Seg2510 Management Principles for Engineering Managers Lecture 5 Present-Worth Analysis Department of Systems Engineering and Engineering Management The Chinese University of Hong Kong 1
Part I Review of the Last Lecture 2
Some Terminologies Consumer Price Index (CPI) Average inflation rate General inflation rate Actual dollars Constant dollars Market interest rate (i) Inflation-free interest rate (i ) Present worth of actual dollars: Deflation method Adjusted-discount method i = i' + f + i' f P n = [(1 + An = (1 + i) An f )(1 + i')] n n 3
Part II Alternatives Analysis 4
Measures of investment worth Screening projects to ensure the right investments are made Financial evaluations of investment proposals Present-worth analysis Annual cash flow analysis Rate-of-return analysis Use equivalence techniques to measure economic worth To compare alternatives on an equal basis To select the wisest alternative from an economic standpoint Note: Assumptions: We will focused on before-tax situations. net cash flows can be viewed as before-tax values or after-tax values Before-tax situation is common for organizations not subjected to tax (e.g. government and nonprofit organizations). Cash flows are in actual dollars. Interest rates are market interest rates. 5
Loan versus Project Cash Flows Loan cash flow interest plus repayment of the principal Project cash flow Future earnings, capital expenditures and annual expenses 6
Initial Project Screening Methods To screen capital investments before progressing to other analysis Whether & when the money invested can be recovered Payback method How long it takes for the net receipts to equal investment outlays Payback period Determined by adding the expected cash flows for each year until the sum is equal to, or greater than, zero. Payback point The cumulative cash flow equals zero» Cash inflows exactly match or pay back the cash outflows Conventional-payback method Ignoring time-value-of-money Discounted-payback method Including time-value-of-money A project does not merit consideration unless its payback period is shorter than some specified period of time. If the payback period is within the acceptable range, a formal project evaluation may begin. 7
Payback Method An Example Find Conventional-payback period 8
Payback Screening Benefits Simplicity Reduces the information search by focusing on that time at which the firm expects to recover the initial investment Reducing the need to make further analysis efforts on those eliminated alternatives Drawbacks Fails to measure profitability It assumes that no profit is made during payback period Fails to recognize the difference between the present and future value of money It ignores differences in the timing of cash flows It does not allow for the possible advantages of a project with a longer economic life It ignores all proceeds after the payback period 9
An Example n Project 1 Project 2 0 -$10,000 -$10,000 1 $1,000 $9,000 2 $9,000 $1,000 3 $1,000 $1,000 Payback period 2 years 2 years 10
Discounted-Payback Period Consider the time value of money Cost of funds (interest) used to support the project Discounted-payback period The number of years required to recover the investment from discounted cash flows 11
An Example Assuming the firm s cost of funds to be 15%, find the discounted-payback period Period Cash Flow Cost of Funds (15%) (unrecovered beginning balance x interest rate) 0 -$85,000 1 $15,000 2 $25,000 3 $35,000 4 $45,000 5 $45,000 Cumulative Cash Flow 6 $35,000 $76,778 12
Present-Worth Analysis 13
Present-Worth Analysis Discounted cash flow techniques (DCFs) Net-present-worth (PW) method PW criterion The present worth of all cash inflows associated with an investment project is compared with the present worth of all cash outflows associated with the project. Net Present Worth (PW) The difference between the present worths Determines if a project is an acceptable investment Select the best project by comparing the PW figures 14
Net-Present-Worth Criterion Evaluation of a Single Project Step 1: Determine the interest rate the firm wishes to earn on its investments The rate at which the firm can always invest the money in its investment pool Required rate of return or Minimum attractive rate of return (MARR) Assume single rate of interest over the life of a project when calculating PW Step 2: Estimate the service life of the project Step 3: Estimate the cash inflow for each period over the service life Step 4: Estimate the cash outflow for each period over the service life Step 5: Determine the net cash flows for each period Net cash flow = cash inflow cash outflow 15
Net-Present-Worth Criterion Evaluation of a Single Project (cont d) Step 6: Find the present worth of each net cash flow at the MARR. The project s PW Sum of the present-worths PW ( i) = A0 (1 + i) 0 + A1 (1 + i) 1 + A2 (1 + i) 2 +... + AN (1 + i) N = N n= 0 An (1 + i) n = N n= 0 A ( P / F, i, n) n Where PW(i) = PW calculated at i, A n = net cash flow at the end of period n, i = MARR (or cost of capital), and n = service life of the project A n will be positive if the period has a net cash inflow, and negative if the period has a net cash outflow. 16
Net-Present-Worth Criterion Evaluation of a Single Project (cont d) Step 7: Decision rule If PW(i) > 0, accept the investment. If PW(i) = 0, remain indifferent. If PW(i) < 0, reject the investment. 17
Net-Present-Worth Criterion Comparing More Than One Alternative Guidelines For selecting the best alternative, select the one with the highest PW (if all have the same service lives) Perform cost-only basis for comparison of mutually exclusive alternatives with the same revenues Accept the project with the smallest PW of costs, or the least negative PW (minimizing costs) 18
An Example Tiger Machine Tool Company is considering the acquisition of a new metal-cutting machine. The cash flows for a three-year project are: What is the economic merit of the acquisition? End of Year 0 1 2 3 Net Cash Flow -$75,000 $24,000 $27,340 $55,760 Answer: $3,553 19
Selecting a MARR The decision to accept or reject an investment is influenced by the choice of MARR, so it is crucial to estimate the MARR correctly. Return To evaluate how investments in financial assets or projects are doing in relation to each other and to the performance of investments in general. Rate of return Function of the following components: Risk-free real return Inflation factor Risk premium(s) 20
Selecting a MARR 21
Net Present Worth Investment-Pool Concept A firm s treasury Funds in this investment pool can earn interest at the MARR Borrowed-Funds Concept 22
Investment-Pool Concept An Example Tiger Machine Tool Company is considering the acquisition of a new metal-cutting machine. The cash flows for a three-year project are: End of Year 0 1 2 3 Net Cash Flow -$75,000 $24,000 $27,340 $55,760 If a firm did not invest in the project & left the $75,000 in the investment pool for 3 years, the fund will have grown to: $ 75,000( F / P,15%,3) = $114,066 23
Investment-Pool Concept An Example If a firm did invest in the project, the cash inflows can return to the investment pool to earn interest at a rate of 15%. The returns after reinvestment are: $24,000( F / P,15%,2) = $32,269 $27,340( F / P,15%,1) = $31,441 $55,760( F / P,15%,0) = $55,760 Total = $119,470 The additional cash accumulation after 3 years from investing in the project is: $ 119,470 $114,066 = $5,404 24
Borrowed-Funds Concept An Example Assume a firm borrows all its capital from a bank at an interest rate of 15%, and invests in the project and uses the proceeds from the investment to pay off the principal & interest on the bank loan. How much is left over at the end of the project period? N 0 1 2 3 Beginning Balance -$75,000 -$61,850 -$43,788 Interest (15%) -$11,250 -$9,278 -$6,568 Payment Received -$75,000 +$24,400 +$27,340 +$55,760 Project Balance -$75,000 -$61,850 -$43,788 $5,404 (net future worth) PW(15%) = $5,404(P/F, 15%, 3)=$3,553 25
Comparing Mutually Exclusive Alternatives Mutually exclusive alternatives: Any one of the alternatives will fulfill the same need The selection of one alternative implies that the others will be excluded A fundamental principle Mutually exclusive alternatives must be compared over an equal time span Case 1 Analysis period equals project lives Case 2 Analysis period differs from project lives 26
Do-nothing alternative If project aims to replace an existing asset or system or is a new endeavor, a do-nothing may alternative exist. If none of the proposals are feasible, then do nothing. For the option of retaining an existing asset or system Generate cash flows of the new proposals relative to those of the do-nothing alternative Find the incremental costs relative to those of the do-nothing alternative For the replacement problem Incremental cash flows Subtracting the do-nothing cash flows from those of each new alternative For new endeavors Incremental cash flows Absolute amounts associated with each alternative (since the do-nothing values are all zero) Note: otherwise stated, it is assumed that doing nothing is not an option. 27
Service Projects versus Revenue Projects Service projects Projects that generate revenues that do not depend on the choice of project, but must produce the same amount of output (revenue). Minimize expenditures PW criterion Choose an alternative with the lower present-value production cost over the service life Revenue projects Projects that generate revenues that depend on the choice of alternative Select the alternative with the largest net gains (output-input) PW criterion Select the one that bring in the largest net present worth 28
Analysis Period Equals Project Lives Compute the PW for each project Select the one with the highest PW 29
An Example Ansell, Inc., a medical-device manufacturer, has two options in addressing the problem of the leaks in the current system. The leaks cause the compressor to run 70% of the time, and hence requires excessive usage of 260 kwh of electricity at a rate of $0.05/kWh (the plant runs 250 days a year, 24 hours a day). Option 1: Continue current operation The compressor s run time will increase by 7% per year for the next 5 years. After 5 years, the system has to be replaced. Option 2: Replace old piping now Cost $28,570 The compressor will run for the same number of days, but run 23% less. The interest rate is 12% compounded annually, is it worth fixing the system now? Answer: Option 2 30
Analysis Period Differs from Project Lives Case 1 Project s life is longer than analysis period Case 2 Analyze each project only for as long as the required service period Unused portion of the equipment Salvage value Salvage value The amount of money for which the equipment could be sold after its service to the project has been rendered or the dollar measure of its remaining usefulness Project s life is shorter than analysis period Consider how, at the end of the project lives, the rest of the required service period can be satisfied Replacement projects may be needed Additional projects to be implemented when the initial project has reached the limits of its useful life 31
Case 1: An Example Waste Management Company (WMC) has won a contract of two years. There are two possible models of machines that WMC could purchase to perform the task. Model A costs $300,000, and will be operable for three years, and at the end of that time, the salvage value will be $50,000. Operating cost: $80,000 per year Model B costs $480,000, and will be operable for 6 years. The estimated salvage at the end of year 6 is $60,000. Operating cost: $45,000 per year After two years, the model A could be sold for $90,000, and model B for $250,000. Assuming the firm s MARR is 15%. Which option is acceptable. Answer: Model A 32
Case 2: An Example The Smith Novelty Company, a mail-order firm, wants to install an automatic mailing system. The firm has a choice between two machines. Model A costs $12,500 and will last 3 years. Model B costs $15,000 and will last 4 years. Neither of the models may be able to handle the Expanded volume of business at the end Of year 5. At MARR=15%, which model should the firm select? Suppose the remaining of required service period 5 can be met by leasing comparable equipment that has an annual lease payment of $5,000, with operating cost of $6,500. N 0 1 2 3 4 Model A -$12,500 -$5,000 -$5,500 -$6,000 +$2,000 Model B -$15,000 -$4,000 -$4,500 -$5,000 -$5,500 +$1,500 Answer: Model B 33
References Chan S. Park, Fundamentals of Engineering Economics. Prentice Hall Lecture 9: Inflation, by Gabriel Fung