Discounted Cash Flow Analysis

Transcription

1 Discounted Cash Flow Analysis Lecture No.16 Chapter 5 Contemporary Engineering Economics Copyright 2016

2 Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. For Single Project Evaluation: Accept the project if the net surplus is positive. For Comparing Multiple Alternatives: Select the alternative with the largest net present worth.

3 Net Present Worth Measure We will first summarize the basic procedure for applying the net-presentworth criterion to a typical investment project. Determine the interest rate that the firm wishes to earn on its investments.(*) Estimate the service life of the Project. Estimate the cash inflow for each period over the service life. Estimate the cash outflow over each service period. Determine the net cash flows (net cash flow = cash inflow - cash outflow). Find the present worth of each net cash flow at the MARR. Add up all the presentworth figures; their sum is defined as the project s NPW

4 Example 5.1 Rev. Single Project Evaluation- Uniform Flow

5 Example 5.1 Rev. Contd.

6 Example 5.5: Tiger Machine Tool Company Given: Cash flow and i= 12% Find: Net present worth

7 Solution $35,560 $37,360 $31,850 $34,400 outflow 0 $76, inflow PW(12%) $35,560( P / F,12%,1) $37,360( P / F,12%,2) inflow PW(12%) $76,000 outflow $31,850( P / F,12%,3) $34,400( P / F,12%,4) $ PW(12%) $106,065 $76,000 $30,065 0, Accept

8 Excel Solution A B C 1 Period Cash Flow 2 0 ($76,000) 3 1 $35, $37, $31, $34,400 7 PW(12%) $30,065 =NPV(12%,B3:B6)+B2

9 PW at Varying Interest Rates

10 Can You Explain What $30,065 Really Means? Project Balance Concept Investment Pool Concept

11 Project Balance Concept o Suppose that the firm has no internal funds to finance the project, so will borrow the entire investment from a bank at an interest rate of 12%. o Then, any proceeds from the project will be used to pay off the bank loan. o Then, our interest is in seeing how much money would be left over at the end of the project period.

12 Calculating Project Balances End of Year (n ) Beginning Project Balance $ (76,000) $ (49,560) $ (18,147) $ 11,525 Interest Charged (12%) $ (9,120) $ (5,947) $ (2,178) $ 1,383 Payment $ (76,000) $ 35,560 $ 37,360 $ 31,850 $ 34,400 Ending Project Balance $ (76,000) $ (49,560) $ (18,147) $ 11,525 $ 47,308

13 Project Balance Diagram: Four Pieces of Information $75,000 $50,000 Net profit (surplus) if you kept the project until its life $25, $25,000 -$50,000 Profit potential after recovering the investment Exposure to financial risk Discounted payback period -$75,000 -$100,000

14 Investment Pool Concept o Suppose the company has $76,000. It has two options: (1) take the money out and invest it in the project or (2) leave the money in the pool and continue to earn a 12% interest. o If you take option 1, any proceeds from the project will be returned to the investment pool and earn 12% interest yearly until the end of the project period. o Let s see what the consequences are for each option.

15 Option A If $76,000 were left in the investment pool for 4 years Option B If $76,000 withdrawal from the investment pool were invested in the project Year Amount $35,650 $37,360 $31,850 $34,400 $76,000(F/P,12%,4) $119,587 $35,650(F/P,12%, 3) $37,360(F/P,12%,2) $31,850(F/P,12%, 1) $34,400(F/P,12%, 0) Investment Pool PW(12%) = $47,309(P/F,12%,4) = $ $166,896 $119,587 1 $49,959 2 $46,864 3 $35,672 4 $34,400 $166,896 $47,309 The net benefit of investing in the project

16 Cost of capital MARR Risk premium Selecting an MARR in Project Evaluation Cost of capital o The required return necessary to make an investment project worthwhile. o Viewed as the rate of return that a firm would receive if it invested its money someplace else with a similar risk Risk premium o The additional risk associated with the project if you are dealing with a project with higher risk than normal project

17 Practice Problem An electrical motor rated at 15HP needs to be purchased for $1,000. The service life of the motor is known to be 10 years with negligible salvage value. Its full load efficiency is 85%. The cost of energy is $0.08 perkwh. The intended use of the motor is 4,000 hours per year. Find the total present worth cost of owning and operating the motor at 10% interest.

18 1HP=0.7457kW Solution 15HP = = kW Required input power at 85% efficiency rating: kW kW 0.85 Required total kwh per year kW 4,000 hours/year =52,638 kwh/yr Total annual energy cost to operate the motor 52,638kWh $0.08/kWh =$4,211/yr The total present worth cost of owning and operating the motor PW(10%) $1,000 $4,211( P / A,10%,10) = $26,875

19 Cash Flow Series Associated with Owning and Operating the Motor $1,000 $4,211 PW(10%) $1,000 $4,211( P/ A,10%,10) $26,875