yrs, Current Book Value? BVsl(n)=P-n((P-S)/N) – BVsl (4) = 380000-4[(380000-30000)/6] (Straight Line Dep.)
2) Depreciation
2) New server farm for $250 000. Six year life, $10 000 salvage value. Depreciation 6 yrs; Dsl (6) = (250k-10k)/6 = 40k; Ddb(6)=(250k(0.5848)^5)(0.41252) = 77099.61 | BV(4)? BVsl = 250k-4((250k-10k)/6)=90k | BVdb = 250k (1 − 0.4152)^4
2) Declining Balance Depreciation
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. 24k(0.75)^4x(0.25) = 1898
2) Effective Monthly Interest Rate
Find the effective monthly interest rate when the nominal rate is 7% with weekly compounding. m = ((1 + 0.07/52)^52/12) – 1 = 0.0058464| 30 years on a $100,000 on 6% weekly vs 6% monthly; F = 100,000(1 + 0.06/52)^52*30 – 100,000(1 + 0.06/12)^12*30 = 2079.80
2) Future Value of an Annuity
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; F = 2000(1.0075)^12*3 – 1000(1.0075)^12*2
2) Future Value of an Investment
Milo inherited $6,500 and purchased an investment certificate. Compounded that the interest rate that the bank pay on investment certificate would allow him to accumulate $7,600 he would need over 4 YRS. What I.R the bank pay? F=P(1+i)^N, 7600=6500(1+i)^4, AN= 4%
2) Present Value of a Loan
It is known the total interest paid over a 5-years is $2,081.13 = X. What was the principal amount borrowed at a 6% nominal interest rate compounded quarterly? PV+X=PV(1+i)^N, Rearrange For PV, AN=6000
2) Future Value of an Investment
Howmuchmorewouldyouearnover30yearsona$100,000investmentifyour earningrateis6%compoundedweeklyinsteadof6%compoundedmonthly? F = 100,000(1 + 0.06/52) 52*30 – 100,000(1 + 0.06/12) 12*30,ans = 2079.80
3) Future Value of an Annuity
You want to save for your child’s education. You plan to invest $X a year. The first deposit is the day they are born, and the final
deposit will be on their 18th birthday. You plan to take them to Italy for their 5th birthday so they can see Nona and Nono, so you won’t make a deposit on that birthday. Find X so they have $50,000 saved up for their 18th birthday. The answer is within 10 of which of the following? MARR = 6%
3) U need a replacement computer, costing $3k, in 5 years. A bank acc earns 8% annual interest, 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? 3000(A/F,8%,5), AN=511
3) Peter and Piper purchased a pickled pepper business for $1.28 million on September 30th, 2022. They put down a deposit of $280,000 and took out a loan for the balance. Their loan payments are due at the end of each month (October 31, 2022 was the date of the first payment) and are based on the assumption that they will amortize the debt over 25 years. Annual nominal interest is 12 percent compounded monthly. It is now April 1, 2023, and Peter and Piper’s last payment was the day before on March 31, 2023. How much do they still owe? A = 1m(A/P,1%,25*12 = 300) = 1m(0.0105) = 10,500|F = 1m(F/P,1%, 6) – 10,500(F/A, 1%,6),ans=996,904
3) You deposit $1,000 into an account at the end of one year, and then you increase the deposits by $1,000 each year until the 20th year (so in the 20th year you deposit $20,000). You miss the 10th year’s deposit. The interest rate is 6%. How much is in the account immediately after the 20th payment? The answer is within $2000 of which of the following? F = 1000(F/A,6%,20) + 1000(A/G,6%,20)(F/A,6%,20) – 10,000(F/P,6%,10),ans=298,639
3) You are responsible for managing the construction of a new skateboard park. One design being considered has an installation cost of $700,000. As far as maintenance, the park will need to be repaved every 5 years starting 15 years after construction, at a cost of $200,000 each time it is paved. Assuming the park will essentially have an infinite life, what is the present worth of the costs associated with this park design? The interest rate is 6%.i5 year= (1.06)5 – 1 = 0.3382| P = 700k + (200k/0.3382) * (P/F,6%,10),ans=1,030,215.72
3) You want to withdraw$1000 every second year indefinitely from an account paying 6% per year. The first withdrawal will be 4 years from today. What is the present value of this indefinite withdrawal plan (MARR = 6%). Find the answer within $20. i2-year= (1.06)2 – 1 = 0.1236| P = [1000/0.1236](P/F,6%,2),ans=7200
3) If money can earn 12% compounded semi-annually, a bond maturing in 15 years with a face value of $5000, and a semi-annual coupon with a nominal coupon rate of 7%, has a present value of:
P = 5000(P/F,6%,30) + (5000 x 0.035) (P/A, 6%,30) = 5000 (P/F,6%,30) + 175 (P/A, 6%,30) = 5000(0.17411) + 175(13.765),ans=3279.43
3) Immediately before the 10 th payment, the economy-wide interest rate has decreased to 8% compounded semi- annually. What is the new value of the bond?
P = 5000(P/F,4%, 20) + 175 + 175(P/A, 4%,20) = 5000(0.45639) + 175 + 175(13.590),ans=4835.20
3) If money is worth 6 percent compounded semiannually, how much is a bond maturing in seven years with a face value of $10000 and a coupon rate of 9 percent worth today?
P=10000(P/F,0.06/2,14)=6611| CouponPayment=10000*0.045=450| P=450(P/A,0.03,14)=5083|PV=5083+6611,ans=11694
4) You are considering two assets with different lifespans. Asset A provides its service for three years, has a first investment of $1000 followed by 3 years of annual $400 maintenance costs. Asset B provides a similar service for 7 years, has a first investment cost of $2000 following by 7 years of annual $200 maintenance costs. You will need the services of only one of the assets for the foreseeable future. Over the next 21 years, in present worth, how much more will asset A cost relative to asset B? The answer is within $5 of which of the following? MARR = 6.
AA = 1000(A/P,6%,3) + 400 = 774.11| AB = 2000(A/P,6%,7) + 200 = 558.28| AA – AB = 215.83| PA-B = 215.83(P/A,6%,21),ans=2539
4) Construction of a ferry between Vancouver Island and the B.C. mainland is estimated to cost BC Ferries Inc. a first cost of $60 million dollars. Ferry tickets are projected to cost passengers $10. Projections also estimate 60,000 passengers per week. The operating and maintenance cost will be 70% of the ticket revenue. Since management of the ferry is contracted out, the O&M costs and the ticket revenues are only collected by B.C. Ferries as a lump sum payment at the end of each year. The payback period is? 60m/[(0.30)*$10*60,000 per week*52 weeks],ans=6.41 years
4) (In keeping with the information provided in the previous question): Construction of a ferry between Vancouver Island and the B.C. mainland is estimated to cost BC Ferries Inc. a first cost of $60 million dollars. Ferry tickets are projected to cost passengers $10. Projections also estimate 60,000 passengers per week. The operating and maintenance cost will be 70% of the ticket revenue. Since
management of the ferry is contracted out, the O&M costs and the ticket revenues are only collected by B.C. Ferries as a lump sum payment at the end of each year. MARR = 6%. The discounted payback period is? 60m
4) Building a bridge will cost $64 million. A round-trip toll of $11 will be charged to all vehicles. Traffic projections are estimated to be 6000 per day. The operating and maintenance costs will be 15 percent of the toll revenue. Find the payback period (in years) for this project. Assume that the savings are earned throughout the year, not just at year end. (Assume there are 365 days per year.)
PaybackPeriod=64m/(11*6000*365*0.85),ans=3.13
4) An advertising campaign will have a first cost of $577,000 and it is expected to permanently increase costs by $6,000 per year. Revenues will be gained in the pattern of an arithmetic gradient with $240,000 in the first year, declining by $40,000 per year to zero in the final year of revenues. If the company’s MARR is 10 percent. The IRR is approximately? P = -577k – 6k/i* + [240k– 40k(A/G,i*,6)](P/A,i*,6) = 0 (Note, N =7 will work as well only if it is used in both factors. This is necessary since the cash flow in year 7
is zero) | If i* = 3% then P = 86.51, which is approximately zero given the size of the cash flows.
4) Consider the following two mutually exclusive projects. Project A involves an investment of $1000 followed by 5 years of $400 returns. Project B involves an investment of $1400 followed by 5 years of the following returns: $400 at the end of the first year, $450 in the second year, $500 in the third year and $550 in the fourth year and $600 in the fifth year. Find the incremental IRR.
Project B – A = -400 + 50(A/G,i*,5)(P/A,i*,5) = 0 >>> i* = 6% gives P = -3.2762 | At i* = 5%, P = -273.53 | At i* = 6%, P = -3.28
5) Consider the following cash flows. MARR = 10% t=0, 1000 at t=1, -500 at t=2, -800 at t=3, 150. Find the precise ERR.
PB1 = 1000(1.10) – 500 = 600 | PB2 = 600(1.10) – 800 = -140 | 0 = -140(1+i e*) + 150 >>> ERR = 0.07
5) Consider the following cash flows. MARR = 10% t=0, 1000 at t=1, -500 at t=2, -800 at t=3, 150. Find the approximated ERR. 1000(1.1) 3 + 150 = 500(1+i*ea) 2 + 800(1+i*ea) | i*ea = 9.8%
5) I plan to purchase a Toyota Rav 4 for $50,000 to use as a Lyft driver. After 5 years of use, I expect to be able to sell the car for $20,000. During that time, I expect my annual expenses to total $5000 per year in the first year and then increase by $1000 per year thereafter (so $6000 in second year and $7000 in third year, etc.). I expect to drive 2000 passengers per year and 10 km per passenger on average. What should I charge per km to cover my costs? The answer is within 4 cents of which of the following? MARR = 6%. 50k (A/P,6%,5) – 20k(A/F,6%,5) + 5k + 1k(A/G,6%,5) = 2000(10km)*P. ans=0.76
6) Total Assets – Total Liabilities = Owners Equity(company value) | Assets: items to which the organization has legal title. – Current assets can be converted into cash within one
year. – Fixed assets (capital assets) have a life greater than one year. Liabilities: debt owed (suppliers, employees, government). –Current Liabilities–debt
normally paid within a year (accounts payable, taxes, wages payable, bank loan payable…). –Long-term liabilities–debt normally paid beyond current year.
6) Salvador Industries bought land and built its plant 15 years ago. The depreciation on the building is calculated using the straight-line method, with a life of 30 years and a salvage value of $52,000. Land is not depreciated. The depreciation for the equipment, all of which was purchased at the same time the plant was constructed, is calculated using declining balance at percent 20%. Salvador currently has two outstanding loans: one for $44,000 due December 31,2020, and another one for which the next payment is due in four years. During April 2020, there was a flood in the building because a nearby
river overflowed its banks after unusually heavy rains. Pumping out the water and cleaning up the basement and the first floor of the building took a week. Manufacturing was suspended during this period and some inventory was damaged. Due to inadequate insurance, this unusual and unexpected event cost the company $96,000, net. Find Retained earnings.
Building less Dep: 224k – 15(224k-52k)/30 = 138,000 | Equipment less Dep: 460k(0.80)^15 = 16,185 | Current Assets = 344k + 2860k + 2002k +162k = 5,368k |
Fixed Assets = 138k + 16185 + 525k = 679,185 | Total Assets = 6,047,185 | Current Liabilities = 949,303 + 44k + 30k = 1023,303 | LT Liabilities = 1,220k _+ 323k = 1,543,000 | Total Liabilities = 2,566,303
Owner’s Equity = Total Assets – Total Liabilities = 3,480,882 | Retained Earnings = Owner’s Equity – Common Shares,ans=1,600,882
7) The following pair of assets differ in the MARR. All assets decline in value by 20 percent of current value each year. Installation costs are zero for all assets.
Find EAC 2 for project A, within $1000.
A | FirstCost = $120,000 Initial OperatingCost=$30,000 RateofOperatingCost= $5000 per year MARR=6%
B | FirstCost = $120,000 Initial OperatingCost =$30,000 RateofOperatingCost= 13.5%/year MARR=25%
EAC2 = 120k(A/P,6%,2) – 120k(0.8 ^2)(A/F,6%,2) + [30k/1.06 + 35k/1.06^ 2](A/P,6%,2),ans=60,598
7) Find Economic life of Project A in previous question. EAC1 = 120k(1.06) – 120k(0.80) + 30k = 61,200
EAC2 = 120k(A/P,6%,2) – 120k(0.8^ 2)(A/F,6%,2) + [30k/1.06 + 35k/1.06^ 2](A/P,6%,2) = 60,598
EAC 3 = 120k(A/P,6%,3) – 120k(0.8^3 )(A/F,6%,3) + [30k/1.06 + 35k/1.06^2 +40k/1.06^3 ](A/P,6%,3) = 60,400 = EAC* ANSWER = 3 years since this is where EAC is the least after this it increases.
EAC4 = 120k(A/P,6%,4) – 120k(0.8^ 4)(A/F,6%,4) + [30k/1.06 + 35k/1.06^ 2 + 40k/1.06 ^3 + 45k/1.06 ^4](A/P,6%,4) = 60,531
7) Suppose your current asset was purchased 3 years ago for $100,000. It has been depreciating at 20% per year and continues to do so. Its O&M cost in its first year was $20,000. O&M increases by $10,000 per year. MARR = 6%. You are considering replacing the asset. Find the MC for the asset’s 4th year so you can make the replacement decision. The answer is within 1000 of which of the following?
Applying one-year principle: 100k(0.8 3)(1.06) + 50k – 100k(0.8)4,ans=63312
7) Bora likes driving giant monster trucks and buys a new one whenever he wrecks the older one (which happens frequently). Typically, he trades in his old truck for a new one costing about $28,000. A new truck warranty covers all repair costs for the first 2 years. After that, his records show an average repair expense (over standard maintenance) of $5300 in the third year (at the end of the year), increasing by 40 percent per year thereafter. A 20 percent declining-balance depreciation rate is used to estimate salvage values and interest is 6 percent. Find the difference between the EAC associated with N=5 and the one associated with N=4.
EAC4 = 28k(A/P, 6%,4) –28k(0.8 ^4)(A/F,6%4) + 5300[1/(1.06)^3 + (1.4)/(1.06)^4](A/P.6%,4) = 8439.23
EAC5 = 28k(A/P, 6%,5) –28k(0.8^ 5)(A/F,6%5) + 5300[1/(1.06)^3 + (1.4)/(1.06)^4 + (1.4 ^2)/(1.06^ 5) ](A/P.6%,5) =9314.07
EAC5 – EAC4,ans=875
7) (In keeping with the information from the previous question) Bora likes driving giant monster trucks and buys a new one whenever he wrecks the older one (which happens frequently). Typically, he trades in his old truck for a new one which then costs about $28,000. A new truck warranty covers all repair costs for the first 2 years. After that, his records show an average repair expense (over standard maintenance) of $5300 in the third year (at the end of the year), increasing by 40 percent per year thereafter. A 20 percent declining-balance depreciation rate is used to estimate salvage values and interest is 6 percent. Find the marginal cost of keeping the machine for its 5th year. The answer is within 100 of which of the following?
C4 = (0.8)^4 (28,000)(1.06) + 5300(1.4)^2 – (0.8)^5(28,000),ans=13,369.89
9) Suppose housing prices in Toronto are increasing at 10% per year and expected to continue increasing at that rate for the indefinite future. The current cost of a house is $1m. The average general rate of inflation is expected to be higher than normal at 5% for the next two years but then fall back to 2% over the following 8 years after that. What will the real cost of a house be in 10 years? The answer is within 50,000 of which of the following? R10 = 1m(1.1)^10/[(1.05 ^2)(1.02 ^8)],ans=2,007,922
9) You invest $12,000 into an investment contract which stipulates that you are to receive a $2,000 cash payment at the end of each year for the next 7 years. The
inflation rate is expected to be stable at 2% per year. What is the real rate of return on your investment? Choose the best answer.
P = -12k + 2k(P/A,i*,7) = 0 >>> (P/A,i*,7) = 6 at i* = 4
i* = i*’ + f + i*’ x f (the cross-product will be negligible here)
4% = i*’ + 2% >>> i*’ = 2%
9) Ela is the financial controller for a dollhouse distribution company. Ela’s current compensation contract is for $95,000 per year for 4 years (end of year payments). She believes inflation will be 6 percent this year and the next, 4 percent in the third year, and then will stay at 2 percent per year afterward. Ela’s real MARR is 6%. What is the real dollar value of Ela’s final year’s compensation (her 4th year’s
paycheck)? The answer is within $50 of which of the following? R4 = 95k/(1.06 ^2 * 1.04 * 1.02) = 79,703.68
9) (In keeping with the information provided in the previous question): Ela is the financial controller for a dollhouse distribution company. Ela’s current compensation contract is for $95,000 per year for 4 years (end of year payments). She believes inflation will be 6 percent this year and the next, 4 percent in the third year, and then will stay at 2 percent per year afterward. Ela’s real MARR is 6%. What is the present worth of Ela’s compensation contract? The answer is within 1000 of which of the following?
R1 = 95k/1.06 = 89,622.64 | R2 = 95k/1.06 ^2 = 84,549.66 | R3 = 95k/(1.06 ^2 * 1.04) = 81,297.75 | R4 = 95k/(1.06 ^2 * 1.04 * 1.02) = 79,703.68
P = R1/1.06 + R2/1.06^ 2 + R3/1.06^ 3 +R4/1.06^ 4, ans= 291,192
10) The city of Toronto is considering an upgrade to its current waterfront development project. The present equivalent cost of the construction is expected to be $1.2 million. Construction is expected to last for 5 years. Annual O&M costs are expected to be approximately $50k per year once the construction is completed. The O&M costs are expected to continue indefinitely. The project is expected to cause $30,000 per year in disruption to local businesses and residents during the 5 years of construction but afterwards, the annual benefit is expected to be $100k per year indefinitely. The MARR is 6% and the social discount rate is 3%.Find the Benefit-Cost ratio. B/C = [100k/.03*(P/F,3%,5) – 30k(P/A,3%,5)] / [1.2m + 50k/.03(P/F,3%,5)],ans=1.038
10) The Engineering department at TMU wants to build a recreation centre for alumni and students to enjoy. It has narrowed the choices down to tennis courts or a swimming pool. The swimming pool will cost $2.6 million to construct and will cost $300,000 per year to operate, but it will bring benefits of $475,000 per year over its 28-year expected life. Both projects are assumed to have a salvage value of zero. The appropriate MARR is 7 percent. Find the benefit cost ratio of the swimming pool? The answer is within 0.05 of which of the following? Choose the best answer.
BCR^SP = 475k/ [2.6m(A/P,7%,28) + 300k], ans=0.9237
10) The Engineering department at TMU wants to build a recreation centre for alumni and students to enjoy. It has narrowed the choices down to tennis courts or a swimming pool. Tennis courts would cost $210,000 to build, cost $21,000 per year to operate, and bring $63,000 per year in benefits over their nine-year life. Both projects are assumed to have a salvage value of zero. The appropriate MARR is 7
percent. Find the modified benefits cost ratio of the tennis courts. The answer is within 0.05 of which of the following? Choose the best answer.
MBCR^TC = [63k – 21k] / 210k(A/P,7%,9),ans=1.3029