Solution Manual Construction Equipment Management for Engineers, Estimators, and Owners

1. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 CHAPTER 2 Time Value of Money 1.A contractor borrowed $250,000 from a bank at an interest rate of 6% to purchase new construction equipment. What annual payment must the contractor make if the loan is to be paid off in 10 years? ? = ($250,000)(?? ⁄ ,6%,10) = ($250,000)(0.136) = $??,??? Chapter 2, Problem 1 Receipts 0 1 2 3 4 5 6 7 8 9 10 Time (Periods) Disbursements ??? ??? ??? ??? ??? ??? ??? ??? ??? ??? $250k 2 - 1 complete document is available on https://solumanu.com/ *** contact me if site not loaded

2. smtb98@gmail.com Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 smtb98@gmail.com Email: smtb98@gmail.com Telegram: https://t.me/solumanu 2.A grader costs $350,000 to purchase and is expected to have a useful life of 8 years. Annual operating, maintenance, and labor costs are estimated to be $35,000 per year, and the salvage value after 8 years of use is estimated to be $50,000. At an interest rate of 9%, what is the present worth equivalent cost to the contractor of owning and operating the grader? ? = ($350,000) + ($35,000)(?? = ($350,000) + ($35,000)(5.535) − ($50,000)(0.502) = $350,000 + $193,725 − $25,100 = $???,??? ⁄ ,9%,8) − ($50,000)(?? ⁄ ,9%,8) Chapter 2, Problem 2 $50k Receipts 0 1 2 3 4 5 6 7 8 Time (Periods) Disbursements $35k $35k $35k $35k $35k $35k $35k $35k $350k 2 - 2 complete document is available on https://solumanu.com/ *** contact me if site not loaded

3. Email: smtb98@gmail.com Telegram: https://t.me/solumanu Contact me in order to access the whole complete document. WhatsApp: https://wa.me/+12342513111 3.Using an interest rate of 12%, find the equivalent uniform annual cost for a piece of construction equipment that has an initial purchase cost of $80,000, an estimated economic life of 8 years, and an estimated salvage value of $10,000. Annual maintenance will amount to $600 per year and periodic overhauls costing $1,000 each will occur at the end of the 2nd, 4th, and 6th years. ? = ($80,000)(?? + ($1,000)(?? ⁄ ,12%,8) + ($1,000)(?? = ($80,000)(0.201) + ($600) + ($1,000)(0.797)(0.201) + ($1,000)(0.636)(0.201) + ($1,000)(0.507)(0.201) − ($10,000)(0.081) = $16,080 + $600 + $160 + $128 + $102 − $810 = $??,??? ⁄ ,12%,8) + ($600) + ($1,000)(?? ⁄ ,12%,4)(?? ⁄ ,12%,6)(?? ⁄ ,12%,2)(?? ⁄ ,12%,8) ⁄ ,12%,8) − ($10,000)(?? ⁄ ,12%,8) Chapter 2, Problem 3 $10k Receipts 0 1 2 3 4 5 6 7 8 Time (Periods) Disbursements $0.6k $0.6k $0.6k $0.6k $0.6k $0.6k $0.6k $0.6k $80k $1k $1k $1k 2 - 3 complete document is available on https://solumanu.com/ *** contact me if site not loaded

4. smtb98@gmail.com smtb98@gmail.com 4.A company owns a fleet of trucks and operates its own maintenance shop. A certain type of truck, normally used for 5 years, has an initial cost of $45,000 and a salvage value of $75,000. Maintenance costs are $7,000 for the first year and increase by $2,000 each year. Assuming interest at 10%, find the equivalent annual cost of owning and maintaining the truck. ? = ($45,000)(?? + ($9,000)(?? ⁄ ,10%,5) + ($11,000)(?? ⁄ ,10%,5) + ($13,000)(?? ⁄ ,10%,5) + ($15,000)(?? = ($45,000)(0.264) + ($7,000)(0.909)(0.264) + ($9,000)(0.826)(0.264) + ($11,000)(0.751)(0.264) + ($13,000)(0.683)(0.264) + ($15,000)(0.621)(0.264) − ($7,500)(0.164) = $11,880 + $1,680 + $1,963 + $2,181 + $2,344 + $2,459 − $1,230 = $??,??? Commented [GM1]: There is a typo in the book. The salvage value should be $7,500. We used the correct value in the solution ⁄ ,10%,5) + ($7,000)(?? ⁄ ,10%,2)(?? ⁄ ,10%,3)(?? ⁄ ,10%,4)(?? ⁄ ,10%,5)(?? ⁄ ,10%,1)(?? ⁄ ,10%,5) ⁄ ,10%,5) − ($7,500)(?? ⁄ ,10%,5) Chapter 2, Problem 4 $7.5k Receipts 0 1 2 3 4 5 Time (Periods) $7k Disbursements $9k $11k $13k $15k $45k 2 - 4 complete document is available on https://solumanu.com/ *** contact me if site not loaded

5. 5.A contractor knows that she must replace her forklift in 12 years at an estimated cost of $195,000. She plans to start making deposits into a bank account at the beginning of the year 3 years from now and continue until the end of the eighth year (7 equal payments). How much money should she deposit in each of the 7 years at an interest rate of 6% to have the needed $195,000 at the end of the 12 years? ($195,000)(?? ($195,000)(?? (?? ($195,000)(0.497) (0.890) + (0.840) + (0.792) + (0.747) + (0.705) + (0.665) + (0.627)=$96,915 ⁄ ,6%,12) = ?(?? ⁄ ,6%,2) + ?(?? ⁄ ,6%,3) + ?(?? ⁄ ,6%,4) + ?(?? ⁄ ,6%,5) + ?(?? ⁄ ,6%,6) + ?(?? ⁄ ,6%,7) + ?(?? ⁄ ,6%,8) ⁄ ,6%,12) ⁄ ,6%,5) + (?? ? = ⁄ ,6%,2) + (?? ⁄ ,6%,3) + (?? ⁄ ,6%,4) + (?? ⁄ ,6%,6) + (?? ⁄ ,6%,7) + (?? ⁄ ,6%,8) = $??,??? = 5.266 2 - 5 complete document is available on https://solumanu.com/ *** contact me if site not loaded

6. smtb98@gmail.com smtb98@gmail.com 6.A contractor knows that he must replace a mobile crane in 10 years at an estimated cost of $600,000. How much should he deposit today and at the beginning of each of the next 7 years (8 equal payments) at an interest rate of 10% to have the needed $600,000 at the end of the 10 years? ($600,000)(?? = ? + ?(?? + ?(?? ($600,000)(?? 1 + (?? ($600,000)(0.386) (1) + (0.909) + (0.826) + (0.751) + (0.683) + (0.621) + (0.564) + (0.513)=$231,600 ⁄ ,10%,10) ⁄ ,10%,1) + ?(?? ⁄ ,10%,7) ⁄ ,10%,2) + ?(?? ⁄ ,10%,3) + ?(?? ⁄ ,10%,4) + ?(?? ⁄ ,10%,5) + ?(?? ⁄ ,10%,6) ⁄ ,10%,10) ⁄ ,10%,4) + (?? ? = ⁄ ,10%,1) + (?? ⁄ ,10%,2) + (?? ⁄ ,10%,3) + (?? ⁄ ,10%,5) + (?? ⁄ ,10%,6) + (?? ⁄ ,10%,7) = $??,??? = 5.867 Chapter 2, Problem 6 ??? ??? ??? ??? ??? ??? ??? ??? Receipts 0 1 2 3 4 5 6 7 8 9 10 Time (Periods) Disbursements $600k 2 - 6

7. 7.A contractor has purchased 5 elevating scrapers for $650,000 each and plans to use them for about 12,500 hours of operation. He expects to be able to sell the used scrapers for 10% of the purchase price after 12,500 hours of use. Tires cost about $20,000 per set to replace (estimated to occur after 2,500 hours of use). What is the estimated ownership cost of each scraper (in $/hour) if its annual usage is estimated to be 1,250 hours per year? 12,500 ℎ???? 1,250 ℎ???? ??? ????= 10 ????? ? = ($650,000)(?? + ($20,000)(?? + ($20,000)(?? = ($650,000)(0.163) + ($20,000)(0.826)(0.163) + ($20,000)(0.683)(0.163) + ($20,000)(0.564)(0.163) + ($20,000)(0.467)(0.163) + ($20,000)(0.386)(0.163) − ($65,000)(0.063) = = $105,950 + $2,693 + $2,227 + $1,839 + $1,522 + $1,258 − $4,095 = $???,??? ?????? ??????ℎ?? ???? = ?????? ??????ℎ?? ???? ???? ??? ???? ????? ?? ???? = ⁄ ,10%,10) + ($20,000)(?? ⁄ ,10%,6)(?? ⁄ ,10%,10)(?? ⁄ ,10%,2)(?? ⁄ ,10%,10) + ($20,000)(?? ⁄ ,10%,10) − ($65,000)(?? ⁄ ,10%,10) + ($20,000)(?? ⁄ ,10%,8)(?? ⁄ ,10%,4)(?? ⁄ ,10%,10) ⁄ ,10%,10) ⁄ ,10%,10) = $111,394 1,250 ℎ???? ??? ????= $89.11/ℎ??? = Chapter 2, Problem 7 $65k Receipts 0 1 2 3 4 5 6 7 8 9 10 Time (Periods) $20k $20k $20k $20k $20k Disbursements $650k 2 - 7

8. smtb98@gmail.com smtb98@gmail.com 8.A contractor is considering the purchase of a new truck for $40,000 which has an estimated useful life of 8 years. He believes that he can sell the used truck for $8,000 at the end of the 8 years. Annual operating costs are estimated to be $2,000 per year. As an alternative, the contractor can purchase a used truck for $20,000 with an estimated useful life of 4 years. Annual operating costs for the used truck are estimated to be $2,800 per year, and the salvage value should be $2,000 at the end of the 4 years. At an interest rate of 8%, which alternative should the contractor select? ????= ($40,000)(?? = ($40,000)(0.174) + $2,000 − ($8,000)(0.094) = $6,960 + $2,000 − $752 = $8,208 ?????= ($20,000)(?? = ($20,000)(0.302) + $2,800 − ($2,000)(0.222) = $6,040 + $2,800 − $444 = $8,396 The annual cost of the new truck would be lower, so it should be selected in absence of other selection criteria. ⁄ ,8%,8) + $2,000 − ($8,000)(?? ⁄ ,8%,8) ⁄ ,8%,4) + $2,800 − ($2,000)(?? ⁄ ,8%,4) Chapter 2, Problem 8 $8k NEW TRUCK Receipts 0 1 2 3 4 5 6 7 8 Time (Periods) Disbursements $2k $2k $2k $2k $2k $2k $2k $2k $40k USED TRUCK $2k Receipts 0 1 2 3 4 5 6 7 8 Time (Periods) Disbursements $2.8k $2.8k $2.8k $2.8k $20k 2 - 8

9. 9.The question arises whether it is more economical to replace the engine on a tractor with a new one, or rebore the cylinders of the old engine and thoroughly recondition it. The original cost of the engine 10 years ago was $7,000. To rebore and recondition it now will extend its useful life for an estimated 5 years and will cost $2,800. A new engine will have an initial cost of $6,200 and will have an estimated life of 10 years. It is expected that the annual cost of fuel and lubricants with the reconditioned engine will be about $2,000 and that this cost will be 15% less with the new engine. It is also believed that repairs will be about $250 a year less with the new engine than with the reconditioned one. Assume that neither engine has any realizable value when retired. At an interest rate of 6%, which alternative should be selected? ???????= ($2,800)(?? ⁄ ,6%,5) + $2,000 + $250 = ($2,800)(0.237) + $2,000 + $250 = $664 + $2,000 + $250 = $2,914 ????= ($6,200)(?? ⁄ ,6%,10) + ($2,000)(1 − 0.15) = ($6,200)(0.136) + ($2,000)(0.85) = $6,960 + $1,700 = $8,660 The annual cost of the reconditioned engine would be lower, so it should be selected in absence of other selection criteria. Chapter 2, Problem 9 NEW ENGINE Receipts 0 1 2 3 4 5 6 7 8 9 10 Time (Periods) Disbursements $1.7k $1.7k $1.7k $1.7k $1.7k $1.7k $1.7k $1.7k $1.7k $1.7k $6.2k RECONDITION OLD ENGINE Receipts 0 1 2 3 4 5 6 7 8 Time (Periods) $2k $2k $2k $2k $2k Disbursements $250 $250 $250 $250 $250 $2.8k 2 - 9

10. smtb98@gmail.com smtb98@gmail.com 10.A contractor has purchased a small backhoe for $120,000 that he plans to use in excavating ditches for utilities construction. He plans to use the backhoe for 8 years and sell the used machine for $40,000. He must replace the tires on the backhoe after each 3,000 hours of use at a cost of $15,000. Annual operating and labor costs are estimated to be $20,000 per year. The contractor estimates that the backhoe will be used about 1,000 hours per year. If the minimum attractive rate of return is 12%; what is the hourly owning and operating cost for the backhoe? Based on the information provided, tires need to be replaced every 3 years. Therefore, a $15,000 investment should be accounted for at the end of years 3 and 6. As the used backhoe will be sold at the end of year 8, tires will have an estimated remaining life of 1,000 hours at time of sale. Therefore, we are assuming that the salvage value takes into account that tires’ estimated remaining life and a 3rd change of tire will not be budgeted for. ? = ($120,000)(?? + $20,000 − ($40,000)(?? = ($120,000)(0.201) + ($15,000)(0.712)(0.201) + ($15,000)(0.507)(0.201) + $20,000 − ($40,000)(0.081) = = $24,120 + $2,147 + $1,529 + $20,000 − $3,240 = $??,??? ?????? ?&? ???? = ?????? ?&? ???? ???? ??? ???? ⁄ ,12%,8) + ($15,000)(?? ⁄ ,12%,3)(?? ⁄ ,12%,8) = ⁄ ,12%,8) + ($15,000)(?? ⁄ ,12%,6)(?? ⁄ ,12%,8) + $ $44,556 1,000 ℎ???? ??? ????= $44.56/ℎ??? = 2 - 10

11. 11.A used tracked dozer costs $135,000. It will be operated for 5 years, when it will be sold for an estimated value of $65,000. It will cost $80,000 to maintain and operate for the first year. This will increase by $5,000 per year over the life of the machine. If the prevailing rate for interest, insurance, taxes, and storage is 14%; what is the equivalent annual cost of owning and operating the machine? ? = ($135,000)(?? + ($90,000)(?? + ($100,000)(?? = ($120,000)(0.201) + ($15,000)(0.712)(0.201) + ($15,000)(0.507)(0.201) + $20,000 + ($40,000)(0.081) = = $24,120 + $2,147 + $1,529 + $20,000 − $3,240 = $??,??? ?????? ?&? ???? = ?????? ?&? ???? ???? ??? ???? ⁄ ,14%,5) + ($80,000)(?? ⁄ ,14%,1)(?? ⁄ ,14%,5) + ($95,000)(?? ⁄ ,14%,5)(?? ⁄ ,14%,5) + ($85,000)(?? ⁄ ,14%,2)(?? ⁄ ,14%,5) ⁄ ,14%,5) ⁄ ,14%,3)(?? ⁄ ,14%,4)(?? ⁄ ,14%,5) = ⁄ ,14%,5) − ($65,000)(?? $44,556 1,000 ℎ???? ??? ????= $44.56/ℎ??? = 2 - 11

12. smtb98@gmail.com smtb98@gmail.com 12.A project has a first cost of $120,000 and an estimated salvage value after 25 years of $20,000. Estimated average annual receipts are $25,900; estimated average annual disbursements are $15,060. Assuming that annual receipts and disbursements will be uniform, compute the prospective rate of return. $120,000 + ($15,060)(?? $120,000 = ($25,900)(?? $120,000 = ($10,840)(?? Now, we must use a trial and error method to find the solution. Let’s try an interest rate of 5%. $120,000 = ($10,840)(14.094) + ($20,000)(0.295) = $152,779 + $5,900 = $158,679 Let’s try an interest rate of 7%. $120,000 = ($10,840)(11.654) + ($20,000)(0.184) = $126,329 + $3,680=$130,009 Let’s try an interest rate of 8%. $120,000 = ($10,840)(10.675) + ($20,000)(0.146) = $115,717 + $2,920=$118,637 Now we can determine the prospective rate of return by interpolation: ? = 7% +$130,009 − $120,000 $130,009 − $118,637= 7% + 0.9% = 7.9% ⁄ ,?,25) = ($25,900)(?? ⁄ ,?,25) + ($20,000)(?? ⁄ ,?,25) ⁄ ,?,25) − ($15,060)(?? ⁄ ,?,25) + ($20,000)(?? ⁄ ,?,25) ⁄ ,?,25) + ($20,000)(?? ⁄ ,?,25) $25.9k $25.9k $25.9k $25.9k Receipts Years 4 to 24 0 1 2 3 25 Time (Periods) Disbursements $15.06k $15.06k $15.06k $15.06k $120k 2 - 12

13. 13.A contractor has purchased a small excavator for $280,000 that she plans to use for 10 years. She anticipates that the used excavator can be sold for $25,000 after 10 years of use. Average annual earnings generated by the excavator are estimated to be $125,000, and the average operating and labor costs are estimated to be $75,000. What is the estimated rate of return that the excavator will generate for the contractor? $280,000 + ($75,000)(?? $280,000 = ($125,000)(?? $280,000 = ($50,000)(?? Now, we must use a trial and error method to find the solution. Let’s try an interest rate of 5%. $280,000 = ($50,000)(7.722) + ($25,000)(0.614) = $386,100 + $15,350 = $401,450 Let’s try an interest rate of 7%. $280,000 = ($50,000)(7.024) + ($25,000)(0.508) = $351,200 + $12,700 = $363,900 Let’s try an interest rate of 10%. $280,000 = ($50,000)(6.145) + ($25,000)(0.386) = $307,250 + $9,650 = $316,900 Let’s try an interest rate of 12%. $280,000 = ($50,000)(5.650) + ($25,000)(0.322) = $282,500 + $8,050 = $290,550 Let’s try an interest rate of 14%. $280,000 = ($50,000)(5.216) + ($25,000)(0.270) = $260,800 + $6,750 = $267,550 Now we can determine the prospective rate of return by interpolation: ? = 12% + ($290,550 − $280,000 $290,550 − $267,550)(2%) = 12% + 0.9% = 12.9% ⁄ ,?,10) = ($125,000)(?? ⁄ ,?,10) + ($25,000)(?? ⁄ ,?,10) ⁄ ,?,10) − ($75,000)(?? ⁄ ,?,10) + ($25,000)(?? ⁄ ,?,10) ⁄ ,?,10) + ($25,000)(?? ⁄ ,?,10) 2 - 13

14. smtb98@gmail.com smtb98@gmail.com 14.A contractor has purchased a wheeled loader for $130,000 and expects to use the loader an average of 1,500 hours per year. Tires cost $6,000 to replace (estimated to occur after each 4,500 hours of use) and major repairs will be needed every 6,000 hours at a cost of $5,000. The contractor expects to be able to sell the loader for $10,000 after she has used it for 15,000 operating hours. Fuel, oil, and minor maintenance cost about $19.75 for each hour the loader is used. Interest, insurance, and taxes total about 16%. How much should the contractor charge per hour for use of the loader to recover her costs? 15,000 ℎ???? 1,500 ℎ???? ??? ????= 10 ????? ?????? ????,??? ??? ????? ??????????? = ($19.75 ????? ?? ???? = )(1,500ℎ???? ????) = $29,625/???? ℎ? ? = ($130,000)(?? ⁄ ,16%,10) + ($6,000)(?? + ($6,000)(?? + ($5,000)(?? ⁄ ,16%,3)(?? ⁄ ,16%,10) + ($5,000)(?? ⁄ ,16%,8)(?? ⁄ ,16%,10) + ($6,000)(?? ⁄ ,16%,4)(?? ⁄ ,16%,10) + ($29,625)− ($10,000)(?? ⁄ ,16%,6)(?? ⁄ ,16%,10) ⁄ ,16%,10) = ⁄ ,16%,10) ⁄ ,16%,9)(?? = ($130,000)(0.207) + ($6,000)(0.641)(0.207) + ($6,000)(0.410)(0.207) + ($6,000)(0.263)(0.207) + ($5,000)(0.552)(0.207) + ($5,000)(0.305)(0.207) + ($29,625) − ($10,000)(0.057) = = $26,910 + $796 + $509 + $327 + $2,758 + $316 + $29,625 − $570 = $??,??? ?????? ?&? ???? = ?????? ?&? ???? $60,671 1,500 ℎ???? ??? ????= $40.45/ℎ??? = ???? ??? ???? 2 - 14