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In PW method, we do renew the shorter life project if the project is not renewable.

Find EUAW of a device that has a useful life of 6 years; initial cost of $1,500; annual benefit of $650. Use 10 years for the analysis and i= 12%?

A mechanical device will cost $20,000 when purchased. Maintenance will cost $1000 per year. The device will generate revenues of $5000 per year for 5 years. The salvage value is $7000. Is the below cash flow is the correct net representation of the above statement.

PW method and the annual worth method will always give you the same conclusion of selecting the best project to implement, even though the life of each project may be different

Find the final F value from P=1500 if the i% keeps changing in the next 15 years as following: 5% for the first 4 years, 8% for the next 5 years and 10% for the last 6 years.

What is the future value of a $15,000 investment today if for the next 4 years the interest rate is assumed to be 5%? Use interest table.

What equal amount (Q) must be deposited at the beginning of each year for the next 5 years in a saving account with 10% interest in order to get a future sum of $1000 at the end of year 6.  The Value of Q is most nearly:

You borrowed $5000 with 10% simple interest per year, then the total repayment will be =$6050 by the end of 2 years.

F=2000, n=3 years and i=12% compounded monthly. P is most nearly:

You are paying 2% per month for the unpaid balance. Therefore, the effective rate of interest/year that you are paying now is: