IE463 Chapter 3. Objective: INVESTMENT APPRAISAL (Applications of Money-Time Relationships)

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1 IE463 Chapter 3 IVESTMET APPRAISAL (Applications of Money-Time Relationships) Objective: To evaluate the economic profitability and liquidity of a single proposed investment project. CHAPTER 4 2 1

2 Equivalent measures of a project s profitability Present Worth (PW) Future Worth (FW) Annual Worth (AW) Internal Rate of Return (IRR) External Rate of Return (ERR) CHAPTER 4 3 Measures of liquidity Simple Paybac Method (Ø) Discounted Paybac Method (Ø') CHAPTER 4 4 2

3 What are We Measuring? PW, AW, FW, and IRR are based on estimated cash flows and are equivalent measures of profitability. Simple paybac (Ø) and discounted paybac (Ø') are measures of a project s liquidity and are used to supplement profitability measures. CHAPTER 4 5 MIIMUM ATTRACTIVE RATE OF RETUR (MARR) A proposed investment project is profitable if it generates sufficient cash flow to pay bac the initial investment and earn an interest rate greater than or equal to the MARR. CHAPTER 4 6 3

4 The value of the MARR will depend on: availability of funds number and type (essential vs. elective) of investment opportunities perceived ris type of organization CHAPTER 4 7 Example Problem for Project Evaluation Methods Cost/Revenue Estimates Initial Investment: $50,000 Annual Revenues: 20,000 Annual Operating Costs: 2,500 Salvage EOY 5: 10,000 Study Period: 5 years MARR 20% per year CHAPTER 4 8 4

5 Draw a cash flow diagram (an important first step) Study Period = 5 years CHAPTER 4 9 Draw a cash flow diagram (an important first step) Initial Investment = $50,000 CHAPTER

6 Draw a cash flow diagram (an important first step) Annual Revenues = $20,000 CHAPTER 4 11 Draw a cash flow diagram (an important first step) Annual Operating Costs = $2,500 CHAPTER

7 Draw a cash flow diagram (an important first step) Salvage Value = $10,000 CHAPTER 4 13 Present Worth (PW) Method Compute the present equivalent of the estimated cash flows using the MARR as the interest rate. If PW (MARR) 0, then the project is profitable. If PW (MARR) < 0, then the project is not profitable. CHAPTER

8 PW Method PW(20%) = 50,000 +(20,000 2,500)(P/A,20%,5) +10,000(P/F,20%,5) = $6, Since PW(20%) 0, the project is profitable. CHAPTER 4 15 PW = $6, tells us: We have recovered our entire $50,000 investment, We have earned our desired 20% on this investment, We have made a lump sum equivalent profit of $6, beyond what was expected (required). CHAPTER

9 FW Method FW(20%) = 50,000(F/P,20%,5) +(20,000-2,500)(F/A,20%,5) +10,000 = $15,813 Since FW(20%) 0, the project is profitable. CHAPTER 4 17 Annual Worth (AW) Method AW(i%) = R E CR(i%) where R = annual equivalent revenues E = annual equivalent expenses CR = annual equivalent capital recovery cost CHAPTER

10 Capital Recovery Cost CR is the equivalent uniform annual cost of capital invested. CR includes the loss in value of the asset and interest (MARR) on invested capital. CR(i%) = I (A/P, i%, ) S (A/F, i%, ) i% = MARR per interest period (usually years) I = investment, and S = salvage value CHAPTER 4 19 Calculating CR(i%) Assume a uniform loss in asset value over the 5 year Study period. (50,000-10,000)/5 = $8,000 loss in value each year CHAPTER

11 Calculating CR(i%) Year Investment at BOY Loss in Value for Year Interest CR Cost for Year 1 $50,000 $8,000 $10,000 $18, ,000 8,000 8,400 16, ,000 8,000 6,800 14, ,000 8,000 5,200 13, ,000 8,000 3,600 11,600 CHAPTER 4 21 Calculating CR(i%) PW of CR cost = $18,000(P/F,20%,1) +16,400(P/F,20%,2) $11,600(P/F,20%,5) = $45, Uniform Annual Equivalent of CR Cost = $45,980.72(A/P,20%,5) = $15,376 CHAPTER

12 AW Method CR(i%) = I (A/P,i%,) S (A/F,i%,) = $50,000(A/P,20%,5) $10,000(A/F,20%,5) = $16,720 $1,344 = $15,376 AW(20%) = R E CR(20%) = $20,000 - $2,500 - $15,376 = $2,124 Since AW(20%) 0, project is profitable CHAPTER 4 23 Equivalent Worth Methods If PW 0, then FW 0 and AW 0. From our example, PW = $6, therefore, FW = 6,354.50(F/P, 20%,5) = $15,812 and AW = 6,354.50(A/P, 20%,5) = $2125 CHAPTER

13 IRR (Internal Rate of Return) The Internal Rate of Return (IRR) method solves for the interest rate that equates the equivalent worth of a project's cash outflows (expenditures) to the equivalent worth of cash inflows (receipts or savings). CHAPTER 4 25 IRR (Internal Rate of Return) IRR is the interest rate that maes the PW, AW, and FW of a project's estimated cash flows equal to zero. That is, PW(i') of cash inflow = PW(i') of cash outflow. CHAPTER

14 IRR (Internal Rate of Return) We commonly denote the IRR by i'. PW(i' %) = 0 AW(i' %) = 0 FW(i' %) = 0 In general, we must solve for i' by trial and error. CHAPTER 4 27 Evaluating Projects with the IRR Once we now the value of the IRR for a project, we compare it to the MARR to determine whether or not the project is acceptable with respect to profitability. IRR = i' MARR project is acceptable IRR = i' < MARR project is unacceptable (reject) CHAPTER

15 Difficulties with the IRR Method: The IRR Method assumes that recovered funds are reinvested at the IRR rather than the MARR Computational intractability Possible multiple IRRs CHAPTER 4 29 Why should you learn the IRR Method? The majority of companies favor the IRR method for evaluating capital investment projects. CHAPTER

16 IRR method Find i'% such that the PW(i'%) = 0. 0 = $50,000+$17,500(P/A, i'%,5)+$10,000(p/f, i'%,5) PW (20%) = tells us that i' > 20% PW (25%) = > 0, tells us that i'% > 25% PW (30%) = -4, < 0, tells us that i'% < 30% 25% < i' < 30% Use linear interpolation to estimate i'%. CHAPTER 4 31 IRR - Linear Interpolation i% PW (a) (f) (b) i' 0 (g) (c) (h) CHAPTER

17 IRR - Linear Interpolation CHAPTER 4 33 IRR - Linear Interpolation b a f g = c a f h CHAPTER

18 IRR - Linear Interpolation b = a + f f g h ( c a) CHAPTER 4 35 IRR - Linear Interpolation i = 25 + (30 25) 25.3% > MARR ( ) CHAPTER

19 ERR (External Rate of Return) The External Rate of Return (ERR) method solves for the interest rate that equates the present equivalent worth of a project's net cash outflows (expenditures and investment cost) to the future equivalent worth of net cash inflows (receipts and salvage (maret) value). CHAPTER 4 37 ERR (External Rate of Return) In other words, the ERR is the interest rate that maes the PW (MARR%) of costs equivalent to the FW(MARR%) of revenues. PW (ε%) of cash outflow (F/P, i'%, ) = FW(ε%) of cash inflow. ε% = MARR% (unless recovered funds are reinvested at a different interest rate rather than MARR). CHAPTER

20 The ERR method involves three steps to calculate the rate of return: Step 1: All net cash outflows are discounted to time 0 (the present) at ε% per compounding period. = 0 E ( P / F, ε %, ) E = excess of expenditures over receipts in period. = project life or number of periods for the study. ε = external investment rate per period. CHAPTER 4 39 The ERR method involves three steps to calculate the rate of return: Step 2: All net cash inflows are compounded to period (the future) at ε%. = 0 R ( F / P, ε %, ) R = excess of receipts over expenses in period. = project life or number of periods for the study. ε = external investment rate per period. CHAPTER

21 The ERR method involves three steps to calculate the rate of return: Step 3: The external rate of return that establishes equivalence between the two quantities, is determined. E ( P / F, ε %, )( F / P, i %, ) = R ( F / P, ε %, ) R = excess of receipts over expenses in period. E = excess of expenditures over receipts in period. = project life or number of periods for the study. = 0 = 0 ε = external investment rate per period. CHAPTER 4 41 ERR Method 2 = 0 R ( F / P, ε %, ) = 0 E ( P / F, ε %, ) (F / P, i' %, ) 3 CHAPTER

22 ERR Method 2 = 0 R ( F / P, ε %, ) = 0 E ( P / F, ε %, ) (F / P, i' %, ) 3 (F / P, i'%, ) = (1+ i') CHAPTER 4 43 ERR Method 2 = 0 R ( F / P, ε %, ) = 0 E ( P / F, ε %, ) (F / P, i' %, ) 3 E + P / F, ε %, )(1 i ) = = 0 = 0 ( R ( F / P, ε %, ) CHAPTER

23 23 CHAPTER 4 45 ERR Method = = = + F P E P F R i 0 0 ) %,, / ( ) %,, / ( ) (1 ε ε = F P E 0 ) %,, / ( ε = P F R 0 ) %,, / ( ε (F / P, i' %, ) CHAPTER 4 46 ERR Method 1 ) %,, / ( ) %,, / ( 0 0 = = = F P E P F R i ε ε = F P E 0 ) %,, / ( ε = P F R 0 ) %,, / ( ε (F / P, i' %, ) 1 3 2

24 Evaluating Projects with the ERR Once we now the value of the ERR for a project, we compare it to the MARR (not to ε) to determine whether or not the project is acceptable with respect to profitability. ERR = i' MARR project is acceptable ERR = i' < MARR project is unacceptable (reject) CHAPTER 4 47 ERR method Find i'% such that [PW(ε%) of costs](f/p, i'%, ) = [FW(ε%) of revenues] FW(20%) [revenues] = (20,000 2,500)(F/A,20%,5) +10,000 = $140,228 PW(20%) [costs] = $50,000 $50,000(1+ i') = $140,228 CHAPTER

25 ERR method $50,000(1+ i') 5 = $140,228 $140,228 i = 5 1 $50,000 i' = 22.9% > MARR, therefore project is acceptable CHAPTER 4 49 Measures of Liquidity Simple Paybac Period (Ø) how many years it taes to recover the investment (ignoring the time value of money). Discounted Paybac Period (Ø') how many years it taes to recover the investment (including the time value of money). CHAPTER

26 Simple Paybac Period Simple Paybac EOY Cumulative PW (i = 0%) 0 $50, ,500 = $50, , ,000 = 32, , ,500 = 15, , ,000 = 2, , ,500 = 2, , ,000 Paybac Period Ø = 3 years CHAPTER 4 51 Discounted Paybac Period Discounted Paybac EOY Cumulative PW (i = MARR = 20%) 0 $50, ,417 = 50, ,500(P/F, 20%, 1) 2 23,264 = 35, ,500(P/F, 20%, 2) 3 13,137 = 23, ,500(P/F, 20%, 3) 4 4,697 = 13, ,500(P/F, 20%, 4) 5 + 6, = 4, ,500(P/F, 20%, 5) Discounted Paybac Period Ø' = 5 years CHAPTER

27 Investment problem Investment $40,000 Annual Expenses $3,000 Annual Revenues (next year) $10,000 decreasing by $500/year thereafter to Yr. 10 Study Period 5 years Maret Yr. 5 $15,000 MARR 12% CHAPTER 4 53 Evaluate this investment using PW IRR ERR methods. CHAPTER

28 Cash flow diagram of the investment CHAPTER 4 55 Present worth of the investment PW (12%) = (P/A,12%,5) 500(P/G,12%,5) (P/F,12%,5) = $9455 < 0 Reject CHAPTER

29 IRR of the investment 0 = (P/A, i'%,5) 500(P/G, i'%,5)+15000(p/f, i'%,5) CHAPTER 4 57 IRR of the investment CHAPTER

30 IRR of the investment i' = 3 + (553 / 1339) = 3.413% per year, due to linearity assumption real IRR = 3.407% CHAPTER 4 59 ERR of the investment PW(costs) = FW(revenues) = 7000(F/A,ε%,5) 500(A/G, ε%,5)(f/a, ε%,5) If ε% = MARR = 12% ERR = [53,833/40,000] 1/5 1 = 6.12% CHAPTER

31 IRR vs ERR IRR (by linear interpolation) = 3.413% IRR (true) = 3.407% CHAPTER 4 61 BEEFIT / COST RATIO The benefit/cost ratio method involves the calculation of a ratio of benefits to costs. The B/C ratio is defined as the ratio of the equivalent worth of benefits to the equivalent worth of costs. Generally, PW or AW is used as equivalent worth measure in B/C ratio. CHAPTER

32 B/C using PW method: B / C = I + PW ( B) PW ( O & M ) PW(x) = present worth of x B = benefits of the proposed project I = initial investment in the proposed project O&M = operating and maintenance costs of the proposed project. CHAPTER 4 63 B/C using AW method: B / C = AW B ( I ) + ( O & M ) AW(x) = annual worth of x CHAPTER

33 B/C with salvage value B / C = B / C = PW ( B) [ I + PW ( O & M )] PW ( S ) [ AW ( I ) + ( O & M )] AW ( S ) S = Salvage value (maret value) B CHAPTER 4 65 EVALUATIG PROJECT WITH B / C RATIO METHOD The resulting B/C ratios for all above formulations should give the results of either B/C > 1.0 or B/C = 1.0 to find a project acceptable. If B/C 1.0, then the project is profitable If B/C < 1.0, then the project is not profitable. CHAPTER

34 B / C RATIO EXAMPLE A Estimated cash flows of the investment: Construction cost = $850,000 Purchasing cost of land = $350,000 Investment cost = $1,200,000 Annual maintenance cost = $26,500 Annual operating cost = $75,000 Monthly salary = $8,000x12 Annual expenses = $197,500 Annual revenues = $490,000 CHAPTER 4 67 B / C RATIO EXAMPLE A Apply the B/C ratio method with a study period of 20 years for MARR of 10% to determine whether the project will be implemented or not without considering the salvage value. CHAPTER

35 B/C solution using PW = 1,200, ,500(P/A, 10%, 20) = $2,881,436 PW(benefit)@10% = 490,000(P/A, 10%, 20) = $4,171,664 B/C = (4,171,664 / 2,881,436) = 1.45 > 1.0 the project is profitable. CHAPTER 4 69 B/C solution using AW AW(cost)@10% = 1,200,000(A/P, 10%, 20) + 197,500 = $338,450 AW(benefit)@10% = $490,000 B/C = (490,000 / 338,450) = 1.45 > 1.0 the project is profitable. CHAPTER

36 B / C RATIO EXAMPLE B Suppose investment cost is $15000 of a project and it is estimated that this project earns 27% of its initial investment EOY for 10 years. When the company expects 20% rate of return on this investment and desires to have a B/C ratio equal to 1.2, what should salvage value be? CHAPTER 4 71 Cash flow diagram Annual revenue = (0.27) = $4050 S = Salvage Value CHAPTER

37 B/C using PW B/C = 1.2 PW(costs) = S(P/F, 20%, 10) = $15000 (0.1615)S PW (revenue) = 4050(P/A, 20%, 10) = $16,980 CHAPTER 4 73 Solve for S: B $16,980 / C = = 1.2 $15,000 (0.1615) S 1.2( S) = 16,980 S = ( ) / = $5,263 CHAPTER