ENSC 31 Future Value and Present Worth PDF

Future Value and ENSC 31 Present Worth Engr. Kelvin Michael A. Crystal Engr. John Paulo M. Perido Department of Agricultural and Food Engineering College of Engineering and Information Technology Cavite State University Indang, C...

Future Value and ENSC 31 Present Worth Engr. Kelvin Michael A. Crystal Engr. John Paulo M. Perido Department of Agricultural and Food Engineering College of Engineering and Information Technology Cavite State University Indang, Cavite Future Value Future value (F or FV) is the value of a current asset at a future date based on an assumed rate of growth. The future value is used to estimate how much an investment made today will be worth in the future. Factors such as inflation, can negatively affect the future value of the asset by eroding its value Present Worth The Present Worth or value (P or PV) is the sum of money which if invested now at a given rate of compound interest will accumulate exactly to a specified amount at a specified future date. Future cash flows are discounted at the discount rate, and the higher the discount rate, the lower the present value of the future cash flows.  SYMBOLIC NOTATIONS Notation Meaning i effective interest rate per interest period n number of compounding (interest) periods present sum of money; the equivalent value of one or more cash flows P at a reference point in time called the present future sum of money; the equivalent value of one or more cash flows at F a reference point in time called the future end-of-period cash flows (or equivalent end-of-period values) in a A uniform series continuing for a specified number of periods, starting at the end of the first period and continuing through the last period. Compound Interest 𝐧 Whenever the interest charge 𝐅𝐧 = P (1 + i) for any interest period is based on the remaining Fn is the total amount of principal amount plus any money accumulated or owed accumulated interest charges after n interest periods. up to the beginning of that period. Relating Present and Future Values of Single Cash Flows Finding P Given F Single payment, present worth factor 𝟏 Factor Functional Symbol P =F [ 𝒏 ] 𝟏+𝒊 (P/F, i%,n) P = F (P/F, i%, n)  Example A firm desires to have Php 1,000 six years from now. What amount should be deposited now to provide for it? Interest rate is 10%. P = 1,000(P/F, 10%, 6) −6 P = 1,000(1 + 0.1) P = 1,000(0.56447) P = Php 564.47 Relating a Uniform Series (Annuity) to Its Present and Future Equivalent Values Annuity a series of payments made at equal intervals. Annuity Four general problems in solving annuity Finding F given A Finding P given A Finding A given F Finding A given P Finding F Given A Uniform series, compound 𝐧 amount factor 𝟏+𝐢 −𝟏 𝐅=𝐀 Factor Functional Symbol 𝐢 (F/A, i%,n) F= A (F/A, i%, n)  Example If three annual deposits of Php 2000 each are placed in an account, how much money has accumulated immediately after the last deposit? Interest rate is 10%. n 1+i −1 F=A i 3 1 + 0.1 − 1 F = 2000 0.1 𝐅 = 𝐏𝐡𝐩 𝟔, 𝟔𝟐𝟎  How did that happen? Using the same example: If three annual deposits of P2000 each are placed in an account, how much money has accumulated immediately after the last deposit? Interest rate is 10%. F1 = A1 1 + i n 2 F1 = 2000 1 + 0.1 𝐅𝟏 = 𝟐𝟒𝟐𝟎 F2 = A2 1 + i n A1 A2 A3 1 F2 = 2000 1 + 0.1 𝐅𝟐 = 𝟐𝟐𝟎𝟎 F3 = A3 1 + i n F = F1 + F2 + F3 Just bring the annuities to 0 F3 = 2000 1 + 0.1 𝐅 = 𝐏𝐡𝐩 𝟔, 𝟔𝟐𝟎 the future! 𝐅𝟑 = 𝟐𝟎𝟎𝟎  Example A uniform amount of Php 5000 is paid every year in 7 years. Calculate the future value of this amount with interest rate 5%. n 1+i −1 F=A i 7 1 + 0.05 − 1 F = 5000 0.05 𝐅 = 𝐏𝐡𝐩 𝟒𝟎, 𝟕𝟏𝟎. 𝟎𝟒 Finding P Given A Uniform series, present 𝐧 worth factor 𝟏+𝒊 −𝟏 𝐏=𝐀 𝐧 Factor Functional Symbol 𝒊 𝟏+𝒊 (P/A, i%,n) P= A (P/A, i%, n)  Example How much should be deposited in a fund now to provide for three end-of-year withdrawals of P10000. Interest rate is 10% 1+i n −1 P=A i 1+i n 3 1 + 0.1 − 1 P = 10000 3 0.1 1 + 0.1 𝐏 = 𝐏𝐡𝐩 𝟐𝟒, 𝟖𝟔𝟖. 𝟓𝟐  How did that happen? End of year 1 𝐹1 = 24,868.52 1 + 0.1 1 𝐹1 = 27,355.372 𝐹1 = 27,355.372 − 10,000 𝑭𝟏 = 𝟏𝟕, 𝟑𝟓𝟓. 𝟑𝟕𝟐 End of year 2 End of year 3 1 1 𝐹2 = 17,355.372 1 + 0.1 𝐹3 = 9,090.9092 1 + 0.1 𝐹2 = 19,090.9092 𝐹3 = 10,000.00 𝐹2 = 19,090.9092 − 10,000 𝐹3 = 10,000 − 10,000 𝑭𝟐 = 𝟗, 𝟎𝟗𝟎. 𝟗𝟎𝟗𝟐 𝑭𝟑 = 𝟎  Example A car is to be bought at 50% down payment and will be paid at P 8,000 monthly for 5 years. If the interest rate is 7%, how much is the car today if it is to be paid in full? n 1+i −1 P=A n i 1+i  Example Solution (compute using the monthly periods) Total compounding periods: 12 periods/ year x 5 years = 60 Monthly interest rate = Annual Nominal rate/ 12 months Monthly interest rate = 7%/ 12 months = 0.005833  Example n 1+i −1 P=A n i 1+i 60 1 + 0.005833 −1 P = (8000) 60 0.005833 1 + 0.005833 P = Php 404,015.948 Full amount = P 404,015.948 + DP (50% of original cost) Full amount = Php 808,031.896 Finding A Given F Uniform series, sinking fund factor 𝒊 𝐀=𝑭 𝐧 Factor Functional Symbol 𝟏+𝒊 −𝟏 (A/F, i%,n) A= F (A/F, i%, n)  Example What uniform annual amount should be deposited each year in order to accumulate P10,000 at the time of the third annual deposit? Interest rate is 10%. i A=F n 1+i −1 0.1 A = 10000 1 + 0.1 3 − 1 𝐀 = 𝐏𝐡𝐩 𝟑, 𝟎𝟐𝟏. 𝟏𝟓 𝐩𝐞𝐫 𝐲𝐞𝐚𝐫  Example When you retire in 25 years from now, you would like to have $500,000 in your retirement account. If you can earn an annual rate of 8%, how much should you deposit at the end of each quarter in order to reach your goal? i A=F n 1+i −1  Example Solution: Total compounding periods: 4 x 25 = 100 Quarterly interest rate: 8%/ 4 = 0.02 i A=F 1+i n−1 0.02 A = 500,000 1 + 0.02 100 − 1 A = $ 1601.37 per quarter Finding A Given P Uniform series, capital 𝐧 recovery factor 𝒊 𝟏+𝒊 𝐀=𝑷 𝐧 Factor Functional Symbol 𝟏+𝒊 −𝟏 (A/P, i%,n) A= P (A/P, i%, n)  Example What is the size of four equal annual payments to repay a loan of Php 1000? The first payment is due 1 year after receiving loan. i = 10% n i 1+i A=P n 1+i −1 4 0.1 1 + 0.1 A = 1000 4 1 + 0.1 − 1 𝐀 = 𝐏𝐡𝐩 𝟑𝟏𝟓. 𝟒𝟕𝟏  Example A man wishes to purchase a car with a total cost of 450,000. He made a down payment of 50,000 and the balance is payable in 24 monthly installments. If the effective interest rate is 12% for each year computed on the total balance to be paid by installment. How much would each installment payment be? n i 1+i A=P 1+i n−1  Example Solution: take note of the term ‘effective interest rate.’ The use of ia will be essential if the computations are on a yearly basis. Interest rate per month = I = (1 + r/M)M – 1 0.12 = (1 + r/12) – 1 12 r/12 = 0.00949 or 0.949% per month 24 0.00949 1 + 0.00949 A = 450,000 – 50,000 1 + 0.00949 24 − 1 𝐀 = 𝐏𝐡𝐩 𝟏𝟖, 𝟕𝟏𝟓. 𝟎𝟎 per month Finding n Given F,P, and A When given F and P When given A and P F A ln ln P A − iP n= n= ln (1 + i) ln (1 + i) When given F and A iF ln +1 A n= ln (1 + i) Summary of Discrete Compounding -Interest Factors and Symbols Sullivan, W.G. et. al. (2014). Engineering Economy 16th ed.  Example What is the future worth of P600 deposited at the end of every month for 4 years if the interest is 12% compounded quarterly? In order to solve this, we need to know the equivalent nominal rate when the compounding is monthly. Take note that the effective annual interest rate will always be the basis since this will never change across any annual compounding periods.  Example a) Compute for the effective interest rate 4 0.12 𝑖𝑎 = 1 + − 1 = 0.125509 4 b) Compute for the nominal rate using the answer from the effective interest rate. 𝑟 12 0.125509 = 1 + − 1 = 0.118820 12 What this means is that 12% compounded quarterly is equivalent to 11.88% compounded monthly. We can now use the 11.88% to compute for F given A  Example You can also equate the nominal and effective rate for faster computation as: 4 12 0.12 𝑟 1+ −1= 1+ −1 4 12 Quarterly compounding Monthly compounding 4 12 0.12 𝑟 1+ = 1+ 4 12 r = 0.118820  Example What is the future worth of Php 600 deposited at the end of every month for 4 years if the interest is 12% compounded quarterly? 4 12 0.12 𝑟 1+ = 1+ 4 12 i or r = 11.88% compounded monthly  Example n 1+i −1 F=A i 4∗12 0.1188 1+ −1 12 F = 600 0.1188 12 𝐅 = 𝐏𝐡𝐩 𝟑𝟔, 𝟔𝟒𝟏. 𝟑𝟐  Example For cases that has cash flows other than F or P A workshop tool costs P18,000. It is expected to have a useful life of 10 years and a salvage value of P3,000. At 15%, what is the capital recovery of the tool?  Example In this case we have a salvage value of 3000 that we need to bring to its present value so that it does not interfere with the computation. To get the present worth of 3000, we use P given F P = 3000 1 + 0.15 −10 P = Php 741.55 Then we deduce this to our Present investment since this is positive, we get the initial investment as Php 18000- Php 741.55 = Php 17,258.45  Example Continuing, we get: N i 1+i A=P 1+i N−1 10 0.15 1 + 0.15 A = 17258.45 1 + 0.15 10 − 1 𝐀 = 𝐏𝐡𝐩 𝟑𝟒𝟑𝟖. 𝟕𝟖 This means that every year for 10 years, we benefit an amount of P3438.78 for the utility of the workshop tool.  Example This can also be solved as Capital recovery = I (A/P,i%,n) – SV (A/F, i%, n) where I is the investment (P) and SV is the salvage value (F) 10 0.15 1 + 0.15 0.15 Capital Recovery =18,000 − 3000 1 + 0.15 10 − 1 1 + 0.15 10 − 1 Capital Recovery =3,586.537125 − 147.7561876 Capital Recovery = Php 3,438.780938  Example Your company has a $100,000 loan for a new security system it just bought. The annual payment is $8,880 and the interest rate is 8% per year for 30 years. Your company decides that it can afford to pay $10,000 per year. How long will the payment be now? How much must be paid off in the last year to complete the payment? P = $100,000 A = $10,000 I = 8%  Example A ln A − iP n= ln (1 + i) 10,000 ln 10,000 − 0.08 x 100,000 n= ln (1 + 0.08) 𝐧 = 𝟐𝟎. 𝟗𝟏 𝐲𝐞𝐚𝐫𝐬 𝐨𝐫 𝟐𝟏 𝐲𝐞𝐚𝐫𝐬  Example How much must be paid off in year 21 to complete the payment? $10,000 (P/A, 8%, 20) + F (P/F, 8%, 21) = $100,000 $10,000 (9.8181) + F (0.1987) = $100,000 F = $9,154.50 Other types of Annuity ANNUITY DUE Annuity Due In annuity due, the equal payments are made at the beginning of each compounding period starting from the first period. Annuity Due The diagram below shows the cash flow in annuity due.  Annuity Due F given A P given A 𝐧 −𝐧 𝟏+𝐢 −𝟏 𝟏− 𝟏+𝐢 𝐅=𝐀 𝟏+𝐢 𝐏=𝐀 𝟏+𝐢 𝐢 𝐢  Example Alan wants to deposit $300 into a fund at the beginning of each month. If he can earn 10% compounded interest monthly, how much amount will be there in the fund at the end of 6 years? N 1+i −1 F=A 1+i i 72 0.1 1+ −1 0.1 12 F = 300 1+ 0.1 12 12 𝐅 = 𝟐𝟗, 𝟔𝟔𝟑 Other types of Annuity DEFERRED ANNUITY Deferred Annuity In deferred annuity the first payment is deferred a certain number of compounding periods after the first. In the diagram, the first payment was made at the end of the kth period and n number of payments was made. The n payments form an ordinary annuity as indicated in the figure. Deferred Annuity  Example When you take your first job, you decide to start saving right away for your retirement. You put $5,000 per year into the company’s retirement plan, which averages 8% interest per year. Five years later, you move to another job and start a new plan. You never get around to merging the funds in the two plans. If the first plan continued to earn interest at the rate of 8% per year for 35 years after you stopped making contributions, how much is the account worth? Example  Example Example: Deferred annuity The total value of the contributions at the end of the five years will be: F5 = $5,000 F/A, 8%, 5 F5 = $5,000 5.8666 𝐅𝟓 = $𝟐𝟗, 𝟑𝟑𝟑  Example In order to get the value of this investment 35 years after you stopped contributing: 𝐹40 = 𝑃5 𝐹/𝑃, 8%, 35 35 𝐹40 = $29,333 1 + 0.8 𝐹40 = $29,333 14.7853 𝑭𝟒𝟎 = $𝟒𝟑𝟑, 𝟔𝟗𝟖 To solve for deferred annuity, bring the annuities to the last year of payment then this future worth (F5) shall be used as present value after 5 years (P5) for the last computation (F40).  Example Suppose that a father, on the day his son is born, wishes to determine what lump amount would have to be paid into an account bearing interest of 12% per year to provide withdrawals of $2,000 on each of the son’s 18th, 19th, 20th, and 21st birthdays  Example Determine the total present worth of 4 withdrawals. 𝑃17 = 𝐴 𝑃/𝐴, 12%, 4 1 + 0.12 4 − 1 𝑃17 = $2,000 0.12 1 + 0.12 4 𝑃17 = $2,000 3.0373 𝑷𝟏𝟕 = $𝟔, 𝟎𝟕𝟒. 𝟔𝟎  Example In order to determine the present worth of P17, well make it the future value as F17. 𝑃0 = 𝐹17 𝑃/𝐹, 12%, 17 1 𝑃0 = $6,074.60( 17 1 + 0.12 𝑃0 = $6,074.60 0.1456 𝑷𝟎 = $𝟖𝟖𝟒. 𝟒𝟔 This means that the father will only invest a total amount of $884.46 at year zero and from year 1 to 17 are deferred. Other types of Annuity PERPETIUTY Perpetuity Perpetuity is an annuity where the payment period extends forever, which means that the periodic payments continue indefinitely. Perpetuity  Perpetuity There is no definite future Present amount of in perpetuity, thus, there is perpetuity, P no formula for the future amount. 𝐀 𝐏= 𝐢  Example If someone is promise a cash flow of $400 per year until they died and they could earn 6% on other investments of similar quality, in present value terms, what is the perpetuity worth? A P= i 400 P= 0.06 𝐏 = $ 𝟔𝟔𝟔𝟔. 𝟔𝟕  There are several ways on how you can solve for the present, future or annuity value of any cash flows. The most important thing is that you understand the equivalence of each value and how time affects them. Store as much decimal places as possible.