ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
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Chapter 5, Problem 68P
To determine
To find:Best alternative from A, B, C, D & E which gives highest benefit.
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3. As supervisor of a facilities engineering department, you consider mobile cranes to be critical
equipment. The purchase of a new, medium-sized truck-mounted crane is being evaluated. The
economic estimates for the two best alternatives are shown in the following table. MARR is at
15% per year. You can use the assumption of repeatability in this case.
Alternative
B
A
Capital investment $272.000 $346.000
Annual expenses
Useful life (years)
| Salvage value
28.800
19,300
6.
25,000
40.000
Show that the same selection is made for the following methods:
a. RORAI method
b. AWC method
c. EUAC method
3. As supervisor of a facilities engineering department, you consider mobile cranes to be critical
equipment. The purchase of a new, medium-sized truck-mounted crane is being evaluated. The
economic estimates for the two best alternatives are shown in the following table. MARR is at
15% per year. You can use the assumption of repeatability in this case.
Alternative
B
A
Capital investment $272.000 $346.000
Annual expenses
Useful life (years)
| Salvage value
28.800
19,300
6.
25,000
40.000
Show that the same selection is made for the following methods:
c. EUAC method
You are considering two alternative machines, Machine A and Machine B, for a manufacturing process. One of the machines, Machine A, costs $40,000 to set up and $6,000 per year to operate, but It must be completely replaced every 7 years, and it has no salvage value. Assuming the cost of capital is 8.7%, what is the equivalent annual cost of the machine?
Chapter 5 Solutions
ENGR.ECONOMIC ANALYSIS
Ch. 5 - Prob. 1QTCCh. 5 - Prob. 2QTCCh. 5 - Prob. 3QTCCh. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7P
Ch. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10PCh. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 64PCh. 5 - Prob. 65PCh. 5 - Prob. 66PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 81PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 98PCh. 5 - Prob. 99PCh. 5 - Prob. 100PCh. 5 - Prob. 101P
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- Two processes are being considered for the production of a machine part. Process 1 requires an equipment cost of P180 000.00, a yearly operating cost of P90 000.00, and has zero salvage value at the end of its 6-year life. Process 2 has corresponding figures of P320 000.00, P72 000.00, P45 000.00 and 9-year life. If the applied MARR is 20%, compare the two processes using PW analysis. (Ans. PT1 = -P693 567.36; PT2= - P718 084.62)- Please use and actual formula NOT AN EXCEL PLEASE. We are students not office workers.arrow_forwardAn equipment is bought for $9,000 and sold at the end of its service life for $800. If the service period is 10 years and the MARR is 10%, then the minimum annual savings that this equipment must yield for it to break even is closest to? a) $1,414 b) $1,701 c) $1,312 O d) $1,514arrow_forwardCB Electronix must buy a piece of equipment to place electronic components on the printed circuit boards it assembles. The proposed equipment has a 10-year life with no scrap value. The supplier has given CB several purchase alternatives. The first is to purchase the equipment for $950,000. The second is to pay for the equipment in 10 equal installments of $125,000 each, starting one year from now. The third is to pay $210,000 now and S85,000 at the end of each year for the next 10 years. Complete parts (a) and (b) below. a. Which alternative should CB choose if its MARR is 11 percent per year? Use an IRR comparison approach. Considering the alternatives in the order of lowest first cost, the best option is the which has an incremental rate of return of %. (Type an integer or decimal rounded to two decimal places as needed. Use an approximate ERR if the IRR cannot be used.)arrow_forward
- A potential project is currently under review. An initial investment of 87000 would be necessary for equipment. The annual revenues and expenses are expected to be 40000 and 19000 each year, respectively, over the 6-year project period. The salvage value of the equipment at the end of the project period is projected to be $17000. Assume a MARR of 9%. Find the PW (to the nearest cent).arrow_forwardAn oil company plans to purchase a piece of vacant land at the corner of two busy streets for $50,000. On properties of this type, the company installs businesses of three different types. Each has an estimated useful life of 15 years. The salvage value for each is estimated to be the $50,000 land cost. Plan Cost (in addition to land cost) Type of business Net annual income A $ 83,000 Conventional gas station $ 26,500 B $ 195,000 Add automatic car wash $ 39,750 C $ 115,000 Add quick car wash $ 31,200 if the oil company expects a 10% rate of return on its investments, which plan (if any) should be selected? Use incremental analysis, before tax.arrow_forwardMatchim's Paving is investing in a new wider paving machine that will cost $131000, which is expected to return $22500 per year of the next 12 years. At the end of the 12 years the paving machine is expected to have a salvage value of $20000. Assuming the cash flow is at the end of the year, what is the Pay Back Period, in years, for the project?arrow_forward
- Problem 05.017 Alternative Comparison - Different Lives Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (i.e., low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 19% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-160,000 $-40,000 $10,000 2 years The present worth for the DDM method is $ The present worth for the LS method is $ The (Click to select) method is selected. LS $-500,000 $-10,000 $33,000 4 yearsarrow_forwardMachine X has an initial cost of P 45,000, annual maintenance of P 2,250 per year, an no salvage value at the end of its four-year useful life. Machine Y costs P 90,000. The first year there is no maintenance cost. The second year, maintenance is P450, and increases P450 per year in the subsequent years. The machine has an anticipated value P22,500 salvage value at the end of its 12-year useful life.If interest is 8%, what is the present worth of cost of 12 years of Machine X? a) P157,357 b) P140,174 c) P52,452 d) P 119,345arrow_forwardYou have access to two N-fertilizer sources. Which is the more economical option? urea (46-0-0): $700/ton UAN (32-0-0): $640 / ton ●arrow_forward
- Current Attempt in Progress Mandy is considering investing in an opportunity that would require an upfront cost of $ 520 but would pay $ 150 per year for each of the next 6 years. If Mandy chooses to invest in this opportunity, what would be the IRR? Click here to access the TVM Factor Table calculator. Carry all interim calculations to 5 decimal places and then round your final answer to 1 decimal place. The tolerance is ±0.5. Should Mandy invest in this opportunity if her personal MARR is 20%?arrow_forwardFor th below two machines and based on AW analysis which machine we should select? MARR=10% Machine A Machine B 26612 12417 4135 Life, years 3 Answer the below question: B- the AW for machine B= First cost, $ Annual cost, $/year Salvage value, $ 140454 7170 infinitearrow_forwardA company is considering buying a new equipment. They have a choice between two models. The company has a MARR of 5%. The salvage value of each model at the end of its service life is zero Model A $ 450 8 years $90 Model B First Cost $180 4 years $100 Life Annual Cost Which alternative should be chosen? Use the IRR method (or ERR if necessary).arrow_forward
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