Report

Applications of MoneyTime Relationships MARR by PW Method All cash inflows and outflows are discounted to the present point in time at an interest rate that is generally the MARR. PW(i) = 0; i = MARR If PW(i) > 0 => The investment is “good”. If the return rate of an investment is greater than MARR then the investment is “profitable”. MARR MARR(hurdle rate) - maximize the economic well-being of a company. Example 1 An investment of $10,000 can be made in a project that will produce a uniform annual revenue of $5,310 for five years and then have a market (salvage) value of $2,000. Annual expenses will be $3,000 each year. The company is willing to accept any project that will earn 10% per year or more, before income taxes, on all invested capital. Show whether this is a desirable investment by using the PW method. PW(10%) = -10,000 + 5,310(P/A, 0.1, 5) + 2,000 (P/F, 0.1, 5) - 3,000(P/A, 0.1, 5) = 0 => This investment is marginally acceptable. Example 2 A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one? PW(20%) = $8,000(P/A,20%,5) +$5,000(P/F,20%,5) - $25,000 = $934.29 Because PW(20%) > 0 => “Investment justifiable” Bond Value Determination Z = face, or par, value C = redemption or disposal price (usually equal to Z) r = bond rate (nominal interest) per interest period N = number of periods before redemption i = bond yield rate per period VN = value (price) of the bond N interest periods prior to redemption VN = C(P/F, i %, N) + rZ(P/A, i %, N) Example 1 Find the current price (PW) of a 10-year bond paying 6% per year (payable semiannually) that is redeemable at par value, if bought by a purchaser to yield 10% per year. The face value of the bond is $1,000. Sol: i6mon = (1.1)1/2 - 1 = 0.049 VN = 1,000 (P/F, 4.9%, 20) + 1,000 (0.03) (P/A, 4.9%, 20) = 384.10 + 377.06 = 761.16 Example 2 A bond with a face value of $5,000 pays interest of 8% per year. This bond will be redeemed at par value at the end of its 20-year life, and the first interest payment is due one year from now. (a) How much should be paid now for this bond in order to receive a yield of 10% per year on the investment? (b) If this bond is purchased now for $4,600, what annual yield would the buyer receive? (a) VN= 5,000(P/F,10%,20) + 5,000(0.08)(P/A,10%,20) = 743 + 3,405.44 = $4,148.44 (b) 4,600 = 5,000(P/F, i, 20) + 5,000(0.08)(P/A, i, 20) i = 8.9% per year. Example 3 A bond that matures in 8 years has a face value of $10,000. The bond stipulates a fixed nominal interest rate of 8% per year, but interest payments are made to the bondholder every three months. A prospective buyer of this bond would like to earn 10% nominal interest per year on his or her investment because interest rates in the economy have risen since the bond was issued. How much should this buyer be willing to pay for the bond? i3mon = (1.1)1/4 - 1 = 0.024 = 2.4% VN = 10,000(0.02) (P/A, 2.4%, 32) + 10,000(P/F, 2.4%, 32) = 4,369.84 + 4,537.71 = 8,907.55 Thus, the buyer should pay no more than $8,907.55 when 10% nominal interest per year is desired MARR by FW Method All cash inflows and outflows are calculated to the future point in time at an interest rate that is generally the MARR. FW(i) = 0; i = MARR If FW(i) > 0 => The investment is “good”. Example A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one? FW(20%) = $8,000(F/A,20%,5) + $5,000 - $25,000(F/P,0.2,5) = $2324.8 > 0 Because FW(20%) > 0, i.e., “Investment justifiable” The Annual Worth Method (AW) Capital Recovery (CR) - the minimum annual profit to recover the initial investment The capital recovery (CR) amount for a project is the equivalent uniform annual cost of the capital invested. It is an annual amount that covers the following two items: 1. Loss in value of the asset 2. Interest on invested capital (i.e., at the MARR) CR(i) = I(A/P, i, n) - S(A/F, i, n) I = initial investment S = market (salvage) value at the end of project n = project duration The Annual Worth Method (AW) AW(i) = R - E - CR(i) R - Annual Revenue E - Annual Expense If AW(i) > 0 => The project is “good” If i = MARR then PW = FW = AW = 0 Consistency among PW, FW, and AW methods Example 1 A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost is $25,000. and the equipment will have a market value of $5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. If the firm's MARR is 20% per year, is this proposal a sound one? AW = R - E - CR(i), R - E = 8,000 AW(20%) = 8,000 - [25,000 (A/P,20%,5) - 5,000 (A/F,20%,5)] = 312.4 > 0 => The project is good. Example An investment company is considering building a 25unit apartment complex in a growing town. Because of the long-term growth potential of the town, it is felt that the company could average 90% of full occupancy for the complex each year. If the following items are reasonably accurate estimates, what is the minimum monthly rent that should be charged if a 12% MARR (per year) is desired? Land investment cost $ 50,000 Building investment cost $225,000 Study period, N 20 years Rent per unit per month R Upkeep expense per unit per month $35 Property taxes and insurance per year 10% of total initial investment Initial investment cost = 50,000 + 225,000 = 275,000 Taxes and insurance/year = 0.1 (275,000) = 27,500 Upkeep/year = 35(12 x 25)(0.9) = 9,450 CR cost/year = 275,000(A/P,12%,20) - 50,000(A/F,12%,20) = $36,123 Revenue = Rx12x25x0.9 = 270R Equivalent AW (of costs) = R - E - CR(i) = 270R - (9450 +27,500) - 36,123 = 0 => R = 270.64 this is the minimum annual rental required equals 73,073 and with annual compounding (M = 1) the monthly rental amount If M = 12, 73,073 = 0.9R(F/A, 0.01, 12)(25) = 260 Internal Rate of Return (IRR) IRR Calculation N PW = Rk(P/F, i'%, k) - Ek(P/F, i'%, k) = 0 k=0 where Rk = net revenues or savings for the kth year 0 Ek = net expenditures including any investment costs for the kth year N = project life (or study period) OR N FW = Rk(P/F, i'%, N-k) - Ek(P/F, i'%, N-k) = 0 k=0 Negative IRR The IRR will be positive if satisfies two conditions: (1) both receipts and expenses are present in the cash flow pattern and (2) the sum of receipts exceeds the sum of all cash outflows. Be sure to check both of these conditions in order to avoid the unnecessary work involved with finding that the IRR is negative Example A capital investment of $10,000 can be made in a project that will produce a uniform annual revenue of $5,310 for five years and then have a salvage value of $2,000. Annual expenses will be $3,000. The company is willing to accept any project that will earn at least 10% per year, before income taxes, on all invested capital. Determine whether it is acceptable by using the IRR method. SOLUTION Sum of positive cash flows ($13,550) exceeds the sum of negative cash flows ($10,000) => IRR >0. PW= 0 = - 10,000 + (5,310 -3,000)(P/A,i'%,5) + 2,000(P/F, i'%, 5) At i' = 5%: PW = - 10,000 + 2,310(4.3295) + 2,000(0.7835) = 1,568 At i' = 15%: PW = -10,000 + 2,310(3.3522) + 2,000(0.4972) = - 1,262 1568 - 0 = 0.05 - i’ => i’ = 10.5% 1568 - (-1262) 0.05 - 0.15 Due to approximation error, the true i’ = 10% For more accuracy: PW(10.5%) < 0 ,PW (8.5%) >0 => Linear Interpolate 10.5% & 8.5% Example A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. Increased productivity attributable to the equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. Evaluate the IRR of the proposed equipment. Is the investment a good one? Recall that the MARR is 20%. Solution PW = 8,000 (P/A, i',5) + 5,000 (P/F, i', 5) - 25,000 PW(MARR) = PW(0.2) = 934.3 > 0 => IRR > MARR => This investment is a good one. To calculate IRR PW(0.25) = -1847.1 Linear Interpolation 943 - 0 = 0.2 - i’ => i’ = 21.58% 943 - (-18147.1) 0.2 - 0.25 i’ > MARR = 20% => This equipment is economically attractive Example In 1915 Albert Epstein allegedly borrowed $7,000 from a bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert computed his interest rate to be [0.07(7,000) x 4] / 7,000 = 0.28 (28%). (a)What true effective annual interest rate did Albert pay? (b) What, if anything, was wrong with his calculation? Example In 1915 Albert Epstein allegedly borrowed $7,000 from a bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert computed his interest rate to be [0.07(7,000) x 4] / 7,000 = 0.28 (28%). (a)What true effective annual interest rate did Albert pay? (b) What, if anything, was wrong with his calculation? PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter 0.07(P/A, i'%, 50) = 1 Example In 1915 Albert Epstein allegedly borrowed $7,000 from a bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert computed his interest rate to be [0.07(7,000) x 4] / 7,000 = 0.28 (28%). (a)What true effective annual interest rate did Albert pay? (b) What, if anything, was wrong with his calculation? PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter 0.07(P/A, i'%, 50) = 1 => (P/A, i’%, 50) = 14.28 (P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.8 15.77 - 14.28 = 0.06 - i’ => i’ = 6.73% 15.77 - (13.8) 0.06 - 0.07 Example In 1915 Albert Epstein allegedly borrowed $7,000 from a bank on the condition that he would repay 7% of the loan every three months, until a total of 50 payments had been made. At the time of the 50th payment, the $7,000 loan would be completely repaid. Albert computed his interest rate to be [0.07(7,000) x 4] / 7,000 = 0.28 (28%). (a)What true effective annual interest rate did Albert pay? (b) What, if anything, was wrong with his calculation? PW = 7,000 = 0.07 ($7,000) (P/A, i'%.,50) per quarter 0.07(P/A, i'%, 50) = 1 => (P/A, i’%, 50) = 14.28 (P/A, 6%, 50) = 15.77 (P/A,7%,50) = 13.8 15.77 - 14.28 = 0.06 - i’ => i’ = 6.73% 15.77 - (13.8) 0.06 - 0.07 4 Effective Rate = (1 + 0.673) - 1 = 30% Example A finance company advertises a "bargain 6% plan" for financing the purchase of automobiles. To the amount of the loan being financed, 6% is added for each year money is owed. This total is then divided by the number of months over which the payments are to be made, and the result is the amount of the monthly payments. For example, a woman purchases a $10,000 automobile under this plan and makes an initial cash payment of $2,500. She wishes to pay the balance in 24 monthly payments Example A finance company advertises a "bargain 6% plan" for financing the purchase of automobiles. To the amount of the loan being financed, 6% is added for each year money is owed. This total is then divided by the number of months over which the payments are to be made, and the result is the amount of the monthly payments. For example, a woman purchases a $10,000 automobile under this plan and makes an initial cash payment of $2,500. She wishes to pay the balance in 24 monthly payments Purchase price = $10,000 - 2,500 = 7,500 = P0 6% finance charge = 0.06 x 2 years x $7,500 = 900 =>Total to be paid = 8,400 Monthly payments = $8,400/24 = $ 350 What effective annual rate of interest does she actually pay? In Reality ….. P0 = A(P/A, i'%, N) => 7,500 = 350(P/A, i'%,24) (P/A, i'%, 24) = 7,500 / 350 = 21.43 (P/A, 0.75%, 24) = 21.89 and (P/A, 1%, 24) = 21.24 Linear Interpolation 21.43 - 21.24 = i’ - 1% => i’ = 0.93% 21.89 - 21.24 0.75% - 1% 12 Effective Rate = (1 + 0.0093) -1 = 12% Difficulties with IRR Method The PW, AW, and FW methods assume that net receipts less expenses (positive recovered funds) each time period are reinvested at the MARR during the study period, N. IRR method assuming that recovered funds, if not consumed in each time period, are reinvested at IRR (i'%) rather than at the MARR. This assumption may not mirror reality in some problems, thus making IRR an unacceptable method for analyzing engineering alternatives. Multiple IRR External Rate of Return (ERR) The reinvestment assumption of the IRR method noted previously may not be valid in an engineering economy study. For instance, if a firm's MARR is 20% per year and the IRR for a project is 42.4%, it may not be possible for the firm to reinvest net cash proceeds from the project at much more than 20%. N Ek(P/F, %, k)(F/P, i'%, N) = k=0 N Rk(F/P, %, N - k) k=0 where Rk = excess of receipts over expenses in period k Ek = excess of expenditures over receipts in period k N = project life or number of periods for the study = external reinvestment rate per period Example The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. Suppose that = MARR = 20% What is the alternative's external rate of return, and is the alternative acceptable? Example The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. Suppose that = MARR = 20% What is the alternative's external rate of return, and is the alternative acceptable? 25,000(F/P, i'%,5) = 8,000(F/A,20%,5) + 5,000 (F/P,i'%, 5) = 2.5813, i' = 20.88% Because i' > MARR, the alternative is barely justified. Example When = 15% and MARR = 20%, determine whether the project whose total cash flow diagram appear below is acceptable. Notice in this example that the use of an is different from MARR. This might occur if, for some reason, part or all of the funds related to a project are "handled" outside the firm's normal capital structure. Eo = $10,000 (k = 0), E1 = $5,000 (k = 1) Rk = $5,000 for k = 2,3, . . .,6 | - 10,000 - 5,000(P/F,15%, 1)|(F/P, i'%, 6) = 5,000(F/A,15%,5); i'% = 15.3% Advantages of ERR over IRR It can usually be solved directly rather than by trial and error. ERR is unique. The Payback Period Method Suppose that a project where all capital investment occurs at time 0. The Payback Period Method Suppose that a project where all capital investment occurs at time 0. Simple Payback period ( N ) is the smallest integer such that N ( Rk - Ek ) I (Initial Investment) k=0 The Payback Period Method Suppose that a project where all capital investment occurs at time 0. Simple Payback period ( N ) is the smallest integer such that N ( R E ) I (Initial Investment) k k k=0 Discounted Payback period ( N ) is the smallest integer such that N ( Rk - Ek )(P/F, i, k) I ; i = MARR k=0 Example The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. Example The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. Example 4 The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. (8,000) 25,000 => Simple Payback Period = 4 k=0 Example 4 The equipment investment cost is $25,000, and the equipment will have a salvage value of $5,000 at the end of its expected life of five years. The equipment will amount to $8,000 per year after extra operating costs have been subtracted from the value of the additional production. (8,000) 25,000 => Simple Payback Period = 4 k=0 5 8,000(P/A,20%,5) + 5,000(P/F,20%,5) > 25,000 k=0 => Discounted Payback Period = 5 Homework Page 177 # 2, 3, 5, 9, 12, 13, 19, 24, 25, 30, 33, 37, 41, 49, 51 Due Date: Oct. 27, 1998