Engineering Economy

ENGINEERING ECONOMY Engineering Economy – is the analysis and evaluation of the factors that will affect the economic success of engineering projects to the end that a recommendation can be made which will ensure the best use of capital. SET 1A: 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 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) Example: A loan of P100 at simple interest will becomeP150 after 5 years.

Cash flow diagram on the viewpoint of the lender

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 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. P124.93 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 4,133.33 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 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 Example 7: (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% Example 8: 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? Ans. P 1,463.41 Example 9: What amount will be available in eight months if P 15,000.00 is invested now at 10% simple interest per year? Ans. P 16,000.00 Example 10: P 1000.00 becomes P 1500.00 in three years. Find the simple interest rate. Ans. 16.67% Example 11: 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? Ans. 50% Compound Interest Compound Interest – the interest for an interest period is calculated on the principal plus total amount of interest accumulated in previous period.

Cash flow diagram on the viewpoint of the borrower Simple Interest Simple Interest – is calculated using the principal only, ignoring any interest that has been accrued in preceding periods. I = Pni F = P(1 + in) Where: I = interest P = principal or present worth n = number of interest periods i = rate of interest per period F = accumulated amount or future worth For Ordinary Simple Interest: Interest period = 1 year = 360 days For Exact Simple Interest: Interest period = 1 year = 365 days (ordinary year) = 366 days (leap year) SAMPLE PROBLEMS 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

1

Principal at Beginning of Period P

2

P(1+i)

P(1+i)i

P(1+i)2

3

P(1+i)2

P(1+i)2i

P(1+i)3

…

…

…

…

n

P(1+i)n-1

P(1+i)n-1i

P(1+i)n

Interest Period

Interest Earned During Period

Amount at End of Period

Pi

P(1+i)

F = P(1 + i)n F⁄ = (1 + i)n = (F⁄ , i%, n) P P P = F(1 + i)−n P⁄ = (1 + i)−n = (P⁄ , i%, n) F F

Where: F = accumulated amount or future worth P = principal or present worth i = rate of interest per interest period n = number of compounding periods F/P = single payment compound amount factor P/F = single payment present worth factor Nominal Rate of Interest – specifies the rate of interest and the number of interest periods in one year. r i= m n = my r my F = P (1 + ) m Where: i = rate of interest per interest period n = number of compounding periods r = nominal rate of interest m = number of compounding periods per year y = number of years Compounding Period Compounded Quarterly Compounded Semi-annually Compounded Monthly Compounded Bi-monthly

m 4 2 12 6

Effective Rate of Interest – is the actual or exact rate of interest on the principal during 1 year, or simply the ratio of accumulated interest in one year to the principal amount. F−P ER = P r m ER = (1 + i)m − 1 = (1 + ) − 1 m SAMPLE PROBLEMS 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 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 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 Example 4: Find the effective rate of interest corresponding to 8% compounded quarterly. Ans. 8.24% Example 5: Find the nominal rate, which if converted quarterly could be used instead of 12% compounded semiannually? Ans. 11.825% Example 6: 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 Example 7: Five years ago, you paid P 340,000 for a lot. Today you sold it at P 500,000. What is the annual rate of appreciation? Ans. 8% Example 8: John borrowed P50, 000.00 from the bank at 25% compounded semi-annually. What is the equivalent effective rate of interest? Ans. 26.56%

Example 9: Find the present worth of a future payment of P 300,000 to be made 5 years with an interest rate of 8% per annum. Ans. P 204,174.96 Example 10: How long will it take money to double itself if invested at 5% compounded annually? Ans. 14.2 years Example 11: The amount of P 50,000 was deposit in the bank earning an interest of 7.5% per annum. Determine the total amount at the end of 5 years, if the principal and interest were not withdrawn during the period? Ans. P 71,781.47 Example 12: Compute the equivalent rate of 6% compounded semi-annually to a rate compounded quarterly. Ans. 5.96% compounded quarterly Example 13: 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 Example 14: 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 Example 15: 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? Ans. 11.57 years Example 16: 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%. Ans. P 1,549.64 Example 17: 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. Ans. (a) 0.322; (b) 3.107 Continuous Compounding Interest From the compound interest formula for m periods per year: r my F = P (1 + ) m m Let = k, then m = rk, as m increases, so must k: r

(1 +

r my 1 rky 1 k ) = (1 + ) = [(1 + ) ] m k k

ry

1 k

The limit of (1 + ) as k approaches infinity is e, thus: k

F = Pery The effective rate of interest for continuous compounding is: ER = er − 1 Where: F = accumulated amount or future worth P = principal or present worth r = nominal rate of interest y = number of years e = Euler’s number ery = continuous compound amount factor 1⁄ ry = present worth of continuous compounding factor e SAMPLE PROBLEMS 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

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% 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 Example 4: If the effective annual interest rate is 4%, compute the equivalent nominal interest compounded continuously. Ans. 3.922% Example 5: What is the nominal rate of interest compounded continuously for 10 years if the compound amount factor is 1.34986? Ans. 3% 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 Discount Discount – 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. 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)−1 F 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 SAMPLE PROBLEMS 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% Example 2: Find the discount if P 2,000 is discounted for 6 months at 8% simple discount. Ans. P 80 Example 3: Discount 1650 for 4 months at 6% simple interest. What is the discount? Ans. P 32.35

Inflation Inflation – 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: 1+i n 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 If the uninflated present worth is to be determined: F F P= = (1 + i)n (1 + f)n (1 + icf )n icf = i + f + if SAMPLE PROBLEMS 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 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

SET 1B: ANNUITIES Annuity – is a series of equal payments occurring at equal periods of time. Ordinary Annuity Ordinary Annuity – a type of annuity were equal payments are made at the end of each period.

P = A(1 + i)−1 + A(1 + i)−2 + A(1 + i)−3 + ⋯ + A(1 + i)−(n−1) + A(1 + i)−n → Eq. 1 Multiplying this equation by (1 + i), the equation becomes: P + Pi = A + A(1 + i)−1 + A(1 + i)−2 + ⋯ + A(1 + i)−n+2 + A(1 + i)−n+1 → Eq. 2 Subtracting Eq. 1 from Eq. 2: Pi = A − A(1 + i)−n A A (1 + i)n − 1 P = [1 − (1 + i)−n ] = [ ] (1 + i)n i i n P⁄ = [(1 + i) − 1] = (P⁄ , i%, n) A A i(1 + i)n The functional symbol (P/A, i%, n) is called the “uniform series present worth factor”. n A⁄ = [ i(1 + i) ] = (A⁄ , i%, n) n P P (1 + i) − 1 The functional symbol (A/P, i%, n) is called the “capital recovery factor”. 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 Substituting P = F(1 + i)−n from the equation of P, it becomes: A F = [(1 + i)n − 1] i

n F⁄ = [(1 + i) − 1] = (F⁄ , i%, n) A A i

The functional symbol (F/A, i%, n) is called the “uniform series compound amount factor”. i A⁄ = [ ] = (A⁄F , i%, n) F (1 + i)n − 1 The functional symbol (A/F, i%, n) is called the “sinking fund factor”. 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

SAMPLE PROBLEMS 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 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 Example 3: A man paid 10% downpayment of P 200,000 for a house and lot and agreed to pay the 90% balance on monthly installment for 60 months at an interest rate of 15% compounded monthly. Compute the amount of monthly payment. Ans. P 42,821.87 Example 4: 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 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% Example 6: A man inherited a regular endowment of P 100, 000 every of 3 months for 10 years. However, he may choose to get a single lump sum payment at the end of 4 years. How much is this lump sum if the cost of money is 14% compounded quarterly? Ans. P 3,702,939.73 Example 7: A service car whose cash price was P 540,000 was bought with a down payment of P 162,000 and monthly installment of P 10,874.29 for 5 years. What was the rate of interest if compounded monthly? Ans. 24% compounded monthly Example 8: 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 Example 9: A man purchased a car with a cash price of P 350,000. He was able to negotiate with the seller to allow him to pay only a down payment of 20% and the balance payable in equal 48 end of the month installment at 1.5% interest per month. On the day he paid the 20th installment, he decided to pay the remaining balance. How much is the monthly payment and what is the remaining balance that he paid? Ans. P 8,224.99; P 186,927.02 Example 10: 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? Ans. P 100,375.37 Example 11: 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%? Ans. P 197,008.25 annual deposits

Example 12: A new office building was constructed 5 years ago by a consulting engineering firm. At that time the firm obtained the bank loan for P 10,000,000 with a 20% annual interest rate, compounded quarterly. The terms of the loan called for equal quarterly payments for a 10-year period with the right of prepayment any time without penalty. Due to internal changes in the firm, it is now proposed to refinance the loan through an insurance company. The new loan is planned for a 20- year term with an interest rate of 24% per annum, compounded quarterly. The insurance company has a onetime service charge 5% of the balance. This new loan also calls for equal quarterly payments. a.) What is the balance due on the original mortgage (principal) if all payments have been made through a full five years? b.) What will be the difference between the equal quarterly payments in the existing arrangement and the revised proposal? Ans. (a) P 7,262,747.03; (b) P 120,862 Example 13: 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 Annuity Due Annuity Due – a type of annuity were equal payments are made at the beginning of each period.

P =A+

F=

beginning of each year. How much should he deposit if the fund is invested at 5% compounded annually? Ans. P 6,057.49 Example 4: 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 Example 5: A farmer bought a tractor costing P 25,000 payable in 10 semi-annual payments, each installment payable at the beginning of each period. If the rate of interest is 26% compounded semi-annually, determine the amount of each installment. Ans. P 4,077.20 Example 6: 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? Ans. second bid, Present worth = P 859,727.18 Deferred Annuity Deferred Annuity – a type of annuity were the first payment is made several periods after the beginning of annuity. A (1 + i)n − 1 P= [ ] (1 + i)−m (1 + i)n i

A (1 + i)n−1 − 1 [ ] (1 + i)n−1 i

A [(1 + i)n+1 − 1] − A i

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 SAMPLE PROBLEMS Example 1: 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 Example 2: 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 Example 3: 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

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 SAMPLE PROBLEMS Example 1: The present value of an annuity of R pesos payable annually for 8 years, with the 1st payment at the end of 10 years is P 187,481.25. Find the value of R if money if money is worth 5%. Ans. P 45,000 Example 2: 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 Example 3: 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

Example 4: A man wishes to provide a fund for his retirement such that from his 60th to 70th birthdays he will be able to withdraw equal sums of P18, 000 for his yearly expenses. He invests equal amount for his 41st to 59th birthdays in a fund earning 10% compounded annually. How much should each of these amounts be? Ans. P 2,285.25 Example 5: 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. Ans. P 5,439.18 Example 6: 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 18 th to the 22nd birthday? Ans. P 20,791.64

Continuous Compounding for Discrete Payments 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 ( rn ) e −1 e A (ern − 1) F= r e −1

Perpetuity Perpetuity – a type of annuity in which payments continue indefinitely.

Capitalized Cost Capitalized Cost – is the sum of the first cost and the present worth of all costs of replacement, operation and maintenance for a long period of time of any property. Capitalized Cost = First Cost + Present Worth of Perpetual Operations and Maintenance + Present Worth of Perpetual Replacement by Sinking Fund Method R Present Worth of Perpetual Replacement = (1 + i)L − 1 R = FC − SV Where: R = replacement cost FC = first cost SV = salvage value i = interest rate per interest period L = useful life in years

P=

A i

Where: P = value or sum of money at present A = series of periodic equal amount of payments i = interest rate per interest period SAMPLE PROBLEMS 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 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 Example 3: 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. Ans. P 9,187.87 Example 4: If money is worth 8% compounded quarterly, calculate the present worth of the following: (a) An annuity of P 1,000 payable quarterly for 50 years (b) An annuity of P 1,000 payable quarterly for 100 years (c) A perpetuity of P 1,000 payable quarterly Ans. (a) P 49,047.35; (b) P 49,981.85; (c) P 50,000 Example 5: 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? Ans. P 400,000

SAMPLE PROBLEMS 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

SAMPLE PROBLEMS Example 1: The first cost of a certain equipment is P 324,000 and a salvage value of P 50,000 at the end of its life of 4 years. If money is worth 6% compounded annually, find the capitalized cost. Ans. P 1,367,901.15 Example 2: Find the capitalized cost of a bridge whose cost is P 250M and life is 20 years. If the bridge must be partially rebuilt at a cost of P 100M at the end of each 20 years. i = 6%. Ans. P 295.3076M Example 3: A machine cost P 150,000 and will have a scrap value of P 10,000 when retired at the end of 15 years. If money is worth 4%, find the annual investment and the capitalized cost of the machine. Ans. P 324,793.85 Example 4: A bridge that was constructed at a cost of P 7.5M is expected to last 30 years at the end of which time its renewal cost will be P 4M. Annual repairs and maintenance is P 300,000. What is the capitalized cost of the bridge at an interest of 6%? Ans. P 13,343,260.77 Example 5: Calculate the capitalized cost of a project that has an initial cost of P 3,000,000 and an additional cost of P 1,000,000 at the end of every 10 yrs. The annual operating costs will be P100, 000 at the end of every year for the first 4 years and P160, 000 thereafter. In addition, there is expected to be recurring major rework cost of P 300,000 every 13 yrs. Assume i =15%. Ans. P 4,281,936

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 below:

A = P 1000, n = 5

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: P = PA + PG A (1 + i)n − 1 PA = [ ] (1 + i)n i G (1 + i)n − 1 PG = [ − n] (1 + i)−n i i 1 (1 + i)n − 1 PG⁄ = − n] (1 + i)−n G i[ i Where: PG/G = Gradient to present worth factor SAMPLE PROBLEMS 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 Example 2: An amortization of a debt is in a form of a gradient series. (a) What is the equivalent present worth of the debt if interest is 5%. (b) Determine also the future amount of amortization as well as the equivalent uniform periodic payment. (c) What is the equivalent uniform annual cost? Ans. P 15,178.34; P 18,449.37; P 4,280.47

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?

SET 2: 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. BVm = FC − Dm DL = FC − SV 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 Straight Line Method Straight Line Method –a method which assumes that the loss in value is directly proportional to the age of the property. d1 = d2 = ⋯ = dm = dL = d FC − SV d= L Dm = md DL = Ld 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 SAMPLE PROBLEMS 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 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 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 Example 4: An engineer bought an equipment for P 500,000.00. He spent an additional amount of P 30,000 for installation and other expenses. The salvage value is 10% of the initial first cost. Life = 15 years. Compute the following: (a) Annual Depreciation. (b) Book Value after 6 years. (c) Total depreciation after 10 years. Ans. (a) P 31,800; (b) P 339,200; (c) P 318,000

Example 5: A machine cost P 73,500 and has a life of 8 years with a salvage value of P 3500 at the end of 8 years. Determine the book value at the end of 4 years using straight line method. Ans. P 38,500 Example 6: 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 straightline method. Ans. P 50,400 Example 7: 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? Ans. P 76,500; P 341,000 Example 8: 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 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. d1 = d2 = ⋯ = dm = dL = d (FC − SV)i d= (1 + i)L − 1 d Dm = [(1 + i)m − 1] i d DL = [(1 + i)L − 1] i 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 SAMPLE PROBLEMS 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 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 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

Example 4: 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%. Ans. P 5,944,400 Example 5: A four-stroke motorbike costs P75 000.00. It will have a salvage value of P10 000.00 when worn out at the end of eight years. Determine the annual replacement deposit using the SFM at 5%. Ans. P 6,806.92 Example 6: 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. Ans. P 5,000 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. dm = FC(1 − k)m−1 k BVm = FC(1 − k)m SV = FC(1 − k)L L SV k=1− √ FC

Dm = FC − BVm 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 SAMPLE PROBLEMS 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 Example 2: A machine having a certain 1st cost has a life of 10 years and a salvage value of 6.633% of the first cost of 10 years. If it has a book value of P 58,914 after 6 years, how much is the first cost of the machine using Matheson’s Method? Ans. P 300,049.23 Example 3: A machine has a current price of P 400,000. If its selling price is expected to decline at the rate of 10% per annum, what will be the selling price after 5 years? Ans. P 236,196.00 Example 4: A radio service panel truck initially costs P 56,000. I 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. Ans. P 12,969.60; P 9,965.84 Example 5: 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?

Ans. P 268,793.20 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. 2 m−1 2 dm = FC (1 − ) L L 2 m BVm = FC (1 − ) L Dm = FC − BVm 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 SAMPLE PROBLEMS 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 4 th year? Ans. P 59,062.50; P 14,765.63 Example 2: An equipment costs P 500,000 and has a salvage value of P 25,000 after its 25 years of useful life. Using Double Declining Balance Method, what will be the book value after 8 years? Ans. P 256,609.44 Example 3: XYZ Company has an equipment that cost P 90,000. After 8 years, it will have a salvage value of P 18,000.00. Using Double declining balance method, find the book value at the end of 5 years. Ans. P 21,357 Example 4: Given the following data for a construction equipment: Initial cost = P 1,200,000.00; Economic Life = 12 years; Estimated salvage value = P 320,000.00. (a) What is the book value after seven years? (b) What is the depreciation charge on the 4th year? (c) What is the total depreciation charge at the end of the 10 th year? Ans. P 334,898; P 115,740.74; P 1,006,193.30 Example 5: A machine costing P 550,000 has an estimated scrap value of P 85,000 at the end of its economic life of 8 years. Using DDBM of depreciation: (a) What is the book value after 4 years of service? (b) What is the book value at the end of its life? Ans. P 174,023; P 55,062.10 Sum of the Years Digit Method (FC − SV) dm = (reverse digit) sum of the digits (FC − SV) Dm = (sum of reverse digits) sum of the digits L sum of the digits = (L + 1) 2 reverse digit = L − m + 1 m sum of reverse digits = (2L − m + 1) 2 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

SAMPLE PROBLEMS 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 Example 2: Mr. Q purchased a Bulk Milk Cooler for P 480,000.00. Shipping, tax, and installation costs amounted to P 25,000.00, P 20,000.00 and P 15,000.00. The machine has a useful life of 7 years and salvage value of P 40,000. (a) Determine the book value after four years. (b) Determine the depreciation charge on its last year of service. (c) Determine the total depreciation after 3 years. Ans. P 147,142.86; P 17,857.14; P 321,428.57 Example 3: A telephone company purchased microwave radio equipment for P 6 million, freight and installation charges amounted to 4% of the purchased price. If the equipment will be depreciated over a period of 10 years with a salvage value of 8%, determine the depreciation cost during 5th year using SYD. Ans. P 626,269.10 Example 4: 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. Ans. P 857.14; P 9,000; P 3,714.29 Example 5: An equipment costing P 500,000.00 has a life expectancy of 5 years. Using some-of-the-year’s digit method of depreciation, what must be its salvage value such that its depreciation charge for the first year is P 100,000.00? Ans. P 200,000.00 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. (FC − SV) Depreciation per unit output = T (FC − SV) (Q) Dm = T 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 SAMPLE PROBLEMS Example 1: A television company purchased machinery for P 100,000 on July 1, 1979. It is estimated that it will have a useful of 10 years, scrap value of P 4,000, production of 400,000 units and working hours of 120,000. The company uses the machinery for 14,000 hours in 1979 and 18,000 hours is 1980. The machinery produces 36,000 units in 1979 and 44,000 units in 1980. Compute the depreciation charge for 1980 using each method given below: (a) Straight Line Method (b) Working Hours Method (c) Output method Also compute the total depreciation at the end of 1980 using: (d) Working Hours Method (e) Service Output Method Ans. P 9,600; P 14,400; P 10,560; P 25,600; P 19,200 Example 2: 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

SUPPLEMENTARY PROBLEMS Example 1: A machine costs P 7,000 which last for 8 years with a salvage value at the end of its life of P 350. 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 Ans. (a) P 831.25, P 3,675; (b) P 710.96, P 1,565.25; (c) P 923.61, P 2,197.22; (d) P 540.66, P 4,416.00; (e) P 738.28; P 2,214.84 Example 2: A P 110,000 chemical plant had an estimated life of 6 years and a projected scrap value of P 10,000. After 3 years of operation, an explosion made it a total loss. How much money would have to be raised to put up a new plant costing P 150,000, if the depreciation reserve was maintained during its 3 years of operation by: (a) Straight Line Method; (b) Sinking Fund Method at 6% interest Ans. P 100,000; P 104,359.08 Example 3: A contractor imported a bulldozer for his job, paying P 250,000 to the manufacturer. Freight and insurance charges amounted to P 18,000; customs’, broker’s fees and arresters services amounted to P 8,500; taxes, permits, and other expenses which is 10% of the purchasing cost. If the contractor estimates the life of the bulldozer to be 10 years with a salvage value of P 20,000, determine the book value at the end of 6 years using the: (a) Straight Line Method; (b) Sinking Fund Method with interest at 8%; (c) Matheson’s Formula; (d) SOYD Method Ans. P 132,600; P 158,949.69; P 59,201.53; P 71,181.82 Example 4: An equipment costs P 10,000 with a salvage value of P 500 at the end of 10 years. Calculate the annual depreciation cost by: (a) Straight Line Method; (b) Sinking Fund Method at 4% interest Ans. P 950; P 791.26 Example 5: A radio service panel truck initially costs P 56,000. Its resale value at the end of the 5th year is estimated at P 15,000. (a) Determine the annual depreciation charge by SLM. (b) By means of the Matheson’s Formula, determine the yearly depreciation charge for the first, second and third year. Ans. P 8,200; P 12,969.60, P 9,965.84, P 7,657.75

SET 3: BASIC METHODS FOR ECONOMY STUDIES Rate of Return Method (ROR) Rate of Return – is a measure of the effectiveness of an investment of capital and its financial efficiency. When this method is used, it is necessary to decide whether the computed rate of return is sufficient to justify the investment. net annual profit ROR = capital invested net annual profit = annual revenue − annual cost Annual cost includes depreciation, labor and material cost, overhead, rental, tax and insurances, etc. SAMPLE PROBLEMS Example 1: An investment of P 270,000 can be made in a project that will produce a uniform annual revenue of P 185,400 for 5 years and having a salvage value of 10% of the investment. Out of pocket costs for operation and maintenance will be P 81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. Using ROR method, determine the rate of return of the investment. Is this a desirable investment? Ans. ROR = 23.70%; It is not a desirable investment Example 2: A young mechanical engineer is considering establishing his own small company. An investment of P 800,000 will be required which will be recovered in 15 years. It is estimated that sales will be P 800,000 per year and that operating expenses will be as follows. Materials P 160,000 per year Labor P 280,000 per year Overhead P 40,000 +10% of sales per year Selling expense P 60,000 The man will give up his regular job paying P 216,000 per year and devote full time to the operation of the business; this will result in decreasing labor cost by P40,000 per year, material cost by P 28,000 per year and overhead cost by P32,000 per year. If the man expects to earn at least 20% of his capital, should he invest? Compute for the actual rate of return. Ans. The man should not invest; ROR = 6.6118% Example 3: The ABC Company is considering constructing a plant to manufacture a proposed new product. The land costs P 15,000,000, the building costs P 30,000,000, the equipment costs P 12,500,000, and P 5,000,000 working capital is required. At the end of 12 years, the land can be sold for P 25,000,000, the building for P 12,000,000, the equipment for P 250,000 and all of the working capital recovered. The annual disbursements for labor, materials, and all other expenses are estimated to cost P 23,750,000. If the company requires a minimum return of 25%, what should be the minimum annual sales for 12 years to justify the investment? Ans. P 39,748,563.43 Example 4: A man formerly employed as chief mechanic of an automobile repair shop has saved P 1,000,000.00 which are now invested in certain securities giving him an annual dividend of 15%. He now plans to invest this amount in his own repair shop. In his resent job, he is earning P 25,000.00 a month, but he has to resign to run his own business. He will need the services of the following: 2 mechanics each earning P400.00 a day, and 8 helpers each earning P200.00 a day. These men will work on the average 300 days per year. His other expenses are the following: Rental P30,000.00 a month Miscellaneous P25,000.00 a month Sales tax 3% of gross income Insurance 2% The length of his lease is 5 years. If the average charge for each car repaired by his shop is P 1,000.00. Determine the number of cars he must service in one year so that he will obtain a profit of at least 20% on his investment? Ans. 2112 cars

Annual Worth Method Annual Worth Method – in this method, interest on the original investment (sometimes called minimum required profit) is included as cost. If the excess of annual cash inflows over annual cash outflows is not less than zero, the proposed investment is justified. Annual cash inflow − Annual cash outflow ≥ 0 SAMPLE PROBLEMS Example 1: An investment of P 270,000 can be made in a project that will produce a uniform annual revenue of P 185,400 for 5 years and having a salvage value of 10% of the investment. Out of pocket costs for operation and maintenance will be P 81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. Using annual worth method, determine the annual cash flow. Is this a desirable investment? Ans. –P 3,509; It is not a desirable investment Example 2: A man is considering investing P 500,000 to open a semi-automatic auto washing business in a city of 400,000 population. The equipment can wash, on the average, 12 cars per hour, using two men to operate it and to do small amount of hand work. The man plans to hire two men, in addition to himself, and operate the station on an 8-hour basis, 6 days per week, and 50 weeks per year. He will pay his employees P 25.00 per hour. He expects to charge P 25.00 for a car wash. Out-of-pocket miscellaneous cost would be P 8,500 per month. He would pay his employees for 2 weeks vacation each year. Because of the length of his lease, he must write off his investment within 5 years. His capital now is earning 15%, and he is employed at a steady job that pays P 25,000 per month. He desires a rate of return of at least 20% of his investment. Using annual worth method, determine the excess of annual revenue over the annual cost. Would you recommend the investment? Ans. P 19,040; the man should invest Present Worth Method Present Worth Method – this method is based on the concept of present worth. If the present worth of all net cash flows is equal to or greater than zero, the project is justified. Cash inflow includes annual revenue and salvage or scrap value. Depreciation is excluded in cash outflow. Present Worth of all net cash flows ≥ 0 SAMPLE PROBLEMS Example 1: An investment of P 270,000 can be made in a project that will produce a uniform annual revenue of P 185,400 for 5 years and having a salvage value of 10% of the investment. Out of pocket costs for operation and maintenance will be P 81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. Using present worth method, determine the present worth of all net cash flows. Is this a desirable investment? Ans. –P 9,436.00; It is not a desirable investment Future Worth Method Future Worth Method – this method is exactly comparable to the present worth method except that all cash inflows and outflows are compounded forward to a reference point in time called the future. Future Worth of all net cash flows ≥ 0 SAMPLE PROBLEMS Example 1: An investment of P 270,000 can be made in a project that will produce a uniform annual revenue of P 185,400 for 5 years and having a salvage value of 10% of the investment. Out of pocket costs for operation and maintenance will be P 81,000 per

year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. Using future worth method, determine the future worth of all net cash flows. Is this a desirable investment? Ans. –P 28,796.50; It is not a desirable investment Payback (Payout) Period Method Payback Period – is defined as the length of time required to recover the first cost of an investment from the net cash flow produced by the investment for an interest rate of zero. Depreciation is not included in cash outflow. Investment − Salvage Value Payback Period = Net annual cash flow SAMPLE PROBLEMS Example 1: An investment of P 270,000 can be made in a project that will produce a uniform annual revenue of P 185,400 for 5 years and having a salvage value of 10% of the investment. Out of pocket costs for operation and maintenance will be P 81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. Determine the payback period. Ans. 2.6 years Example 2: A fixed capital investment of P 10,000,000.00 is required for a proposed manufacturing plant and an estimated working capital of P 2,000,000.00. Annual depreciation is estimated to be 10% of the fixed capital investment. Determine the rate of return on the total investment and the payout period is the annual profit is P 2,500,000.00. Ans. ROR = 12.5%; 4.8 years

SET 4: COMPARING ALTERNATIVES Rate of Return on Additional Investment Method The formula for the ROR on additional investment is: annual net savings ROR on additional investment = additional investment annual net savings = difference in annual cost additional investment = difference in capital invested If the rate of return on additional investment is satisfactory, then, the alternative requiring a bigger investment is more economical and should be chosen. SAMPLE PROBLEMS Example 1: A company is considering two types of equipment for its manufacturing plant. Pertinent data are as follows: Type A Type B First cost P 200,000 P 300,000 Annual operating cost P 32,000 P 24,000 Annual labor cost P 50,000 P 32,000 Insurance and property taxes 3% 3% Payroll taxes 4% 4% Estimated Life 10 10 If the minimum required rate of return is 15%, which equipment should be selected? What is the rate of return of return on additional investment? Ans. Type B; 18.79% Annual Cost Method To apply this method, the annual cost of alternatives including interest on investment is determined. The alternative with the least annual cost is chosen. SAMPLE PROBLEMS Example 1: Example 1: A company is considering two types of equipment for its manufacturing plant. Pertinent data are as follows: Type A Type B First cost P 200,000 P 300,000 Annual operating cost P 32,000 P 24,000 Annual labor cost P 50,000 P 32,000 Insurance and property taxes 3% 3% Payroll taxes 4% 4% Estimated Life 10 10 If the minimum required rate of return is 15%. What is the annual cost of Type A? Type B? Which equipment should be selected? Ans. P 129,850; P 126,056; Type B Equivalent Uniform Annual Cost Method In this method, all cash flows (irregular or uniform) must be converted to an equivalent uniform annual cost, that is, a year-end amount which is the same each year. The alternative with the least equivalent uniform annual cost is preferred. When the EUAC method is used, the equivalent uniform annual cost of the alternatives must be calculated for one life cycle only. SAMPLE PROBLEMS Example 1: Example 1: A company is considering two types of equipment for its manufacturing plant. Pertinent data are as follows: Type A Type B First cost P 200,000 P 300,000 Annual operating cost P 32,000 P 24,000 Annual labor cost P 50,000 P 32,000 Insurance and property taxes 3% 3% Payroll taxes 4% 4% Estimated Life 10 10 If the minimum required rate of return is 15%. What is the EUAC of Type A? Type B? Which equipment should be selected? Ans. P 129,850; P 126,060; Type B

Present Worth Cost Method In comparing alternatives by this method, determine the present worth of the net cash outflows for each alternative for the same period of time. The alternative with the least present worth of cost is selected. SAMPLE PROBLEMS Example 1: Example 1: A company is considering two types of equipment for its manufacturing plant. Pertinent data are as follows: Type A Type B First cost P 200,000 P 300,000 Annual operating cost P 32,000 P 24,000 Annual labor cost P 50,000 P 32,000 Insurance and property taxes 3% 3% Payroll taxes 4% 4% Estimated Life 10 10 If the minimum required rate of return is 15%. What is the present worth of Type A? Type B? Which equipment should be selected? Ans. P 651,689; P 632,643; Type B

Break-Even Analysis SAMPLE PROBLEMS Example 1: 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. Ans. 7,143 sets Example 2: A company has a production capacity of 500 units per month and its fixed costs are P 250,000 a month. The variable costs per unit are P 1,150 and each unit can be sold for P 2,000. Economy measures are instituted to reduce the fixed costs by 10% and the variable cost be 20%. (a) Determine the old break-even point. (b) Determine the new break-even point. (c) What is the old monthly profit at 100% capacity? (d) What is the new monthly profit at 100% capacity? Ans. 294 units per month; 208 units per month; P 175,000; P 315, 000 Example 3: Two machines are being considered for the production of a particular part for which there is a long term demand. Machine A costs P 50,000 and is expected to last 3 years and have a P 10,000 salvage value. Machine B costs P 75,000 and is expected to last 6 years and have zero salvage value. Machine A can produce a part in 18 seconds; Machine B requires only 12 seconds per part. The out-of-pocket hourly cost of operation is P 38 for A and P 30 for B. Monthly maintenance costs are P 200 for A and P 220 for B. If interest on invested capital is 25%, determine the number of parts per year at which the machines are equally economical. Ans. 29, 544 parts

Benefit-Cost Ratio SAMPLE PROBLEMS Example 1: A non-profit educational research organization, is contemplating an investment of P 1,500,000 in grants to develop new ways to teach people the rudiments of profession. The grants would extend over a ten-year period and would achieve an estimated savings of P 500,000 per year in professors’ salaries, student tuition, and other expenses. The program would be an addition to ongoing and planned activities, thus an estimated P 100,000 a year would have to be released from the other program to support the educational research. A rate of return of 15% is expected, calculate the B/C ratio. Is this a good program? Ans. B/C = 1.34; It is a good program Example 2: The National Government intends to build a dam and hydroelectric project in the Cagayan Valley at a total cost of P 455,500,000. The project will be financed by soft foreign load with a rate of interest of 5% per year. The annual cost for operation and maintenance, distribution facilities and others would total P 15,100,000. Annual revenues and benefits are estimated to be P 56,500,000. If the structures are expected to last for 50 years, with no salvage value, determine the B/C ratio of the project. Ans. B/C = 1.41