solved problems in engineering economy 2014 Western University ENGINEERIN ES1050

SOLVED PROBLEMS IN ENGINEERING ECONONOMY Charlie A. Marquez, PIE AUGUST 2014 SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 ENGINEERING ECONOMY – the analysis and evaluation of the factors that will affect the econom ...

SOLVED PROBLEMS IN ENGINEERING ECONONOMY Charlie A. Marquez, PIE AUGUST 2014 SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 ENGINEERING ECONOMY – 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. FORMULAS IN ENGINEERING ECONOMY SIMPLE INTEREST – the interest on a loan that is based only on the principal. Usually used for short-term loans where the period is measured in days rather than years. I = Pni (1) F = P + I = P + Pni F = P(1+ni) (2) where: I = interest P = principal or present worth n = number of interest periods i = rate of interest per interest period F = accumulated amount or future worth TYPES OF SIMPLE INTEREST ORDINARY SIMPLE INTEREST - interest is computed on the basis of 12 months of 30 days each which is equivalent to 360 days a year. In this case, the value of n that is used in the preceding formulas may be computed as: where d is the number of days the principal was invested EXACT SIMPLE INTEREST – interest is computed based on the exact number of days in a given year which is 365 days for a normal year and 366 days during a leap year (which occurs every 4 years, or if it is a century year, it must be divided by 400). Note that during leap years, February has 29 days and 28 days only during a normal year. In this case, the value of n that is used in the preceding formulas may be computed as: for a normal year for a leap year SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 DISCOUNT – discount in simple terms is the interest deducted in advance. It is the difference between the amount a borrower receives in cash (present worth) and the amount he pays in the future (future worth). Discount = Future Worth – Present Worth D = F – P (3) Rate of discount is the discount on one unit of principal for one unit of time. d = 1 – (1 + i)-1 (4) i = (5) Where: d = rate of discount i = rate of interest for the same period COMPOUND INTEREST – interest which is based on the principal plus the previous accumulated interest. It may also be defined as ‘interest on top of interest.” This is usually used in commercial practice especially for longer periods. CASH FLOW DIAGRAMS – a graphical representation of cash flows drawn on a time scale. ↑ = receipts (positive cash flow or cash inflow) ↓ = disbursements (negative cash flow or cash outflow) F F = P(1+i)n (6) • • • • P = F(1+i)-n (7) P Where: F = future amount of money P = present worth or principal i = rate of interest per interest period n = number of interest periods (1+i)n = single payment compound amount factor (1+i)-n = single payment present worth factor 0 1 2 3 n SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 RATE OF INTEREST – the cost of borrowing money or the amount earned by a unit principal per unit time. TYPES OF RATES OF INTEREST NOMINAL RATE OF INTEREST – is the basic annual rate of interest. It specifies the rate of interest and the number of interest periods in one year. i = (8) Where: i = rate of interest per interest period r = nominal rate of interest m = number of compounding periods per year EFFECTIVE RATE OF INTEREST – is the actual or the exact rate of interest earned on the principal during a one-year period. ERi = (1+i)m – 1 (9) Where: ERi = effective rate of interest CONTINUOUS COMPOUNDING – based on the assumption that cash payments occur once per year but compounding is continuous throughout the year. F = P Let x = F = P But = e Therefore, F = P (10) EQUATION OF VALUE – this is obtained by setting the sum of the values on a certain comparison or focal date of one set of obligations to the sum of the values on the same date of another set of obligations. nm xnm SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 ANNUITIES – a series of equal payments occurring at equal interval of time. TYPES OF ANNUITIES ORDINARY ANNUITY – this type of annuity is one where the payments are made at the end of each period beginning from the first period. • • • • • A A A A Finding F when A is given: F = A{[(1+i)n – 1] / i} (11) Finding P when A is given: P = A{[1-(1+i)-n ] / i} (12) Where: F = future worth of an annuity A = a series of periodic, equal amounts of money P = present worth of an annuity i = interest rate per interest period n = number of interest periods DEFERED ANNUITY – this type of annuity is one where the first payment is made several periods after the beginning of the annuity. • • • • • • • A A A A A Finding F when A is given: 0 1 2 3.. n P0 Fn 0 1 2 3 n P0 Fn 4 5… SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 F = A{[(1+i)n – 1] / i}(1+i)n (13) Finding P when A is given: P = A{[1-(1+i)-n ] / i}(1+i)-n (14) PERPETUITY – is an annuity wherein the payments continue indefinitely. • • • P = A{[1-(1+i)-n ] / i} = A{[1-(1+i)- ∞ ] / i} P = (15) CAPITALIZED COST – this is one of the most important applications of perpetuity. The capitalized cost of any property is the sum of its first cost and the present worth of all costs for replacement, operation, and maintenance for a long period or forever. Case 1: No Replacement, maintenance and/or operation every period. CC = FC + P (16) Where: CC = capitalized cost FC = first cost P = present worth of perpetual operation and maintenance Case 2: Replacement only, no operation and maintenance CC = FC + X (17) X = S / (1+i)k -1 (18) Where: X = present worth of perpetual replacement S = amount needed to replace the property every k period P n ∞ SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 k = periodic replacement GRADIENT – A series of disbursements or receipts that increases or decreases in each succeeding period by a constant amount = P PA PG P = PA + PG P = A(P/A, i%,n) + G(P/G, i%, n) (19) PG = G(P/G, i%,n) = [(1+i)n -1/i]-n}(1+i)-n Where: PA = present worth of an annuity PG = present worth of gradient CAPITAL FINANCING WITH BONDS BONDS – a financial security note issued by businesses or corporations and by the government as a means of borrowing long-term fund. It may also be defined as a long-term note issued by the lender to the borrower stipulating the terms of repayment and other conditions. BOND VALUE – the value of a bond is the present worth of all future amounts that are expected to be received through ownership of the bond. METHODS OF BOND RETIREMENT 1. The corporation may issue another set of bonds equal to the amount of bonds due for redemption. 2. The corporation may set up a sinking fund into which periodic deposits of equal amounts are made. The accumulated amount in the sinking fund is equal to the amount needed to retire the bonds at the time they are due. 0 1 2 n F n-1 Fr Fr Fr Fr C 0 1 2 3 n SOLVED PROBLEMS IN ENGINEERING ECONOMY 2014 A A A A P A = F/(F/A, i%, n) (20) P = Fr [(1-(1+i)-n )/i] + C(1+i)-n (21) Where: A = periodic deposit into the sinking fund F = amount needed to retire the bonds, face / par value C = redemption price (often equal to F) r = bond rate i = investment rate or yield per period P = purchase price of the bond / value of the bond n periods before redemption. DEPRECIATION – the decrease in the value of a physical property with the passage of time. TYPES OF DEPRECIATION 1. Physical depreciation