Financial Calculations for Investments and Retirement Planning
Find the effective monthly interest rate when the nominal rate is 7% with weekly compounding. [im = ((1 + (0.07/52)^52/12) – 1 = 0.0058464] | 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. [2000(1+(0.09/12))12*3 – 1000(1+(0.09/12))12*2 = 1420.88] | 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. [D(5) = 24,000(0.75)^4 (0.25) = 1898.44] | 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? [ F/A=((((1+i)^n)−1)/i); F = 1000(F/A,6%,5) = 5637.10] | 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? [10,000(F/P,6%,12) = X(F/A,6%,12) + 3000 X = 1014.94] [F/A=((1+i)^n)/i; F/P = (1+i)^n] | 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? [V20 = 30,000/0.06 = 100,000(F/P,6%,20) + X(F/A,6%,20), X = 4874] | 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. [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; A/P = i/(1-(1+i)^-n), P/A =(1-(1+i)^-n)/i] | You invest $1000 into a project which returns $100 at the end of the first year, $200 at the end of the second year, $300 at the end of the third year, continuing to increase the return by $100 each year for 20 years (20 payments). What is the annual worth of the project at 6% per year? [AW = -1000(A/P,6%,20) + 100 + 100(A/G,6%,20) = 773.33; A/G = ((1+g)^n−(1+i)^n)/g-i] | 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 the payment? [i 0 = 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] | 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? [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] | A service project involves a first cost of $5,000 and returns $1500 a year for 5 years. The service will be required for 50 years so the intention is to repeat the project in cycles over the required service life. MARR = 6%. Find the PW of the 50- year service life of the project. [Acycle = -5000(A/P,6%,5) + 1500 = 313 P = 313(P/A,6%,50) = 4933.51] Find the AW of a project that has a first cost of $20,000, a salvage value of $6,000 at the end of 5 years, and annual revenues of $4000. MARR = 6%. [A = -20,000(A/P,6%,5) + 4000 + 6000(A/F,6%,5) = 316.40] | 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 1 therefore option 2 is preferred to do nothing.]
You and your friend are saving for retirement. Your first deposit will be $1000 on your 27 th birthday, and you plan to increase the deposits by $100 per year. Your final deposit will be on your 40th birthday. Today is your 20th birthday. Your friend wants to retire at the same time with the same amount but intends to be make one lump sum deposit today with no additional deposits. How much does that deposit need to be to support their retirement plan? MARR = 4%.[P = [1000(P/A,4%,40-26 = 14) + 100(A/G,4%,14)(P/A,4%,14)](P/F,4%,6) P = 13,245] | You are responsible for managing the construction of a new skateboard park. One design being considered has an installation cost of $70,000. As far as maintenance, the park will need to be repaved every 6 years starting 15 years after construction, at a cost of $20,000 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 5%. [i6 year = (1.05)^6 – 1 = 0.3401 P = 70k + (20k/0.3401) * (P/F,5%,9) = 107,907] | A bond with that pays $720 each half year has a semi-annual coupon rate of 3% per half-year. There are 22 payments remaining before the bond matures and you want to earn 10% compounded semi-annually. The first of those 22 payment is due now. What is the most you should pay for the bond? [Face value *0.03 = 720 >> FV = 24,000 P = 720 + 720(P/A,5%,21) + 24,000(P/F,5%,21) = 18,566] | Suppose housing prices in Toronto are increasing at 12% 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 8% for the next eight years but then fall back to 2% thereafter. What is the expected real cost of a house in Toronto in 10 years [C10 = 1m(1.12)^10 = 3,105,848.21 R 10 = 3,105,848.21/[(1.08^8 )(1.02^2 )] = 1,612,834.63 1m(1+i’*)^10 = 1,612,834. >> i’* = .0490 or 4.9%] | Ela is a financial controller for a dollhouse company. Ela’s current supplier for dolls is shutting down operations, but there are two options for alternative suppliers. Ela needs the service provided by at least one of these suppliers for the indefinite future. Supplier A requires a first cost of $800 followed by 5 years of $200 reimbursements. Supplier B requires a first cost of $1640 followed by 5 years of $400 reimbursements. You need to determine which is the preferred alternative. The MARR is 6%. The incremental IRR is [ Cash flows are -840, 200, 200, 200, 200, 200, 200 P = 0 = -1500 + 200(P/A,i*,6) >>> i* = 6%] | Bora and Nicky are both circus clowns working for the same company. They each sign a contract stipulating that they will each receive a $100,000 bonus after 5 years of performing. Bora lives and works in Mexico where the inflation rate is expected to remain a
stable 10% annually for the indefinite future. Nicky lives and works in Canada where the inflation rate is 7% over the next two years but is expected to fall to 4% thereafter. What is the difference in the real dollar value of their bonus compensation contracts (within $100)? The real MARR = 4% [R 5 Bora = 100k/(1.10^5)= 62,092.13 R 5 Nicky = 100k/[(1.07^2)(1.04^3)]= 77,648.39 77,648.39 – 62,092.13 = 15,556] | Suppose your current asset was purchased 5 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 and would like to know it’s economic life and associated EAC* so you can make the replacement decision. Find the marginal cost for this (6th ) year for the asset so you can make the replacement decision. [Applying one-year principle 100k(0.85)(1.06) + 70k – 100k(0.8)6 = 78,520] | Consider the following cash flows. MARR = 10% Cash flow t=0 t=1 t=2 t=3; 1000 -800 -500 98; Find the approximated ERR. [Approximated ERR: 1000(1.10)3 + 98 = 800(1+iea*)2 + 500(1+iea*). >> Approx ERR = 6%] | 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 3 years. After that, his records show an average repair expense (over standard maintenance) of $3500 in the fourth year (at the end of the year), increasing by 20 percent per year thereafter. A 40 percent declining-balance depreciation rate is used to estimate salvage values and interest is 5 percent. Find the EAC associated with N=6. [EAC6 = 28k(A/P, 5%,6) –28k(0.6^6 )(A/F,5%6) + 3500[1/(1.05)^4 + (1.2)/(1.05)^5 + (1.2^2 )/(1.05^6 ) ](A/P,5%,6) = 7281.14] | The inflation rate is steady at 5%. An investment contract stipulates that you are to receive a $3,150 payment in one year, with payments increasing by 5% per year for each the next 9 years (10 payments in total) in or to protect the investor from the effects of inflation. To sign on to the investment, you need to pay $26,950 today. What is the real rate of return on the investment [Payments from R1 to R10 = $3000 P = 0 = -26,950 + 3000(P/A,i’*,10) >>> i’* =2%] | Consider the following cash flows. MARR = 12% Cash Flow t = 0 t = 1 t = 2 t = 3 t = 4; 3000 0 -4000 0 261 Find the precise ERR. The answer is within 0.1% of which of the following? [PB2 = 3000(1.12)^2 – 4000 = -236.8 P = 0 = -236.8(1+ie *)^2 + 261 >>> i e * = 0.05] | You are considering a project that costs $10,000 up front, with annual end-of year returns estimated to be $800. The MARR is 6%. What is the discounted payback period for the project? [800(P/A,6%,N) > 10,000 for N > 24] | Several new big-box stores have created additional congestion at an intersection in north Barrie, Ontario. City engineers have recommended the addition of a turn-lane, a computer-controlled signal, and sidewalks, at an estimated cost of $1 million. The annual maintenance costs at the new intersection will be $90,000, but users will save $280,000 per year due to reduced waiting time. In addition, accidents are expected to increase, representing a property and medical cost of $12,000 per year. The renovation is expected to handle traffic adequately over a 15-year period. The city uses a MARR of 6 percent. Find the difference between the BCR and BCRM of the project. BCRM – BCR is within 0.05 of [BCR = [280k – 12k]/ [1m(A/P,6%,15) + 90k] = 1.39 BCRM = [280k – 12k – 90k]/ [1m(A/P,6%,15)]= 1.73 BCRM – BCR = 0.34] | You won a sweepstakes. There are two options for how to receive the winnings. You can either receive $1,485,550 in 5 years, or you can receive $250,000 today and an additional payment, $X, after 7 years. How much will the $X payment be in 7 years for the two options to be equivalent at 12% compounded quarterly? [1,485,550 (P/F,i Q = 3%, 5*4 = 20) = 250000 + X(P/F,i Q = 3%, 7*4 = 28) X = 1,309,873] | Asset A has a four-year lifespan and has a first investment of $1000 followed by 4 years of annual $300 maintenance costs. Asset B has a three-year lifespan with a first investment of $2000 following by 3 years of annual $100 maintenance costs. You expect to need the services of only one of the assets indefinitely. Compare the present cost of 12 years of service from each asset. How much more will asset B cost relative to asset A over the 12 years (in present worth)? The answer is within $100 of which of the following? MARR = 6% [AA = 1000(A/P,6%,4) + 300 = 588.59 AB = 2000(A/P,6%,3) + 100 = 848.22 AB – AA = 259.63 PB-A = 259.63(P/A,6%,12) = 2176.69] | The following data is available for two mutually exclusive options to build an extension to the local community centre. The two options provide service for 10 years in duration: First Costs Annual O&M Costs Annual Benefits; Project A 2 million 200,000 700,000; Project B 4 million 200,000 900,000 The MARR is 6% and the social discount rate is 4%. Compute the benefit-cost ratio for the increment (difference) between the projects. [(900k-700k)(P/A,4%,10)] / [4m – 2m] = 0.81] | A project under consideration will lead to revenues that are expected to start at $30,000 and decline by $5000 per year until the 6th year, at which point they will remain stable at $5000 indefinitely. The first cost of the project is $193,521. If the company’s MARR is 10 percent. The IRR is [P = 0 = -193,521 + [30k – 5k(A/G,i*,5)]*(P/A,i*,5) + 5k/i* (P/F,i*,5) If i* = 4% then P = 0] | A construction company owns a set of industrial cranes, many of which are left unused for long periods. The company wants to find out the breakeven daily rate for the cranes. A new crane costs $170,000. The salvage value is determined based on the declining balance method at 20% depreciation per year. Cranes will need to be replaced every 7 years. The company expects end-of-year repair costs to begin at $30,000 and increase by $10,000 per year. They also expect to have the cranes rented 6 days per week, every week of the year. MARR = 6% What daily rate should the company charge to break even? [A = 170k(A/P,6%,7) – 170k(0.8)^7 (A/F,6%,7) + 30k + 10k(A/G,6%,7) = 6*52*X X = 268.85] | Suppose the interest rate is 2.5% per quarter. If you deposit $1000 in an account now, how much will be in the account in 3 years. [1000(1.025)12 = 1344.89] | You deposit $1000 at a rate of 10% nominal interest compounded weekly. Find the account balance after 2.5 years. [i M = (1 + 0.10/52)^(52/12) – 1 = 0.0084 (or 0.84%) F = 1000(1.0084)^2.5*12=30 = 1285.24] | You borrow $10,000 at 6% per year. You pay it off with 10 equal annual payments of. The first payment is 2 years after you receive the loan (instead of 1 year). [X = 10000(F/P,6%,1)(A/P, 6%,10) = 1440.22] | You deposit $1000 into an account today, and increase the deposits by 5% each year for the next 9 years (10 deposit in all). How much is in the account at the time of your final deposit. MARR = 8%? [i 0 = (1.08/1.05) – 1 = 0.0286 (2.86%) V-1 = 1000(P/A,g=5%,i=8%,10) = 1000(8.1824) = 8182.40 F = V 9 = 8182.4(F/P,8%,10) = 17,664.98] | You are considering two different savings options. One involves an up-front payment of $10,000 today. The second involves 5 annual payments of $3000 starting today. Find how much each savings plan accumulates to in 10 years from today. MARR = 10%. What is the difference between the two savings plans? [F1 – F2 = 10,000(F/P,10%,10) – 3000(F/A,10%,5) (F/P10%,6) = 6510.39] | Suppose an investment requires an investment of $1000 and returns $200 a year for 7 years. Find the IRR. [ 0 = -1000 + 200(P/A,i*,7) 5 = (P/A,i*,7) i* = 9% ]