Engineering Economy

Engineering Economy Blank, L T., (2012), Engineering Economy (7th Edition), New York: McGraw Hill, Chapter 1

Description and Role in Decision Making Most decisions involve money, called capital or capital funds, which is usually limited in amount. The decision of where and how to invest this limited capital is motivated by a primary goal of adding value as future, anticipated results of the selected alternative are realized.

Description and Role in Decision Making Since most decisions affect what will be done, the time frame of engineering economy is primarily the future. Therefore, the numbers used in engineering economy are best estimates of what is expected to occur. The estimates and the decision usually involve four essential elements: • Cash flows • Times of occurrence of cash flows • Interest rates for time value of money • Measure of economic worth for selecting an alternative

Description and Role in Decision Making

Performing an Engineering Economy Study The steps in an engineering economy study are as follows: 1. Identify and understand the problem; identify the objective of the project. 2. Collect relevant, available data and define viable solution alternatives. 3. Make realistic cash flow estimates. 4. Identify an economic measure of worth criterion for decision making.

Performing an Engineering Economy Study The steps in an engineering economy study are as follows: 5. Evaluate each alternative; consider noneconomic factors; use sensitivity analysis as needed. 6. Select the best alternative. 7. Implement the solution and monitor the results.

Interest and Money-Time Relationships Interest – is the amount of money paid for the use of borrowed capital (borrower’s viewpoint) or the income produced by money which has been loaned (lender’s viewpoint). F=P+I

Where: I = interest P = principal or present worth F = accumulated amount or future worth

Interest and Money-Time Relationships Cash-Flow Diagrams Cash-Flow Diagram – is a graphical representation of cash flows drawn on a time scale. ↑ - receipt (positive cash flow or cash inflow) ↓ - disbursement (negative cash flow or cash outflow)

Interest and Money-Time Relationships Cash-Flow Diagrams Example: A loan of P100 at simple interest will become P150 after 5 years.

Cash flow diagram on the viewpoint of the lender.

Cash flow diagram on the viewpoint of the borrower.

Interest and Money-Time Relationships Simple Interest The interest earned by the principal is computed at the end of the investment period.

Future Worth, F: F=P+I Interest earned, I: I = Prt Where: P = principal or present worth r = simple interest rate (per year) t = time in years of fraction of a year

Interest and Money-Time Relationships Simple Interest Ordinary simple interest The interest is computed on the basis of one banker’s year (1 banker’s year = 360 days) 𝑫 I = Pr 𝟑𝟔𝟎

Interest and Money-Time Relationships Simple Interest Exact simple interest The interest is based on the exact number of days in a year (ordinary year = 365 days, leap year = 366 days) 𝑫 I = Pr 𝟑𝟔𝟓

Interest and Money-Time Relationships Simple Interest Example 1: Determine the ordinary simple interest on P 20,000 for 9 months and 10 days if the rate of interest is 12%. Ans. P 1,866.67

Interest and Money-Time Relationships Simple Interest Example 2: Determine the (a) ordinary and (b) exact simple interests on P 100,000 for the period January 15 to June 20 2012 if interest is 15%. Ans. (a) P 6,541.67; (b) P 6434.43

Interest and Money-Time Relationships Simple Interest Example 3: Calculated the exact interest on an investment of P 2,000.00 for a period from January 30 to September 15, 2001 if the rate of interest is 10%. Ans. P 124.93

Interest and Money-Time Relationships Simple Interest Example 4: If P 4000 is borrowed for 75 days at 16% per annum. How much will be due at the end of 75 days? Ans. P 4133.33

Interest and Money-Time Relationships Simple Interest Example 5: How long will it take for a deposit of P 1, 500.00 to earn P 186 if invested at the simple interest rate of 7 1/3%? Ans. 1.6909 years

Interest and Money-Time Relationships Simple Interest Example 6: If you borrow money from your friend with simple interest of 12%, find the present worth of P 20,000 at the end of 9 months. Ans. P 18,348.60

Interest and Money-Time Relationships Simple Interest Assignment 1: (CE Board) A deposit of P 110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P 890.36. Find the rate of return annually.

Interest and Money-Time Relationships Simple Interest Assignment 1: (CE Board) A deposit of P 110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P 890.36. Find the rate of return annually. Ans. 11.75%

Interest and Money-Time Relationships Compound Interest The interest earned is computed every end of each interest period (compounding period) and the interest earned for that period is added to the principal.

Future Worth, F: F = P(1+i)n Where: i = effective interest per interest period nominal interest rate i= number of compounding per year

n = total number of compounding

Compound Interest

To compute values of i and n: nominal interest rate = 12% number of years of investment = 6 years a. compounded annually i= 0.12/1 = 0.12 n = 6(1) = 6 b. compounded semi-annually i= 0.12/2 = 0.06 n = 6(2) = 12 c. compounded quarterly i= 0.12/4 = 0.03 n = 6(4) = 24

d. compounded monthly i= 0.12/12 = 0.01 n = 6(12) = 72 e. compounded bi-monthly i= 0.12/6 = 0.02 n = 6(6) = 36

Compound Interest Single Payment Compound Amount Factor, F/P: Fൗ = 1 + i P

n

= FൗP , i%, n

Single Payment Present Worth Factor, P/F: Pൗ = 1 + i F

−n

= PൗF , i%, n

Compound Interest Effective Rate of Interest, ER: ER =

F−P interest earned in one year = = (1+i )n - 1 principal during that year P

Equivalent Rates: ER1 = ER2

Interest and Money-Time Relationships Compound Interest Example 1: The amount of P 20,000 was deposited in a bank earning an interest rate of 6.5% per annum. Determine the total amount at the end of 7 years if the principal and interest were not withdrawn during this period. Ans. P 31,079.73

Interest and Money-Time Relationships Compound Interest Example 2: A man expects to receive P 25,000 in 8 years. How much is that money worth now considering interest at 8% compounded quarterly? Ans. P 13,265.83

Interest and Money-Time Relationships Compound Interest Example 3: How many years will P 100,000 earn a compounded interest of P 50,000 if interest is 9% compounded quarterly? Ans. 4.56 years

Interest and Money-Time Relationships Compound Interest Example 4: A sum of P 1,000.00 is invested now and left for eight years, at which time the principal is withdrawn. The interest has accrued is left for another eight years. If the effective annual interest rate is 5%, what will be the withdrawal amount at the end of the 16th year? Ans. P 705.42

Interest and Money-Time Relationships Compound Interest Example 5: If money is worth 5% compounded quarterly, find the equated time for paying a loan of P 150,000 due in one year and P 280,000 in 2 years. Ans. 1.6455 years

Interest and Money-Time Relationships Compound Interest Example 6: How long will it take money to double itself if invested at 5% compounded annually? Ans. 14.2 years

Interest and Money-Time Relationships Compound Interest Example 7: Compute the equivalent rate of 6% compounded semiannually to a rate compounded quarterly. Ans. 5.96% ≈ 6% compounded quarterly

Interest and Money-Time Relationships Compound Interest Example 8: If P5, 000.00 shall accumulate for 10 years at 8% compounded quarterly. Find the compounded interest at the end of 10 years. Ans. P 6,040.20

Plates Simple Interest 1. A man buys an electric fan from a merchant that charges P1500.00 at the end of 90 days. The man wishes to pay cash. What is the cash price if money is worth 10% simple interest? 2. What amount will be available in eight months if P 15,000.00 is invested now at 10% simple interest per year? 3. P 1000.00 becomes P 1500.00 in three years. Find the simple interest rate. 4. An engineer borrowed a sum of money under the following terms: P 650,000.00 if paid in 90 days, or P 600,000.00 if paid in 30 days. What is the equivalent annual rate of simple interest?

Plates Compound Interest 1. By the condition of a will, the sum of P 2,000 is left to a girl to be held in trust fund by her guardian until it amounts to P 5,000, when will the girl received the money if the fund is invested at 8% compounded quarterly? 2. A student plan to deposit P1, 500 in the bank now and another P3, 000 for the next 2 years. If he plans to withdraw P5, 000 3 years after his last deposit for the purpose of buying shoes, what will be the amount of money left in the bank after one year of his withdrawal? Effective annual interest rate is 10%. 3. If the interest rate of a certain account is 6.5%, compute the (a) single payment present worth factor; and (b) single payment compound amount factor at the end of 18 years.

Interest and Money-Time Relationships Continuous Compounding Interest F = Pert where: ert is continuous compounded amount factor r = nominal rate of interest t = number of years

Effective rate if compounded continuously, ER: ER = er - 1

Interest and Money-Time Relationships Continuous Compounding Interest Example 1: P 100,000 is deposited in a bank that earns 5% compounded continuously. What will be the amount after 10 years? Ans. P 164,872.13

Interest and Money-Time Relationships Continuous Compounding Interest Example 2: Money is deposited in a certain account for which interest is compounded continuously. If the balance doubles in 6 years, what is the annual percentage rate? Ans. 11.55%

Interest and Money-Time Relationships Continuous Compounding Interest Example 3: A man wishes to have P 40,000 in a certain fund at the end of 8 years. How much should he invest in a fund that will pay 6% compounded continuously? Ans. P 24, 751.34

Interest and Money-Time Relationships Continuous Compounding Interest Example 4: If the effective annual interest rate is 4%, compute the equivalent nominal interest compounded continuously. Ans. 3.922%

Interest and Money-Time Relationships Continuous Compounding Interest Example 5: What is the nominal rate of interest compounded continuously for 10 years if the compound amount factor is 1.34986? Ans. 3%

Interest and Money-Time Relationships Continuous Compounding Interest Example 6: Deposits of P35,000.00, P48,000.00 and P25,000.00 were made in a savings account eight years, five years, and two years ago, respectively. Determine the accumulate amount in the account today if a withdrawal of P55,000.00 was made four years ago. The applied interest rate is 11% compounded continuously. Ans. P 113,330.66

Interest and Money-Time Relationships Discount It is the difference between the future worth of a certain commodity and its present worth. 2 Types of Discount: • Trade Discount – discount offered by the seller to induce trading. • Cash Discount – is the reduction on the selling price offered to a buyer to induce him to pay promptly.

Interest and Money-Time Relationships Discount D=F–P

where: D = amount of discount F = accumulated amount or future worth P = principal or present worth

Discount Rate – is the discount on one unit of principal per unit of time. 𝐅−𝐏 𝐝= =𝟏− 𝟏+𝐢 𝐅

−𝟏

Interest and Money-Time Relationships Discount If the commodity is discounted in a certain period of time:

Fd = F − P P = F 1 − d For 1 year 𝐏 = 𝐅 𝟏 − 𝐧𝐝 For n years

The relationship between discount rate and interest rate becomes: 𝐝 𝐢= 𝟏−𝐝 and 𝐢 𝐝= 𝟏+𝐢 where: d = discount rate for the period involved i = rate of interest for the same period

Interest and Money-Time Relationships Discount Example 1: Mr. T borrowed money from the bank. He receives from the bank P 1,340 and promised to pay P 1,500 at the end of 9 months. Compute: (a) Simple interest rate; and (b) Discount Rate. Ans. (a) 15.92%; (b) 13.73%

Interest and Money-Time Relationships Discount Example 2: Find the discount if P 2,000 is discounted for 6 months at 8% simple discount. Ans. P 80

Inflation It is the increase in the prices for goods and services from one year to another, thus decreasing the purchasing power of money. 𝐅𝐂 = 𝐏𝐂 𝟏 + 𝐟

𝐧

where: FC = future cost of a commodity PC = present cost of a commodity f = annual inflation rate n = number of years

In an inflationary economy, the buying power of money decreases as cost increases: 𝐏 𝐅= 𝟏+𝐟 𝐧 If interest is computed as the same time that inflation is occurring: 𝐧 𝟏+𝐢 𝐅=𝐏 𝟏+𝐟 where: F = future worth of today’s present amount P f = annual inflation rate n = number of years i = rate of interest

Inflation If the uninflated present worth is to be determined: 𝐏=

𝐅 𝟏+𝐢

𝐧

𝟏+𝐟

𝐧

𝐅 = 𝟏 + 𝐢𝐜𝐟

𝐢𝐜𝐟 = 𝐢 + 𝐟 + 𝐢𝐟

𝐧

where: f = annual inflation rate i = rate of interest 𝐢𝐜𝐟 = rate of interest that can take care of the cost of money and inflation

Interest and Money-Time Relationships Inflation Example 1: A man invested P 130,000 at an interest rate of 10% compounded annually. What will be the final amount of his investment, in terms of today’s peso, after 5 years, if inflation remains the same at the rate of 8% per year? Ans. P 142,491

Interest and Money-Time Relationships Inflation Example 2: What is the uninflated present worth of a P 200,000 future value in two years if the average inflation rate is 6% and interest rate is 10%. Ans. P 147,107

Plates Continuous Compounding Interest 1. If money is invested at a nominal rate of interest of 8% for a period of 4yrs. (a) What is the effective rate if it is compounded continuously. (b) What is the value of the compound amount factor if it is compounded continuously. 2. What is the nominal rate of interest compounded continuously for 8 years if the present worth factor is equal to 0.6187835. Inflation 1. If the inflation rate is 6%, cost of money is 10%, what interest rate will take care of inflation and the cost of money. 2. In year zero, you invest P 10,000 in a 15% security for 5 years. During that time, the average annual inflation is 6%. How much, in terms of year zero will be in the account at maturity.

Plates Discount 1. An engineer promised to pay P 36,000 at the end of 90 days. He was offered a 10% discount if he pays in 30 days. Find the rate of interest.

Review Quiz 1. What is the principal amount if the amount of interest at the end of 2 ½ years is P45,000 for a simple interest rate of 6% per annum? 2. Which of the following has the greatest effective rate? Show your solution per interest rate given. a. 12.31% compounded quarterly b. 12.20% compounded monthly c. 12.35% compounded annually d. 12.32% compounded semi-annually 3. A certain nominal annual interest rate has an effective rate of 19.722% when compounded continuously. What is its effective rate if compounded bi-monthly?

Annuities Ordinary Annuity Ordinary Annuity – a type of annuity were equal payments are made at the end of each period.

Annuity Due Annuity Due – a type of annuity were equal payments are made at the beginning of each period. Deferred Annuity Deferred Annuity – a type of annuity where the first payment is made several periods after the beginning of annuity.

Annuities Perpetuity Perpetuity – a type of annuity in which payments continue indefinitely. Continuous Compounding for Discrete Payments Continuous Compounding for Discrete Payments – an annuity compounded continuously.

Ordinary Annuity Ordinary Annuity – a type of annuity were equal payments are made at the end of each period.

Ordinary Annuity

Ordinary Annuity The functional symbol P/A, i%, n is called the “uniform series present worth factor”. Pൗ = A

1+i n−1 Pൗ , i%, n = A i 1+i n

The functional symbol A/P, i%, n is called the “capital recovery factor”. n i 1 + i Aൗ = Aൗ , i%, n = P P 1+i n−1

Ordinary Annuity Where: P = value or sum of money at present A = series of periodic equal amount of payments i = interest rate per interest period n = number of interest periods/number of equal payments P/A = uniform series present worth factor A/P = capital recovery factor

Ordinary Annuity Substituting P = F 1 + i

from the equation of P, it becomes: A F= 1+i n−1 i −n

Ordinary Annuity The functional symbol F/A, i%, n is called the “uniform series compound amount factor”. n−1 1 + i Fൗ = Fൗ , i%, n = A A i The functional symbol A/F, i%, n is called the “sinking fund factor”. i Aൗ = Aൗ , i%, n = F F 1+i n−1

Ordinary Annuity Where: F = value or sum of money at some future time A = series of periodic equal amount of payments i = interest rate per interest period n = number of interest periods/number of equal payments F/A = uniform series compound amount factor A/F = sinking fund factor

Annuities Ordinary Annuity Example 1: Find the annual payment to extinguish a debt of P 100,000 payable for 6 years at 12% interest annually. Ans. P 24,322.57

Annuities Ordinary Annuity Example 2: What annuity is required over 12 years to equate to a future amount of P 200,000? i = 8%. Ans. P 10,539.00

Annuities Ordinary Annuity Example 3: Mr. Y bought a house and lot for $ 2,800,000 with a downpayment of $ 300,000. Interest is 5% to be paid for 30 years on a monthly basis. Compute the amount of monthly payment. Ans. $ 13,420.54

Annuities Ordinary Annuity Example 4: An annual payment is made for 10 years with an annual interest rate of 8%. Compute the following: (a) Uniform series present worth factor; (b) Capital recovery factor; (c) Uniform series compound amount factor; (d) Sinking fund factor Ans. (a) 6.710; (b) 0.149; (c) 14.487; (d) 0.069

Annuities Ordinary Annuity Example 5: A piece of machinery can be bought for P 10,000 cash, or for P 2,000 downpayment and payments of P 750 per year for 15 years. What is the annual interest rate of the time payments? Ans. 4.6%

Annuities Ordinary Annuity Example 6: If P500.00 is invested at the end of each year for 6 years, at an annual interest rate of 7%, what is the total peso amount available upon the deposit of the sixth payment? Ans. P 3,576.65

Plates Ordinary Annuity 1. For having been loyal, trustworthy and efficient, the company has offered a superior a yearly gratuity pay of P 20,000.00 for 10 years with the first payment to be made one year after his retirement. The supervisor, instead, requested that he be paid a lump sum on the date of his retirement less interest that the company would have earned if the gratuity is to be paid on yearly basis. If interest is 15%, what is the equivalent lump sum that he could get? 2. In anticipation of a much bigger volume of business after 10 years, a fabrication company purchased an adjacent lot for its expansion program where it hopes to put up a building projected to cost P 4,000,000.00 when it will be constructed 10 years after. To provide for the required capital expense, it plans to put up a sinking fund for the purpose. How much must the company deposit each year if interest to be earned is computed at 15%?

Annuity Due Annuity Due – a type of annuity were equal payments are made at the beginning of each period.

Annuity Due Where: P = value or sum of money at present F = value or sum of money at some future time A = series of periodic equal amount of payments i = interest rate per interest period n = number of interest periods/number of equal payments

Annuities Annuity Due Example 1: A man agrees to make equal payments at the beginning of each 6 months for 10 years to discharge a debt of P 50,000 due now. If money is worth 8% compounded semiannually, find the semiannual payment. Ans. P 3,537.58

Annuities Annuity Due Example 2: To accumulate a fund of P 80,000 at the end of 10 years, a man will make equal annual deposits in the fund at the beginning of each year. How much should he deposit if the fund is invested at 5% compounded annually? Ans. P 6,057.49

Annuities Annuity Due Example 3: Determine the present worth and the accumulated amount of an annuity consisting of 6 payments of P120, 000 each, the payment are made at the beginning of each year. Money is worth 15% compounded annually. Ans. P = P 522,259; F = P 1,208,016

Annuities Annuity Due Example 4: If money is worth 4% compounded semiannually, find the present amount of an annuity due paying P 5,000 semiannually for a term of 3.5 years. Ans. P 33,007.15

Plates Annuity Due 1. A farmer bought a tractor costing P 25,000 payable in 10 semiannual payments, each installment payable at the beginning of each period. If the rate of interest is 26% compounded semiannually, determine the amount of each installment. 2. A certain manufacturing plant is being sold and was submitted for bidding. Two bids were submitted by interested buyers. The first bid offered to pay P 200,000 each year for 5 years, each payment being made at the beginning of each year. The second bid offered to pay P 120,000 the first year, P 180,000 the second year, and P 270,000 each year for the next 3 years, all payments being made at the beginning of each year. If money is worth 12% compounded annually, which bid should the owner of the plant accept?

Perpetuity Perpetuity – a type of annuity in which payments continue indefinitely. Where: P = value or sum of money at present A = series of periodic equal amount of payments i = interest rate per interest period

Annuities Perpetuity Example 1: Find the present worth of perpetuity of P 5,200 payable monthly if the interest is 16% compounded monthly. Ans. P 390,000

Annuities Perpetuity Example 2: Find the present value of a perpetuity of P 15,000 payable semiannually if money is worth 8% compounded quarterly. Ans. P 371,287

Annuities Perpetuity Example 3: If money is worth 12% compounded quarterly, what is the present value of the perpetuity of P1,000 payable monthly? Ans. P100,993.78

Deferred Annuity Deferred Annuity – a type of annuity where the first payment is made several periods after the beginning of annuity. 𝐀 𝟏+𝐢 𝐧−𝟏 𝐏= 𝟏+𝐢 𝐧 𝐢 𝟏+𝐢

−𝐦

Deferred Annuity A 1+i n−1 P= 1+i n i 1+i

−m

Where: P = value or sum of money at present F = value or sum of money at some future time A = series of periodic equal amount of payments i = interest rate per interest period n = number of interest periods/number of equal payments m = number of interest periods when there is no payment made

Deferred Annuity A. Present Worth [ 1+i 5 −1] P1 = (1+i)5i

P=

P1 1+i

3

Deferred Annuity B. Future Worth A[ 1+i 6 −1] F= 1+i 6i

F2 = F(1 + i)4

Annuities Deferred Annuity Example 1: A boy is entitled to 10 yearly endowments of P30,000 each starting at the end of the eleventh year from now. Using an interest rate of 8% compounded annually, what is the value of these endowment now? Ans. P93,241.98

Annuities Deferred Annuity Example 2: A man invested P100,000 every end of the year for 10 years, then waited for another 10 years for his money to grow. If his investments earned 8% after tax compounded annually, what would be the sum of his investments and earnings at the end of the 20th year in pesos. Ans. P 3,127,540.18

Annuities Deferred Annuity Example 3: A parent on the day that child is born wishes to determine what lump sum would have to be paid into an account bearing interest of 5% compounded annually, in order to withdraw P 20,000 each on the child’s 18th, 19th , 20th and 21th birthdays? Ans. P 30,941.73

Annuities Deferred Annuity Example 4: An asphalt road requires no upkeep until the end of 2 years when P60, 000 will be needed for repairs. After this P90, 000 will be needed for repairs at the end of each year for the next 5 years, then P120, 000 at the end of each year for the next 5 years. If money is worth 14% compounded annually, what was the equivalent uniform annual cost for the 12-year period? Ans. P 79,245.82

Plates Perpetuity 1. If money is worth 8%, determine the present value of a perpetuity of P 1,000 payable annually with the 1st payment due at the end of 5 years. 2. It costs P 50,000 at the end of each year to maintain a section of Kennon road in Baguio City. If money is worth 10%, how much would it pay to spend immediately to reduce the annual cost by P 10,000?

Plates Deferred Annuity 1. A lathe for a machine shop costs P 60,000, if paid in cash. On the installment plan, a purchaser should pay P 20,000 downpayment and 10 quarterly installments, the first due at the end of the first year after purchase. If money is worth 15% compounded quarterly, determine the quarterly installment. 2. A man invests P 10,000 now for the college education of his 2 year old son. If the fund earns 14% effective interest rate, how much will his son get each year starting from his 18th to the 22nd birthday?

Uniform Payment Series with Continuous Compounding For an annuity compounded continuously, replace interest rate with the effective rate for compounded continuously. Recall that: ER = er − 1 Replacing the interest rate for the formula of ordinary annuity with ER, the formula becomes: 𝐀 𝐞𝐫𝐧 − 𝟏 𝐏= 𝐫 𝐞 −𝟏 𝐞𝐫𝐧 𝐀 𝐅= 𝐫 𝐞𝐫𝐧 − 𝟏 𝐞 −𝟏

Uniform Payment Series with Continuous Compounding 𝐏= Where:

𝐀 𝐞𝐫𝐧 −𝟏 𝐞𝐫 −𝟏 𝐞𝐫𝐧

P = Present Worth F = Future Worth r = nominal rate of interest n = number of years

and

𝐅=

𝐀 𝐞𝐫 −𝟏

𝐞𝐫𝐧 − 𝟏

Uniform Payment Series with Continuous Compounding Compound amount factor, Sinking Fund Factor,

ern −1 CAF = r e −1

SFF = Present Worth Factor, Capital recovery factor,

er −1 ern −1

1− e−rn PWF = r e −1

er −1 CRF = 1 − e−rn

Annuities Uniform Payment Series with Continuous Compounding Example 1: Determine the accumulated amount to an account paying P 5,000 annually (payments are made at the beginning of each period) for 18 years if money is worth 8% compounded continuously. Also determine the present worth. Ans. P 209,452.57; P 49,625.13

Annuities Uniform Payment Series with Continuous Compounding Example 2: Ryan invest P5,000 at the end of each year in an account which gives a nominal annual interest of 7.5%, compounded continuously. Determine the total worth of his investment at the end of 15 years. Ans. P 133,545.58

Plates Continuous Compounding for Discrete Payments 1. A boy deposited an amount of P600 each year in a local bank that pays a nominal interest of “i”% per annum compounded continuously. If he receives a total amount of P3323.81 after % years, a. Compute the value of “i” b. What is the effective interest rate?

2. For a uniform series of payments, what is the value of (P/A, 8%, 4) if it is compounded continuously.

Review Quiz 1. Find the value after 20 years in pesos of an annuity of P20,000 payable annually for 8 years, with the first payment at the end of 2 years, if money is worth 5%. 2. A man receives P145,000 credit for his old car when buying a new model costing P375,000. What cash payment will be necessary so that the balance can be liquidated by payments of P12,500 at the end of each month for 18 months when interest is charged at the rate of 6% compounded monthly? 3. What quarterly payment is required over 15 years to equate with a future amount P150,000? Assume interest rate of 6% compounded continuously. 4. With interest rate of 9% compounded continuously, what is the present worth of a perpetuity of P8000 payable monthly?

Uniform Arithmetic Gradient Uniform Arithmetic Gradient – is the increase by a relatively constant amount each period.

The cash flow above is equal to the sum of the two cash flows shown.

Uniform Arithmetic Gradient We denote the difference between two preceding amount (increase per period) as G, which is also known as uniform gradient amount, in this case: G = 500, n = 5 The formulas that may be used in this type of cash flow may be analyzed using this formulas:

Annuities Uniform Arithmetic Gradient Example 1: An individual makes 5 deposits that increase uniformly by P 300 every month in a savings account that earns 12% compounded monthly. If the initial deposit is P 4,500, determine the accumulated amount in the account just after the last deposit. Ans. P 25,984.67

Annuities Uniform Arithmetic Gradient Boardwork: A contract has been signed to lease a building at P20,000 per year with an annual increase of P1,500 for 8 years. Payments are to be made at the end of each year, starting one year from now. The prevailing interest rate is 7%. What lump sum paid today would be equivalent to the 8-year lease-payment plan?

Depreciation • Depreciation – is the decrease in value of physical property with the passage of time. • Book Value – is the worth of property as shown on the accounting records of an enterprise • Salvage/Resale Value – is the price that can be obtained from the sale of the property after it has been used. • Scrap Value – the amount of property would sell if disposed of as junk.

Depreciation

Where: BVm = book value of a property at any time m Dm = total depreciation of a property at any time m DL = total depreciation at the end of its useful life FC = first cost SV = salvage or scrap value

Depreciation Straight Line Method Straight Line Method –a method which assumes that the loss in value is directly proportional to the age of the property.

Where: d = depreciation at any year Dm = total depreciation of a property at any time m DL = total depreciation at the end of its useful life L = useful life in years FC = first cost SV = salvage or scrap value

Depreciation Straight Line Method Example 1: A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after 10 years. Using Straight Line Method of Depreciation: (a) What is the annual depreciation? (b) What is the book value after 5 years? (c) What is the total depreciation after 3 years? Ans. (a) P 4,000; (b) P 30,000; (c) P 12,000

Depreciation Straight Line Method Example 2: An Engineer bought an equipment for P 500,000. He spent an additional amount of P 30,000 for installation and other expenses. The salvage value is 10% of the first cost. If the book value at the end of 5 years is P 291,500 using straight line depreciation, compute the life of the equipment in years. Ans. 10 years

Depreciation Straight Line Method Example 3: A machine which cost P 10,000 was sold as scrap after being used for 10 years. The scrap value is P 500. Determine the total depreciation at the end of 5 years. Ans. P 4750

Depreciation Straight Line Method Example 4: An equipment has a salvage value of P1M at the end of 50 years. The straight line depreciation charge is P2M. (a) What is the first cost of the machine? (b) What is the book value after 25 years? (c) At what year will its total depreciation be P30M? Ans. P 101M; P 51M; 15th year

Plates Straight Line Method 1. What is the book value of electronic test equipment after 8 years of use if it depreciates from its original value of P 120,000.00 to its salvage value of 13% in 12 years? Use straight-line method. 2. The initial cost of paint sand mill, including its installation is, P800 000.00. The BIR approved life of this machine is 10 years for depreciation. The estimated salvage value of the mill is P 50,000.00 and the cost of dismantling is estimated to be P 15,000.00. Using straight line depreciation, what is the annual depreciation charge and what is the book value of the machine at the end of six years?

Depreciation Sinking Fund Method Sinking Fund Method – a method which assumes that the sinking fund established in which funds will accumulate for replacement. The total depreciation that has been taken place up to any given time is assumed to be equal to the accumulated amount in the sinking fund at any time.

Where: d = depreciation at any year Dm = total depreciation of a property at any time m DL = total depreciation at the end of its useful life L = useful life in years FC = first cost SV = salvage or scrap value

Depreciation Sinking Fund Method Example 1: Given FC = 100,000, SV = 10,000, L = 10 years, i = 5%. (a) Annual Depreciation, d. (b) Book Value after 3 years. (c) Book Value after 8 years. Ans. P 7,155.41; P 77,442.56; P 31,672.21

Depreciation Sinking Fund Method Example 2: An equipment cost P 100,000 with a salvage value of P 5,000 at the end of 10 years. Using Sinking Fund Method with interest rate= 4%. (a) Compute the annual depreciation cost. (b) Find the book values at years 1 to 4. Ans. (a) P 7,912.64; (b) P 92,087.36; P 83,858.21; P 75,299.90; P 66,399.26

Depreciation Sinking Fund Method Example 3: A plant erected to manufacture socks with a first cost of P 10,000,000 with an estimated salvage value of P 100,000 at the end of 25 years. Find the appraised value to the nearest 100 by sinking fund method at 6% interest rate at the end of a. 10 years b. 20 years Ans. P 7,621,600; P 3,362,200

Plates Sinking Fund Method 1. A factory is constructed at a 1st cost of P 8,000,000 and with an estimated salvage value of P 200,000 at the end of 25 years. Find its appraised value to the nearest 100 at the end of 10 years by using sinking fund of depreciation assuming an interest of 5%. 2. A machine that costs P75 000.00 five years ago now cost P45 864.31, when 7% interest is applied using the sinking fund formula. Determine the salvage value of the machine for an estimated useful life of 10 years.

Depreciation Declining Balance Method (Matheson’s Method) Declining Balance Method – a method which assumes that the annual cost of depreciation is a fixed percentage (k) of the salvage value at the beginning of the year.

Note: This method is not applicable if there is no salvage value.

Where: dm = depreciation at any time m BVm = book value of a property at any time m Dm = total depreciation of a property at any time m L = useful life in years FC = first cost SV = salvage or scrap value k = rate of depreciation

Depreciation Declining Balance Method (Matheson’s Method) Example 1: A machine costing P 720,000 is estimated to have a book value of P 40,545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining value. (a) What is the annual rate of depreciation? (b) What is the book value after 3 years? (c) What is the depreciation charge at the 4th year? (d) What is the total depreciation after 6 years? Ans. (a) 0.25; (b) P 303,750; (c) P 75,937.50; (d) P 591,855.47

Depreciation Declining Balance Method (Matheson’s Method) Example 2: An equipment has a first cost of P500,000 and the cost of installation is P30,000. If the salvage value is 10% of the equipment cost at the end of its useful life of 5 yrs. Compute the book value at the end of the 3rd year. Ans. P 128,526.95

Depreciation Declining Balance Method (Matheson’s Method) Example 3: A machine having a certain first cost has a life of 10 years and a salvage value of 6.633% of the first cost at the end of 10 years. If it has a book value of P58,914 at the end of the 6th year, how much is the first cost of the machine if the constant percentage of declining value is used in the computation for its depreciation. (Sometimes called Matheson’s method) Ans. P 300,000

Depreciation Declining Balance Method (Matheson’s Method) Example 4: A machine costing P720,000 is estimated to have a life of 10 yrs. If the annual rate of depreciation is 25%, determine the total depreciation using a constant percentage of the declining balance method. Ans. P 679,454.27

Plates Declining Balance Method (Matheson’s Method) 1. A radio service panel truck initially costs P 56,000. If resale value at the end of the fifth year is estimated at P 15,000. By means of the Declining Balance Method, determine the yearly depreciation charge for the first and second years. 2. An engineer bought an equipment for P 800,000. Other expenses, including installation, amounted to P 50,000. At the end of its estimated useful life of 10 years, the salvage value will be 10% of the first cost. Using the constant percentage method of depreciation, what is the book value after 5 years?

Depreciation Double Declining Balance Method Double Declining Balance Method – a method which is similar to declining balance method except that the rate of depreciation k is replaced by 2/L.

Where: dm = depreciation at any time m BVm = book value of a property at any time m Dm = total depreciation of a property at any time m L = useful life in years FC = first cost

Depreciation Double Declining Balance Method Example 1: A machine has a first cost of P 140,000 and a life of 8 years with a salvage value of P 10,000 at the end of its useful life. Using double declining balance method: (a) What is the Book Value on the 3rd year? (b) What is the depreciation charge on the 4th year? Ans. P 59,062.50; P 14,765.63

Depreciation Double Declining Balance Method Example 2: A lathe machine has an estimated salvage value of P10,000 at the end of its useful life of 8 yrs. Compute the first cost of the machine if the total depreciation at the end of 4th year using double declining balance method is P957,031.25. Ans. P1,400,000

Depreciation Double Declining Balance Method Example 3: An equipment worth P5,000 has an anticipated salvage value of P500 at the end of its 5 year depreciable life. Compute the depreciation of the equipment on the 3rd year only using DDB method. Ans. P 720

Depreciation Double Declining Balance Method Example 4: A certain machine cost P400,000 and has a life of 5yrs and a salvage value of P50,000 at the end of its expected life. Compute the depreciation on the 3rd year using double declining balance method. Ans. P 57,600

Depreciation Double Declining Balance Method Example 5: A machine cost P100,000 and has a salvage value of P5,000 after its useful life of 25 yrs. Compute the book value after 8 yrs using double declining balance method. Ans. P 51,321.89

Plates Double Declining Balance Method 1. An earth moving equipment that cost P 90,000 will have an estimated salvage value of P18,000 at the end of 8 years. Using double-declining balance method, compute the book value and total depreciation at the end of 5th year. 2. Cost of machine = P1,400,000 Useful life = 8 yrs Salvage value = P10,000 Determine the 4th year depreciation using double declining balance method.

Depreciation Sum of the Years Digit Method

Where: dm = depreciation at any time m Dm = total depreciation of a property at any time m L = useful life in years FC = first cost SV = salvage or scrap value

Depreciation Sum of the Years Digit Method Example 1: An asset is purchased for P 9000. Its estimated life is 10 years, after which is will be sold for P 1,000. Using SOYD (a) Find the book value during the 3rd year. (b) Find the depreciation during the 2nd year. (c) Find the total depreciation after 4 years. Ans. P 5,072.72; P 1,309.09; P 4,945.45

Depreciation Sum of the Years Digit Method Example 2: The first cost of a machine is P1,800,000 with a salvage value of P300,000 at the end of its life at 5 yrs. Determine the total depreciation after 3 yrs using SOYD Method. Ans. P 1,200,000

Depreciation Sum of the Years Digit Method Example 3: A machine cost P7,350 has a life of 8 years and has a salvage value of P350 at the end of 8 years. Determine its book value at the end of 4 years using SOYD Method. Ans. P 2294.44

Depreciation Sum of the Years Digit Method Example 4: A telephone company purchased a microwave radio equipment for P6M. Freight and installment charges amounted to 3% of the purchased price. If the equipment shall be depreciated over a period of 8 yrs with a salvage value of 5%. Determine the depreciation charge during the 5th year using the sum of the years digit method. Ans. P 652,333.33

Depreciation Sum of the Years Digit Method Example 5: ABC Corporation makes its policy that for every new equipment purchased, the annual depreciation should not exceed 20% of the first cost at any time without salvage value. Determine the length of service if the depreciation used is the SOYD Method. Ans. 9 yrs

Plates Sum of the Years Digit Method 1. A company purchases an asset for P 10,000.00 and plans to keep it for 20 years. If the salvage value is zero at the end of the 20th year: (a) What is the depreciation in the third year? (b) What is the total depreciation at the end of 14 years? (c) What is the book value of the asset at the end of 8 years? Use sum-of-the-year’s digits depreciation. 2. A certain equipment costs P7,000 has an economic life of “n” years and a salvage value of P350 at the end of “n” years. If the book value at the end of 4 years is equal to P2197.22, compute the economic life of the equipment using the SOYD Method.

Depreciation Service-Output Method Service-Output Method – a method which assumes that the total depreciation that has taken place is directly proportional to the quantity of output of the property up to that time.

Where: Dm = total depreciation of a property at any time FC = first cost SV = salvage or scrap value T = total units of output up to the end of its life Q = total number of units of output at any time

Depreciation Service-Output Method Example 1: An asphalt and aggregate mixing plant having a capacity of 50 cu.m. every hour costs P 2,500,000. It is estimated to process 800,000 cu.m. during its life. During a certain year it processed 60,000 cu.m. If its scrap value is P 100,000, determine the total depreciation during the year and the depreciation cost chargeable to each batch of 50 cu.m. using the service output method. Ans. P 180,000.00; P 150.00

Depreciation Service-Output Method Example 2: An equipment cost P480,000 and has a salvage value of 10% of its cost at the end of its life of 36,000 operating hours in a period of 5 years. In the first year of service, it was used for 12,000 hours. If at the end of the second year it was used for 15,000 hours, find the depreciation at the end of the second year. Ans. P 180,000

Depreciation Working Hours Method Example 3: A machine was purchased at an original cost of P400,000 with a salvage value of P20,000. Life of this machine is expected to last for 6 years. It was used for 4000 hrs in the first year, 6000 hrs in the second year, and 8000 hrs in the third year. The machine is expected to last for 38,000 hrs in a period of 6 yrs. Find the depreciation at the end of the second year. Ans. P 60,000

Constant Unit Method Example 4: A coin machine costing P200,000 has a salvage value of P20,000 at the end of its economic life of 5 years. The schedule of production per year is as follows: Year Number of Coins 1 100,000 2 80,000 3 60,000 4 40,000 5 20,000 Determine the annual reserve for depreciation for 3rd year only. Ans. P 36,000

Depreciation Modified Accelerated Cost Recovery System, MACRS Assumption: A shift from SL to DDM

( ( (

) ) )

1 2 d1 = FC – 0 2 n 2 d2 = FC – d1 n 2 d3 = FC – d1- d1 n

Depreciation Modified Accelerated Cost Recovery System, MACRS Example 1: The cost of equipment is P500,000 and the cost of installation labor, taxes and miscellaneous expenses is P30,000. If the salvage value is 10% of the cost of equipment at the end of its life of 5 years, compute the book value at the end of third year using MACRS Method. Ans. P 152,640

Depreciation Modified Accelerated Cost Recovery System, MACRS Example 2: A certain machine cost P400,000 and has a life of 5 years and a salvage value of P50,000 at the end of its expected life. Compute the depreciation on the 3rd year using Modified accelerated cost recovery system. Ans. P 76,800

End of Depreciation

Review Quiz (1 whole) A machine costs P 8,000 which last for 7 years with a salvage value at the end of its life of P 355. Determine the depreciation charge during the 4th year and the book value at the end of 4 years by: (a) Straight Line Method; (b) Declining Balance Method; (c) SOYD Method; (d) Sinking Fund Method with interest of 12%; (e) Double Declining Balance Method

BASIC METHODS FOR ECONOMY STUDIES AND COMPARING ALTERNATIVES

Payout Period =

Fixed Capital Annual profit+Annual depreciation

EXAMPLES Example 1: A broadcasting corporation was formed duly approved by the Securities and Exchange office has a working capital of P 20 M and a fixed capital of P 80 M. Annual depreciation amounts to P 5 M and expected annual profit is P 16 M. Determine the recovery period in years. Ans. P 6.25 M

EXAMPLES Example 2: A fixed capital investment of P 10 M is required for a proposed manufacturing plant and an estimated working capital of P 2.5 M. Annual depreciation is 10% of the fixed capital investment. Which of the following gives the payout period in years? Ans. 2.86

EXAMPLES Example 3: A fixed capital investment of P 10 M is required for a proposed manufacturing plant and an estimated working capital of P 2 M. Annual depreciation is 10% of the fixed capital investment. If the annual profit is P 2.5 M, what is the rate of return? Ans. 20.83%

EXAMPLES Example 4: Candy bars are sold in a local store for 50 cents each. The factory has P1000 in fixed cost plus 10 cents of additional expenses for each candy bar made. Assuming all candy bars manufactured can be sold, find the break-even point. Ans. 2500 candy bars

EXAMPLES Example :5 The cost of producing a small transistor radio set consists of P 23.00 for labor and P 37.00 for materials. The fixed charges in operating the plant are P 100,000 per month. The variable cost is P 1.00 per set. The radio set can be sold for P 75.00 each. Determine how many sets must be produced per month to break-even.

Review Problems SITUATION 1 In his request to defer his payment, a man was ask to pay P34,317.60 after 60 days for an appliance whose cash price is P32,580.00. What simple interest rate was charged to him?

SITUATION 2 An engineer promised to pay P380,000 at the end of 120 days. He was offered 12% discount if he pays in 45 days. Find the simple interest rate.

Review Problems SITUATION 3 What is the principal amount if the amount of interest at the end of 2 ½ years is P45,000 for a simple interest rate of 6% per annum?

SITUATION 4 How many years are required for an amount of money to quadruple if it is invested at 4% interest rate?

Review Problems SITUATION 5 A certain nominal annual interest rate has an effective rate of 19.722% when compounded continuously. What is the effective rate if compounded bi-monthly?

SITUATION 6 How often must a nominal rate of 18% be compounded in order to yield 18.81% annually?

Review Problems SITUATION 7 The effective rate on 25% compounded daily is nearest to ____.

SITUATION 8 A person borrowed P500,000 at an interest rate of 18% compounded monthly. Monthly payments of P16,710 are agreed upon. The length of the loan in months is closest to ____.

Review Problems SITUATION 9 What quarterly payment is required over 15 years to equate with a future amount P150,000? Assume interest rate of 6% compounded continuously.

SITUATION 10 A man invests P750,000 in a 6% account today. What uniform annual withdrawal can he make for 6 years starting 15 years from now?

Review Problems SITUATION 11 Find the present worth of perpetuity of P5200 payable monthly if the interest is 16% compounded monthly.

SITUATION 12 A man borrowed P200,000 from his friend and agreed to pay after 6 years at an interest rate of 10% per annum. How much should the man deposit monthly in a bank in order to discharge his depth, if the bank offers 6% annual interest rate?

Review Problems SITUATION 13 Given the following data for construction equipment: Initial Cost = P1,200,000.00 Economic life = 12 years Estimated Salvage Value = P320,000 Determine the book value after seven years using: a) SOYD b) DDBM c) Sinking Fund Method using 6% interest