Engineering-Economy PDF

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ENGINEERING ECONOMYEngineering 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 RELATIONSHIPSInterest – is the a...

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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

Pi

P(1+i)

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

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 r = k, then m = rk, as m increases, so must k: k ry

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

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

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 = 1 − (1 + i)−1 d= 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 ] = [ ] i (1 + i)n 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) P n 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

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

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.

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 i (1 + i)n

A (1 + i)n−1 − 1 P= A+ [ ] (1 + i)n−1 i

A F = [(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 ...