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# economy Chapter5_by louy Al hami

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### economy Chapter5_by louy Al hami

1. 1. Evaluating a Single project 1
2. 2. IntroductionIn this chapter we will answer the following question : Whether a proposed capital and its expenditures can recovered by revenue over time in addition to a return on the capital sufficiently attractive in view of the risks . 2
3. 3. Methods of Evaluating theEconomic Profitability of aProblem Solution Present Worth ( PW ) Future Worth ( FW ) Annual Worth ( AW ) Internal Rate of Return ( IRR ) External Rate of Return ( ERR ) 3
4. 4. Assumptions1. We know the future with certainty2. We can borrow and lend with the same i%. 4
5. 5. The most-used method is thepresent worth method. The present worth (PW) is found by discounting all cash inflows and outflows to the present time at an interest rate that is generally the MARR. A positive PW for an investment project means that the project is acceptable (it satisfies the MARR). 5
6. 6. Present Worth ( PW ) Eq..(5-1) =F0(1+i)0+F1(1+i)-1+F2(1+i)-2+………..+FN(1+i)-N i = effective interest rate or MARR K = index of each compounding period Fk = future cash flow at the end of period k N = # of compounded periods . 6
7. 7.  PW Decision Rule:If PW (i=MARR) ≥ 0 , the project is economically justified 7
8. 8. Note The higher the interest rate and the further into the future a cash flow occurs , the lower its PW is See figure 5-2 PW of 1,000 10 years from now i =5% is \$613.90 PW of 1,000 10years from now i=10% is \$385.5 8
9. 9. Figure 5-2 9
10. 10. Example(PW)Consider a project that has an initialinvestment of \$50,000 and that returns\$18,000 per year for the next four years.If the MARR is 12%, is this a goodinvestment? PW = -50,000 + 18,000 (P/A, 12%, 4) PW = -50,000 + 18,000 (3.0373)PW = \$4,671.40  This is a good investment! 10
11. 11.  Example 5.1: A piece of new equipment has been proposed by engineers to increase the productivity of a certain manual welding operation. The investment cost \$25,000 and the equipment will have a market value of \$5,000 at the end of a study period of five years. Increased productivity attributable to the equipment will amount to \$8,000 per year after extra operating costs have been subtracted from the revenue generated by the additional production. A cash flow diagram for this investment opportunity is given below. If the firm’s MARR is 20% per year, is this proposal a sound one? Use the PW method. 11
12. 12.  Example. 5-1 PW = PW(inflows) – PW (outflows) PW= \$8,000(P/A,20%,5) +\$5,000(P/F,20%,5)-\$25,000 PW= \$934,29 12
13. 13. Bond Value The value of the bond at any time is the PW of future cash receipts Two types of payments1. Series of periodic interest payments (rZ)2. A single payment = C ,When the bond is sold or retired . 13
14. 14. VN = C(P/F, i%, N) + rZ (P/A, i%,N) Eq.5-2  Z= Face ,or parValue,  C = redemption or disposal price (usually = Z).  r = bond rate (nominal interest) per interest period  N= # of periods before redemption  i = bond yield rate per period  VN = PW 14
15. 15. 15
16. 16. Example5-4 16
17. 17. The Capitalized-Worth Method  A special variation of the PW  To find PW for infinite length of time  N ∞ CW = PWN  ∞ = A ( P / A, i%, ∞ ) ( 1+i )N - 1 = A lim ------------------ =A ( 1 / i )=A/i N ∞ i ( 1 + i )N 17
18. 18. Ex. 5-5 page 219 18
19. 19. (a)As we discussed before when N= ∞ (p/A,i%,N)=1/i, then (p/A,8%,N)=1/0.08 = 12.5go to App.C then N= 100 this assumed as (∞ ) 19
20. 20. Ex. 5-5 page 219 (b) 30,000 20,000 A / F ,8%,4) ( CW  (100000 A / P,8%, )) / .08 , ( 0.08  30,000  20,000 ( A / F ,8%,4)  CW  100,000     \$530,475  0.08  20
21. 21. Future Worth ( FW ) Equivalent worth of all cash flows (In& Out) at the end of the planning horizon at MARR Eq.(5-3) i = effective interest rate or MARR K = index of each compounding period Fk = future cash flow at the end of period k N = # of compounded periods . 21
22. 22.  FW Decision Rule:If FW (i=MARR) ≥ 0 , the project is economically justified 22
23. 23.  Ex 5-6 page 221 23
24. 24. Example5-7 24
25. 25. The Annual Worth Method Annual worth is an equal periodic series of dollar amounts that is equivalent to the cash inflows and outflows, at an interest rate that is generally the MARR. The AW of a project is annual equivalent revenue or savings minus annual equivalent expenses, less its annual capital recovery (CR) amount. AW(i%) = R – E – CR(i%) R = Annual equivalent Revenue or saving E = Annual equivalent Expenses CR = Annual equivalent Capital Recovery amount 25
26. 26.  AW Decision Rule:If AW (i=MARR) ≥ 0 , the project is economically justified 26
27. 27. CR (Capital Recovery)  CR is the equivalent uniform annual cost of the capital invested that covers 1. Loss of value of the asset. 2. Interest on invested capital (at MARR) 27
28. 28. CR CR (i%) = I(A/P, i%, N) – S(A/F, i%,N) Eq. 5-5 I = initial investment for project S = salvage (market) value at the end of the study period N = project study period 28
29. 29.  CR (i%) = (I – S)(A/F, i%, N) +I(i%) CR (i%) = (I –S)(A/P, i%, N) + S(i%) 29
30. 30. Example 5-8 page 224 30
31. 31. Example 5-9 31
32. 32. Example 5-10 32
33. 33. Example 5-11 33
34. 34. A project requires an initial investment of\$45,000, has a salvage value of \$12,000 aftersix years, incurs annual expenses of \$6,000,and provides an annual revenue of \$18,000.Using a MARR of 10%, determine the AW ofthis project. Since the AW is positive, it’s a good investment. 34
35. 35. Internal Rate Of Return Method ( IRR ) IRR solves for the interest rate that equates the equivalent worth of an alternative’s cash inflows (receipts or savings) to the equivalent worth of cash outflows (expenditures) IRR is positive for a single alternative only if:  both receipts and expenses are present in the cash flow diagram  the sum of inflows exceeds the sum of outflows 35
36. 36. Internal Rate of Return It is also called the investor’s method, the discounted cash flow method, and the profitability index. If the IRR for a project is greater than the MARR, then the project is acceptable. 36
37. 37. INTERNAL RATE OF RETURN METHOD ( IRR )  IRR is i’ %, using the following PW formula: R k = net revenues or savings for the kth year E k = net expenditures including investment costs for the kth year N = project life ( or study period ) Note: FW or AW can be used instead of PW 37
38. 38. Solving for the IRR is a bit morecomplicated than PW, FW, or AW The method of solving for the i% that equates revenues and expenses normally involves trial-and- error calculations, or solving numerically using mathematical software. 38
39. 39. 39
40. 40. 40
41. 41. Installment Financing An application of IRR method Ex.5-14 will be discussed 41
42. 42. 42
43. 43. 43
44. 44. 44
45. 45. 45
46. 46. 46
47. 47. 47
48. 48. Challenges in Applying the IRRMethod. It is computationally difficult without proper tools. In rare instances multiple rates of return can be found(take a look to page 255 example 5-A-1). The IRR method must be carefully applied and interpreted when comparing two more mutually exclusive alternatives (e.g., do not directly compare internal rates of return). 48
49. 49. The External Rate Of Return Method ( ERR ) ERR directly takes into account the interest rate(  ) external to a project at which net cash flows generated over the project life can be reinvested (or borrowed ). If the external reinvestment rate, usually the firm’s MARR, equals the IRR, then ERR method produces same results as IRR method 49
50. 50. Calculating External Rate of Return ( ERR )1. All net cash outflows are discounted to the present (time 0) at ε% per compounding period.2. All net cash inflows are discounted to period N at ε %.3. Solve for the ERR, the interest rate that establishes equivalence between the two quantities. The absolute value of the present equivalent worth of the net cash outflows at ε % is used in step 3. A project is acceptable when i ‘ % of the ERR method is greater than or equal to the firm’s MARR 50
51. 51. ERR is the i% at which where Rk = excess of receipts over expenses in period k, Ek = excess of expenses over receipts in period k, N = project life or number of periods, and ε = external reinvestment rate per period. 51
52. 52. Calculating External Rate of Return ( ERR ) 2 N  R ( F / P,  %, N - k ) k = 0k 0 1 Time N N E ( P / F,  %, k ) k=0 k 52
53. 53. Calculating External Rate of Return ( ERR ) 2 3 N  R ( F / P,  %, N - k ) k = 0k 0 i ‘ %= ? 1 Time N N  Ek ( P / F,  %, k ) ( F / P, i ‘ %, N ) k=0 53
54. 54. Example 5-17  \$25,000 (F/P,i’%,5)=\$8,000(F/A,20%,5)+\$5,000  (F/P,i’%,5)= \$64,532.80  2.5813  (1  i )5 \$25,000  i’ =20.88% 54
55. 55. Example 5-18 55
56. 56. ExampleFor the cash flows given below, find the ERR when theexternal reinvestment rate is ε = 12% (equal to the MARR).Year 0 1 2 3 4Cash Flow -\$15,000 -\$7,000 \$10,000 \$10,000 \$10,000ExpensesRevenueSolving, we find 56
57. 57. The payback period method  The simple payback period is the number of years required for cash inflows to just equal cash outflows.  It is a measure of liquidity rather than a measure of profitability. 57
58. 58. Payback is simple to calculate. The payback period is the smallest value of θ (θ ≤ N) for which the relationship below is satisfied.For discounted payback future cash flows are discountedback to the present, so the relationship to satisfy becomes 58
59. 59.  A low- valued pay back period is considered desirable 59
60. 60. Finding the simple End End of Net Cumulat Present Cumulati of Cash ive PW worth at ve PW atand discounted Year Year Flow at 0% i = 6% 6%payback period for aset of cash flows. 0 0 -\$42,000 -\$42,000 -\$42,000 -\$42,000The cumulative cash 1 1 \$12,000 -\$30,000 11,320.8 -\$30,679flows in the table werecalculated using theformulas for simple 2 2 \$11,000 -\$19,000 9,790 -\$20,889and discountedpayback. 3 3 \$10,000 -\$9,000 8,396 -\$12,493From the calculations 4 4 \$10,000 \$1,000 7,921 -\$4,572θ = 4 years and θ = 5years. 5 5 \$9,000 6,725.7 \$2,153 60
61. 61. Col.3:EOF1= -42,000-12,000=-30,000 End of Net Cash Cumulative Cumulative End Net Cumulat Present Cumulati Year Cash of Flow ive PW at 0% at PW at 6% PW worth ve PW atEOF2 : 11,000-30,000=-19,000 Year Flow at 0% i = 6% 6%EOF3 : 10,000-19,000=-9,000 00 -\$42,000 -\$42,000 -\$42,000 -\$42,000 -\$42,000 -\$42,000 -\$42,000EOF4 : 10,000-9,000=1,000>0θ=4Col.4: 11 \$12,000 -\$30,000 11,320.8 -\$30,679 \$12,000 -\$30,000 -\$30,67912,000 (p/F,6%,1) =12,000* 0.9434= 113,208 22 \$11,000 -\$19,000 \$11,000 -\$19,000 9,790 -\$20,889 -\$20,88911,000 (p/F,6%,2) = 11,000*0.8900=9,790 33 \$10,000 \$10,000 -\$9,000-\$9,000 8,396 -\$12,493 -\$12,49310,000*(P/F,6%,3) =10,000*0.8396= 8,396 44 \$10,000\$1,000 \$1,000 \$10,000 7,921 -\$4,572 -\$4,57210,000*(P/F,6%,4) =10,000*0.7921 =7,9219,000*(P/F,6%,5) = 55 \$9,000 \$9,000 6,725.7 \$2,153 \$2,1539,000*0.7473 = 6,725.7 61
62. 62. End Net Cumulat Present CumulatiCol.5: of Cash ive PW worth at ve PW at Year Flow at 0% i = 6% 6%EOF1: -42,000 + 11,320.8 = -30,679.2 0 -\$42,000 -\$42,000 -\$42,000 -\$42,000EOF2 : 1 \$12,000 -\$30,000 11,320.8 -\$30,6799,790 -30,679.2 = -20,889.2EOF3 : 2 \$11,000 -\$19,000 9,790 -\$20,8898,396 -20,889.2 = -12,493.2EOF4 : 3 \$10,000 -\$9,000 8,396 -\$12,4937,921-12,493.2 = -4,572.2 <0EOF5 : 4 \$10,000 \$1,000 7,921 -\$4,5726,725.7 -4,572.2 = 2,153.5 >0 5 \$9,000 6,725.7 \$2,153 62
63. 63. Problems with the payback period method. It doesn’t reflect any cash flows occurring after θ, or θ. It doesn’t indicate anything about project desirability except the speed with which the initial investment is recovered. Recommendation: use the payback period only as supplemental information in conjunction with one or more of the other methods in this chapter. 63
64. 64. Suggested Problems 4, 10, 16, 21, 25, 36, 46, 49, 55, 60 64
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