5-2. Sol’n:
Extra Cost = P1,200 Savings = 4 mi/gal For 8 cylinders = 20 mi/gal gasoline Approx. driving = 1,200 mi/month i = 0.5%/month Cost of Gasoline = P2/gal Cost for 8-cylinders 1,200 mi/month = -------------------------- x P2/gal = P120/month 20 mi/gal Cost for 4-cylinders 1,200 mi/month = -------------------------- x P2/gal = P100/month 20 + 4 mi/gal Savings on cost = P120 – P100 = P20/month Equating the present worth: P20(P/A,i%n) = Extra Cost 1-(1.005) -n P20(------------) P20(------------) = P1200 0.005 1.005 -n = 1200(0.005)/20 – 1 -n(ln1.005) = ln(0.7) -n = -0.357/0.005 n = 71.4 months 5-4. Present Worth Outflow Inflow P13,000
Investment N = 15 years Market (Salvage Value) = 3,000((P/F,i%,n) = 3,000(1.12) -15 = 1 –(1.12) -15 Annual Op Expenses = P100(P/A,i%,n) = P100(------------P100(----- ---------------------)) = 0.12 Overhaul Exp. (end of 5 yrs) = P200(1.12)-5 = -10 Overhaul Exp. (end of 10 yrs) = P550(1.12) = Net Present Worth = P548.09 – P13,000 – P681.09 – P113.49 – P117.09
P548.09 P681.09 P113.49 P117.09 = -P13,363.50
5-5.
Investment N = Market value = -P1000(P/F,i%,N) = -P1000(1.15 -5) = 1 – 1.15-5
Present Worth Inflow Outflow P10,000 5 years -P497.18
Annual receipts = P8000(P/A,i%,N) = P8000(----------------) = 0.15 1 – 1.15-5 Annual Expenses = P4000(P/A,i%,N) = P4000(----------------) = 0.15 Net PW = -P497.18 +P26,817.24 – P10,000 – P13,408.62 =
P26,817.24
P13,408.62 P2,911.44
By the Future Method Investment = P10,000(F/P,i%N) = P10,000(1.15) 5 = N = Market value = 1.15-5 - 1 Annual receipts = P8000(F/A,i%,N) = P8000(----------------) = 0.15 1.15-5 - 1 Annual Expenses = P4000(F/A,i%,N) = P4000(----------------) = 0.15 Net FW = -P1000 +P53,939.05 – P20,113.57 – P26,969.53 By the Annual Worth Method
P53,939.05
P26,969.53 =
P5,855.95 Annual Worth Inflow Outflow
1.155 - 1 Investment = P10,000(A/P,15%%,5) = P10,000(--------------) = P2,983.16 0.15 N = 0.15 Market value = -P1000(A/F,15%,5) = -P1,000(--------------) = 1.155 - 1 Annual receipts = Annual Expenses = Net FW = -P148.32 + P8,000 – P2,983.16 – P4,000
Future Worth Inflow Outflow P20,113.57 5 years -P1000
=
5 years -P148.32 P8000 P4000 P868.52
5-8. Present Worth Inflow Outflow Land, Building, Equipment & Additional Working Capital = P300,000 + P600,000 + P250,000 +P100,000 = P1,250,000 Annual Sales = P750,000(P/A,15%10) = P750,000(5.01877) = P3,764,076.47 -10 Market Value = (400,00+350,000+50,000)(1.15) = P197,747.76 -10 Annual Expenses = P475,000(P/A,15%,10) = P475,000(1-1.15 )/0.15 = P2,383,915.10 Net PW = P3,764,076.47 + P197,747.76 – P1,250,000 – P2,383,915.10 = P327,909.13
5-12.
a).
N = 10 x 4 = 40 periods r = 8%/4 = 2% Z = P10,000 C = P10,000 r = 8%/4 = 2% = 0.02 N = 10 x 4 = 40 i = 12%/4 = 3%/period = 0.03 Market value 5-27. Referring to the accompanying cash-flow diagram, complete the following statements P2,000 P2,000 P1,000 P1,000 P1,000 0
1
2
3
4
5
6
P1,000 EOY P3,000 a. As i oo, the PW equals ______ b. the discounted payback period (θ‘) is ___ years Let MARR = 12% per year. c. If the cash flow at the end of year six had been –P2,000 instead of +P2000, AW (0%) = _____
by using PW,AW, Payback Period 5-30. To purchase a used automobile, you borrow P8,000 from JohnAgs Enterprises. They tell you the interest rate being charged is 1% per month for 35 months. They also charge you P200 for a credit investigation, so you leave with P7,800 in your pocket. The monthly payment they calculated for you is P8,000(0.01)(35) + P8,000 ------------------------- = P308.57/month 35 If you agree to these terms and sign their contract, what is the actual APR (annual percentage rate) that you are paying? Sol’n: Computing the annual worth: for i = 0.01 Net AW = P8,000(A/P, 1%,35) – 308.57 P272.029 – P308.57 = -P36.54 for i = 0.02 Net AW = P8,000(A/P,2%,35) – 308.57 P320.00 – P308.57 = P11.45 i – 0.01 ----------0.02 – 0.01
0 – (-36.54) = ----------------- ; 11.45 – (-36.54) 12 i’ = (1.0176 – 1)100% = 23.28%
i = 0.0176
5-33. 5-34.A small business owner estimates that the heat loss through the exterior walls of her warehouse will cost approximately P4,000 next year. A local company is offering insulation that can reduce the heat loss by 80%, with an insulation cost of P17,000 now. She intends to keep the warehouse for 10 years. If the cost of the heat loss increases by P300 per year, after next year, what is the IRR? 5-38. Determine the single (and unique) IRR in each of these situations: a. EOY Cash Flow b. EOY Cash Flow
0 -3 4 5 6 7 8 9
0 -P1,000 300 300 300 300 300
0
-P1,800 1 2 3
-700 1,830 1,830
c. EOY
Cash Flow 0 0 1 -P3,000 2 1,000 3 1,900 4 -800 5 2,720 5-41. A P20,000 ordinary life insurance policy for a 22-year old female can be obtained for annual premiums of approximately P250. This thype ofpolicy (ordinary life) would pay a death benefit of P20,000 in exchange for annual premiums of P250 that are paid during the lifetime of the insured person. If the average life expectancy of a 22-year old female is 77 years, what interest rate establishes equivalence between cash outflows and inflows for this type of insurance policy? Assume that all th
premiums are paid on a beginning of a year basis and that the last premium is paid on the female’s 76
birthday. 5-43. Refer to the following cash-flow diagram: P500 P500 P500 P500 0
1
P200 P500
2
3
4
5
6
EOY
a.. What is the breakeven life (θ’) of this project
--- by the use of Payback b. What is the breakeven interest rate? --- IRR 5-49. The prospective exploration for oil in the outer Cebu sea by a small, independent drilling company has produced a rather curious pattern of cash flows, as follows: End of Year Net Cash Flow 0 -P520,000 1-10 +200,000 10 -1,500,00 The P1,500,000 expense at the end of year 10 will be incurred by the company in dismantling the drilling rig. a. Over the 10-year period, plot PW versus the interest rate (i) in an attempt to discover whether multiple rate of return exist. -- IRR b. Based on the projected net cash flows and results in (a), what would you recommend regarding the pursuit of this project? Customarily the company expects to earn at least 20% per year on invested capital before taxes. Use the ERR method (є = 20%)
-- ERR 5-63. Willie purchased a used car for P100,000. She wrote a check for P20,000 as downpayment for the car and financed the P80,000 balance. The annual percentage rate (APR) is 9% compounded monthly,
and the loan is to be repaid in equal monthly installments over the next four years. Willie’s monthly car
payment is? ---AW Comparison of Alternatives 6-5. Which mutually exclusive design should be chosen from the lists? A 10-year study period is to be used, and MARR is 10% per year. All market values are negligible. Design A Design B Design C Capital Investment 170,000 330,000 300,000 Annual receipts 114,000 147,000 130,000 Annual expenses 70,000 79,000 64,000 Use the FW method. Confirm your recommendation, using the PW and AW methods. 6-9. Consider the following mutually exclusive alternatives: Alternative A Alternative B Capital Investment 780,000 1,840,000 Net Annual receipts 138,060 311,000 Both alternatives have a useful life of 1o years and no market value at that time. The MARR is 10% per year. Use the FW method to identify the most appropriate course of action. Confirm your recommendation using AW and PW methods. 6-49. The Tree Top Airline (TTA) is a small 6-50. Complete the following analysis of investment alternatives and select the preferred alternative. The study period is three years and the MARR=15% per year. A B C Capital investment 11,000 16,000 13,000 Annual revenues 4,000 6,000 5,540 Annual Costs 250 300 400 Market value at EOY3 5,000 6,150 2,800 PW(15%) 850 ??? 577 6-51. Complete the following analysis of cost alternatives and select the preferred alternative. The study period is 10 years and the MARR=12%. Which alternative is most economical? A B C D Capital Investment 11,000 16,000 13,000 18,000 Annual Costs 250 300 400 100 Market Value at EOY 10 1,000 1,300 1,750 2,000 Use ROR method.
6-53. For the following table, assume a MARR of 15% per year, and a useful life for each alternative of eight years. The rank order of alternatives from least capital investment to greatest capital investment is Z-Y-W-X. Complete the incremental analysis by selecting the preferred alternative. Z-Y Y-W W –X Δ Capital investment -250 -400 -550 Δ Annual cost savings 70 90 15 Δ Market value 100 50 200 Δ PW (15%) 97 20 ???
Depreciation 7-10.A Company purchased a machine for P15,000. It is paid sales taxes and shipping costs of P1,000 and nonrecurring installation costs amounting to P1,200. At the end of three years, the company had no further use for the machine, so it spent P500 to have the machine dismantled and was able to sell the machine for P1,500. a.. What is the cost basis for this machine? b.. The Company had depreciated the machine on an SL basis, using an estimated useful life of five years and P1,000 SV. By what amount did the depreciation deductions fail to cover the actual depreciation? 7-14. During the current year, a pharmaceutical company purchased a mixing tank that had a fair market price of P120,000. It replaced an older, smaller mixing tank that had a BV of P15,000. Because a special promotion was underway, the old tank was used as a trade-in for the new one, and the cash price (including delivery and installation) was set at P99,500. The new mixing tank will have a useful life of 9 years. If 200% DB depreciation had been applied to this problem, what would be the cumulative depreciation through the end of year four? 7-15. A special-purpose machine is to be depreciated as a linear function of use (units-of-production method). It costs P25,000 and is expected to produce 100,000 units and then be sold for P5,000. Up to the end of the third year, it had produced 10,000 units and during the fourth year it produced 10,000 units. What is the depreciation deduction for the fourth year and the BV at the end of the fourth year. Replacement Studies 9-7. A city water and waste-water department has a four year old sludge pump that was initially purchased for P65,000. This pump can be kept in service for an additional four years, or in can be sold for P35,000 and replaced by a new pump. The purchase price of the replacement pump is P50,000. The projected market values and O&M costs over the four –year planning horizon as shown in the table. Assuming that the MARR is 10%, determine (a) the economic life of the challenger, (b) when the defender should be replaced. Defender Challenger Year MV at EOY O&M Cost MV at EOY O&M Cost 1 25,000 18,500 40,000 13,000 2 21,000 21,000 32,000 15,500 3 17,000 23,500 24,000 18,000 4 13,000 26,000 16,000 20,500
Prob. A. An asset for drilling was purchased and replaced in service b a petroleum production company. Its cost basis is P60,000 and it has an estimated MV of P12,000 at the end of an estimated useful life of 14 years. Compute the depreciation amount in the third year and the BV at the end of fifth year of life by each of these methods. a). SL method; b). Sinking Fund method with compound interest of 6% Prob. B. JGA Corp. purchased a machine with a basic cost of P180,000. With additional options costing P15,000, the cost basis for depreciation purposes is P195,000. Its market value at the end of eight years is estimated as P40,000. Compute the depreciation charge and book value at the end of a). 5 years by
the Declining Balance Method. b). 6 years by the Double Declining Method Method
c). 7 years by the SYD
Prob. 1. Loader loans Maria P5,000 with interest compounded at a rate of 6% per year. How money will Maria owe Loader if she repays the entire loan at the end of five years? Prob. 2. If P25,000 is deposited now into a savings account that earns 6% per year, what uniform annual amount could be withdrawn at the end of each year for 10 years so that nothing would be left in the account after the 10 th withdrawal? Prob. 3. Maintenance costs for a small bridge with an expected 50-year life are estimated to be P1,000 each year for the first 5 years, followed by a P10,000 expenditure in the year 15 and P10,000 expenditure in year 30. If i = 10% per year, what is the equivalent uniform annual cost over the entire 50-year period? ME BOARD APRIL 1998 A machine has an initial cost f P50000 and a salvage value of P10000 after 10 years. What is the book value after five years using straight line depreciation?Ans. P30000 ME Board April 1998 An asset is purchased for P500000. The salvage value in 25 years is 100000. What are the depreciations in the first three years using straight line method?Ans. P48000 CE Board Nov 1997 The cost of equipment is P500000 and the cost of installation is P30000.If the salvage value is 10% of the cost of equipment at the end of 5 years, determine the book value at the end of the fourth year.use straight line method.Ans. P146000 ECE Board Nov 1999 A machine costs P8000 and an estimated life of 10 years with a salvage value of 500. What is the book value after 8 years using straight line method?Ans. P2000 An engineer bought an equipment for P500000. He spent an additional amount of P30000 for installation and other expenses. The salvage value is 10% of the first cost. If the book value at the end of 5 years will be 291500 using straight line method of depreciation, compute the useful life of the equipment in years? Ans 10 years ME Board April 1992 A unit of welding machine cost P45000 with an estimated life of 5 years.Its salvage value is P2500 find its depreciation rate of sinking fund method. Assuming that will deposit the money to a bank giving 8.5%. Solve for the depreciation.Ans. 7172.54 EE board April 1980 A equipment cost p10000 with a salvage value of P500 at the end of 10 years. Calculate the annual depreciation by sinking fund method at 40% interest. CHE Board Oct 1980 A 110000 chemical plant had an estimated life of 6 years and a projected scrap value of P10000. After 3 years made it a total loss . How much money would have to be raised to put up a new plant costing
P150000, if a depreciation reserved has been maintained during its 3 years of operation by sinking fund method 6%.104359 ME board Sept 1971 A dump truck was bought for 30000 six years ago. It will have a salvage value of P3000 four years from now. It is sold now for P8000. What is the sunk cost if the depreciation method used is sinking fund method at 6%. 7712 CE Board 1999 The corporation purchased a machine for P1 million. Freight and installation charges amounted to 3% of the purchased price. If the machine shall be depreciated over a period of 8 years with a salvage value of 12% of the first cost. Determine the depreciation charged during the 5th year using the SYD method.100711 ME Board April 1998 An Asset is purchased for P9000. Its estimated life is 10 years, after which it will be sold for P1000. Find the book value during the third year if SYD depreciation is used.5072 ECE Board 1998 ABC corporation makes its policy that for every new equipment purchased the annual depreciation cost should not exceed 20% of the first cost at anytime without salvage value. Determine the length of service if the depreciation used is the SYD method.9 years ME board April 1996 An asset is purchased for 120000. Its estimated life is 10 years after which it will be sold for 12000. Find the depreciation for the second year using the SYD method.17673 XXxxxxx Instruction: Choose the BEST answer. 1. It is a depreciation method based on the assumption that the value of a property is directly proportional to its age. a. Straight Line Method b. Sinking Fund Method c. Declining Balance Method d. Service Output Method 2. Which of the following is NOT a requirement of a depreciation method? a. It should be simple b. It should recover the capital c. BIR approved d. The book value should be higher than the market value 3. The decrease in the purchasing power of money a. Devaluation b. Depreciation c. Inflation d. Depletion
4. Equal periodic payment of a debt. a. Annuity b. Amortization c. Depreciation d. Bond 5. Minimum number of years required for a bond redemption. a. 5 years b. 1 year c. 10 years d. 15 years 6. Amount of money spent on a business operation which cannot be recovered due to certain reason. a. Fixed cost b. Sunk cost c. First cost d. Increment cost 7. Represents share in the ownership of the capital corporation. a. Bond b. Mutual fund c. Stock d. Burse 8. It is the actual interest earned by the principal in ONE YEAR. a. Effective interest b. Exact interest c. Simple interest d. Compounded interest 9. A type of annuity where the payment goes on indefinitely a. Ordinary annuity b. Annuity due c. Perpetuity d. Deferred annuity 10. An economic situation when there is only one vendor of a product or service. a. Oligopoly b. Monopoly c. Communism d. Socialism xxxxx Below are the answer to the questions posted on the subject Engineering Economy. 1. It is a depreciation method based on the assumption that the value of a property is directly proportional to its age. A. Straight Line Method
2. Which of the following is NOT a requirement of a depreciation method? D. The book value should be higher than the market value 3. The decrease in the purchasing power of money C. Inflation 4. Equal periodic payment of a debt. B. Amortization 5. Minimum number of years required for a bond redemption. C. 10 years 6. Amount of money spent on a business operation which cannot be recovered due to certain reason. B. Sunk cost 7. Represents share in the ownership of the capital corporation. C. Stock 8. It is the actual interest earned by the principal in ONE YEAR. A. Effective interest 9. A type of annuity where the payment goes on indefinitely B. Perpetuity 10. An economic situation when there is only one vendor of a product or service. B. Monopoly Xxxxxxx Here are some questions for the board exam on the topic Engineering Economy. Instruction: Choose the BEST answer. 1. It is the quantity of production when the income equals total cost. a. Satiation point b. Breakeven point c. Optimal production point d. Point of production 2. The length of time required to recover the first cost of an investment. a. Break-even period b. Payback period c. Return of investment period d. Payout period 3. Common term for the certificate of indebtedness a. Stock b. Loan c. Bond d. Mortgage
4. It represents o rdinary share in ownership of the corporation’s capital without special guarantees of return. a. Preferred stock b. Special stock c. Common stock d. Stock option 5. A type of bond where the name of the owner is recorded in the books of the corporation. Interests are periodically sent to him without claim action. a. Coupon bond b. Registered bond c. Bond paper d. Corporate bond 6. A type of business organization where two or more persons associate to engage for a business. a. Single proprietorship b. partnership c. Cooperative d. Corporation 7. A type of business organization which is a distinct legal entity capable of business transactions like a real person. a. Cooperative b. Single proprietorship c. Corporation d. Partnership 8. In a partnership type of business, what will happen if one of the owners dies? a. The partnership will continue b. The partnership will continue as long as the dead member has been replaced. c. The partnership automatically dies d. The partnership will become corporation. 9. It is the decrease in value of a property due to the passage of time. a. Degradation b. Depletion c. Depression d. Depreciation 10. It is the length of time during which a property may be used for profit. a. Useful life b. Physical life c. Fiscal life d. Economic life
5.1Your uncle has almost convinced you to invest in his peach farm. It would require a $10,000 initial investment on your part. He promises you revenue (before expenses) of $1,800 per year the first year, and increasing by $100/year thereafter. Your share of the estimated annual expenses is $500. You figure you would invest for 6 years. Your uncle has promised to buy out your share of the business at that time for $12,000 (at which point you would use it to buy a new car). You have decided to set a personal MARR of 15% per year. Evaluate the investment using the AW method (show entire AW calculation). Use the AW to calculate the PW. AW (15%) = -$10,000(A/P, 15%, 6) + 12,000(A/F, 15%, 6) + (1800- 500)(P/F, 15%, 1) + [1300 + 100(A/G, 15, 5) = -$2642 + 1370.4 + 1130.48 + [1300 + 172.28] = 1331.16 PW (15%) = 1300 (P/A, 15%, 6) + 100(P/A, 15%, 5) + 12000(P/F, 15%, 6) – 10,000 = 442.67
Calculate the IRR for this investment using linear interpolation. State what your final decision should be. Be sure to set up correctly and to bracket the IRR within 5%. PW (15%) = 1300 (P/A, 15%, 6) + 100(P/A, 15%, 5) + 12000(P/F, 15%, 6) – 10,000 = 442.67 PW (20%) = 1300 (P/A, 20%, 6) + 100(P/A, 20%, 5) + 12000(P/F, 20%, 6) 10,000 = -1358.00 Line BA = Line dA Line BC Line de 20% - 15%/ 442.67 – (-1358.99) = i% - 15%/ 442.67 – 0 5%/ 1801.66 = i% - 15%/ 442.67 442.67(5%) = 1801.66(i% - 15%) 292.37 = 1801.66i i = 16.22 %
