car B -20000 car A first cost annual operating cost salvage life, years -32000 -4000 4000 You must decide between two different cars. Using an interest rate of 8% per year, the PV of car A is closest to -3500 1000 3 6 $-121396 .A O $-101396 .B O $-111396.C O $-112396.D O

car B -20000 car A first cost annual operating cost salvage life, years -32000 -4000 4000 You must decide between two different cars. Using an interest rate of 8% per year, the PV of car A is closest to -3500 1000 3 6 $-121396 .A O $-101396 .B O $-111396.C O $-112396.D O

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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A3
Q5: The machine using PW with first cost $75000, savings the annual operating cost $11000 per year and salvage value $55000 at the end of 4 years find the (P/F, i*,6).
Problems 43 and 44 are based on the following statement: Used New machine machine Initial Cost 15,000 40,000 ($) Annual 8,000 2,000 operating cost ($/year) Salvage 5,000 10,000 value ($) Life (years) 3 6 Use an interest rate of 10% per year 43. The present worth of the new machine is closest to: (A) $43,630 (B) $43,066 (C) $45,210 (D) $49,660 44. To compare the machines on the basis of a present worth analysis, the present worth values to use are: (A) PWused $31,139, PW new = $43,066 (B) PWused $31,139, PW new = $37,461 (C) PWused $54,533, PW new = $43,066 (D) PWused $54,533, PW new = $37,461
In the choice of mechanical equipment you are required to submit your recommendation as to which of the two steam shovels for earth excavation is better to buy. The basis is 20 year service & interest at 7%. Taxes & insurance are the same in each case: 5% & 1% respectively. Use Annual Cost & CC. 50-ton shovel 70 ton shovel First Cost P30k P38k 20 years 20 years P2k Life Salvage value Annual Operation Annual Upkeep Annual Output P3k P15k P17k P1.5k P2k 150k yards 200k yards
QI\ a contractor want to invest his money in one of projects A or B if you know that the interest rate is 10%, choose the best altemative A or B using net present worth method ? A. B. Initial Investment 120,000 CU 135,000 CU Annual revenue Annual expenses Salvage vahie |Project life Cost of machines replaced every 3 years 15,000 CU\year 10,000 CU year 12,000 CU 12 years 17,000 CU year 11,000 CU year 20,000 CU 6 усars 3500 CU 3000 CU Q2 A company is planning to purchase a new equipment to increase the production and revemies. The cost of the equipment is 451105 CU and the equipment installation cost is 17% of its cost. Find the payback period for this equipment Year 1 Year 2 120,000 CU 130,000 CU 90,000 CU Year 3 Year 4 |100,000 CU Year 5 115,000 CU 90,000 CU Year 6 Q3 \ Machinery costs (60000 CU) and its useful life is (5 year). compute the schedule of depreciation and book value at the end of each year for the machinery by using the Sum of Year Digit method , if you know that the…
Question 1) Nascar Company has two alternatives for a new truck investment: Buy or lease. Buying option: The purchasing cost of the truck is 230.000 dollars. The economic life will be five years and at the end fifth year the truck will have a salvage value, 40.000 dollars Repair and insurance costs: 1.year 10.000 dollars 2.year 15.000 3.year 20.000 4.year 25.000 5.year 28.000 Leasing option: Lease period: 5 years Annual lease payments: 60.000 dollars (payments will be made at the beginning of each year) Also, at the beginning of contract period Nascar Co. must pay a deposit, 20.000 dollars. This deposit amount will be refunded at the end of fifth year. According to the leasing contract all repair and insurance expenses are under the responsibility of leasing company. Make your decision according to Net Present Value Method. (Assume the discount rate of the company is 12%) Present value of 1 dollar Period Discount rate (12%) 1 0,893 2 0,797 3…
9. A local company is analyzing 5 alternatives for a new machine the information is given in the table below. Option Cost Savings per Year Salvage value FV Vehicle life n A -$10,000.00 $ 2,050.00 $ 500.00 B -$12,500.00 $ 3,405.00 -$15,300.00 $ 3,680.00 -$20,500.00 $ 5,281.50 -$30,125.00 $ 6,785.00 C D E $1,000.00 $3,000.00 $ 5,000.00 $5,000.00 5 5 5 5 5 a) which alternative should be selected without considering salvage value? The company is considering a MARR of 7%. b) which alternative should be selected with considering salvage value? The company is considering a MARR of 10%.
urgent please in one hour
Q/ A contractor purchase an equipment with a delivered price of (750000$) to operate it in one of his projects. He estimate the cost of this equipment according to the following justification - Total interest rate (including bank interest, storage, insurance, .) =15 % Consumption of (oil, fuel, and grease) =(8.25 S/hr) - Cost of maintenance and repair = 60% of the annual straight line depreciation - salvage value = 150000 S Useful life = (10 years) with (1400 hours used per year). a- Determine the probable cost per hour according to the previous justification. b- After (7 years) of operating the equipment, the contractor recorded all the costs of the equipment as shown in the table blow. He notice the increment in the average consumption and maintenance in the last two years (the 6", and 7th years). Calculate the average cost per hour of the equipment for the first five years, according to the salvage value of the equipment at the end of the 5th year = (200000$). Fuel oil Grease…
Question 5 A machine that have the following cost is under consideration for a new manufacturing process. The interest rate is 6% compounded semiannually. First cost = $70,000 Semiannual operating cost = 6,000 Semiannual income = 18,000 Salvage value = 9,000 Life in years = 5 years a. PW = $29,059 b. PW = $39,059 O c. PW = -$39,059 d. PW = $19,059
At i = 10% per year, a machine has a first cost of $9,000 and estimated operating costs and year-end salvage values as shown. The annual equivalent cost if you operate the machine for 3 years is closest to which value? Operating Salvage Cost, $ Value, $ 7000 Year -1000 -1200 5000 -1500 4200 -2000 3000 -3000 2000 O-$-3,568 = $-3,770 -1500 4200 -2000 3000 -3000 2000 O =$-3,568 O = $-3,770 = 2-3,636 %3D O =$-3,454 O = $-3,832 123 n 345

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