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ENGG 401 X2 Fundamentals of Engineering Management Spring 2008 Chapter 7: Present Worth Analysis Dave Ludwick Dept. of Mechanical Engineering University.

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Presentation on theme: "ENGG 401 X2 Fundamentals of Engineering Management Spring 2008 Chapter 7: Present Worth Analysis Dave Ludwick Dept. of Mechanical Engineering University."— Presentation transcript:

1 ENGG 401 X2 Fundamentals of Engineering Management Spring 2008 Chapter 7: Present Worth Analysis Dave Ludwick Dept. of Mechanical Engineering University of Alberta http://members.shaw.ca/dave_ludwick/

2 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 2 ENGG 401 X2 – Fundamentals of Engineering Management Present Worth present worthEquivalence says that we can shift any sum to an equivalent sum at some other point in time… the equivalent sum at the present time (right now) is called present worth (PW). –PW is the comparable equivalent value at the present time of a future sum or set of sums. –PW or NPV can be thought of as the difference between a future set of cash inflows and cash outflows shifted to the present. This is typical of any investment. You make a cash outlay expecting future cash payments net present valuePW is also called net present value (NPV), although that term is more often used when referring to the total present worth of a series of sums.

3 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 3 ENGG 401 X2 – Fundamentals of Engineering Management Present Worth Analysis Present worth analysisPresent worth analysis is when we compare the net present value of multiple mutually exclusive options. Present worth analysis considers only future incomes and expenditures. sunk costs –Costs incurred in the past are called sunk costs and are irrelevant when comparing future cash flows. –Sunk costs are in the past. You can’t undo them! FinancinginvestmentFinancing and investment activities are considered separate activities and then their NPVs combined. –The goal is to maximize the PW of investment benefits, minimize the PW of financing costs, and maximize the NPV of a combination of financing and investment activities.

4 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 4 ENGG 401 X2 – Fundamentals of Engineering Management Present Worth Analysis Example You can consider two types of equipment for your company, each with different costs, net annual benefits, useful lives, and salvage values: #1 #2 Purchase cost$ 200k$ 700k Annual benefit$ 95k$ 120k Useful life 6 yrs 12 yrs Salvage value$ 50k$ 150k Which option is best (assuming a 10% interest rate)?

5 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 5 ENGG 401 X2 – Fundamentals of Engineering Management Capitalized Costs Capitalized cost analysisCapitalized cost analysis is a special kind of present worth analysis that chooses between several alternatives with infinite analysis periods. capitalized cost –The capitalized cost of an option is the NPV of a perpetual series of cash flows. –Alternatively, capitalized cost is the present sum needed to provide a perpetual series of cash flows that will support a capital project. The capitalized cost, P, that is needed to support a uniform perpetual annuity, A, is:

6 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 6 ENGG 401 X2 – Fundamentals of Engineering Management Capitalized Costs Example Your municipality decides to invest in a long-term project that will cost $8 000 000, and need to be replaced every 70 years (at the same cost). If you estimate a 7% interest rate, what is the project’s capitalized cost? 070140 $8M 01234570 AAAAAA $8M 210240 070 $8M

7 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 7 ENGG 401 X2 – Fundamentals of Engineering Management Present Worth Analysis using Spreadsheets We can use (P/A, i, n) to calculate the present worth of a series of sums, but another method of doing so is with spreadsheets (e.g., MS Excel). Example: A = $1000, i = 10%, n = 10. Previous Method:

8 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 8 ENGG 401 X2 – Fundamentals of Engineering Management Spreadsheet Example #1 How much would you pay for a contract that returns $580 per year for 15 years if you have the option of putting your investment into an account that pays 13% interest?

9 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 9 ENGG 401 X2 – Fundamentals of Engineering Management Spreadsheet Example #2 What uniform annual payments would pay back a $25000 loan over 20 years with a 7% interest rate? goal seek –Hint: We could use trial and error, but using the “goal seek” feature in MS Excel will make this much easier.

10 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 10 ENGG 401 X2 – Fundamentals of Engineering Management Problem 7.3(a) from Course Notes You are contemplating the purchase of a company that has a $600,000 “balloon payment” on a new 4-year loan with 8.5% interest. As part of your plan to purchase this company, you are contemplating paying off the debt early (i.e., today) in order to take out a larger long term debt with longer repayment provisions, at the current long term interest rate of 7.5%. Questions: –What payment would be a fair offer to repay the balloon payment, assuming that you are going to refinance the new long term debt from the same financial institution at current rates? –What if the company had an equivalent 4-year loan with uniform annual payments?

11 Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Present Worth Analysis Summer 2008 11 ENGG 401 X2 – Fundamentals of Engineering Management Problem 7.3(a) from Course Notes (2)


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