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Rate of return calculations usually involve trial and error solutions Equations can be written in terms of P, F, or A Example: An investment of $20,000.

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Presentation on theme: "Rate of return calculations usually involve trial and error solutions Equations can be written in terms of P, F, or A Example: An investment of $20,000."— Presentation transcript:

1 Rate of return calculations usually involve trial and error solutions Equations can be written in terms of P, F, or A Example: An investment of $20,000 in a certain machine will generate income of $7000 per year for 3 years, at which time the machine can be sold for $8000. If the company’s MARR is 15% per year, should it buy the machine? Solution: The ROR equation is: 0 = -20,000 + 7000(P/A,i,3) + 8000(P/F,i,3) Solve for i by trial and error or Excel:i = 18.2% per year Thus, since i >MARR, the investment should be made

2 Multiple values of i may exist when there is more than one sign change in net cash flow(CF). Such cash flow is called non-conventional. Two rules apply to multiple i values: Descarte’s rule states that the number of real i values is less than or equal to the number of sign changes in net cash flow Norstrom’s criterion states that if the cumulative cash flow starts off negatively and has only one sign change, there is only one i value

3 Determine the maximum number of i values for the cash flow below Year Expense Income 0 -12,000 - 1 -5,000 + 3,000 2 -6,000 +9,000 3 -7,000 +15,000 4 -8,000 +16,000 5 -9,000 +8,000 Solution: Therefore, there is only one i value( i = 4.7%) Net cash flow -12,000 -2,000 +3,000 +8,000 -1,000 +8,000Cumulative CF -12,000 -14,000 -11,000 -3,000 +5,000 +4,000 The cumulative cash flow begins negatively with one sign change The sign on the net cash flow changes twice, indicating two possible i values

4 Bonds have conventional cash flows(1 sign change). Therefore, have one i value Example: A $10,000 bond with 6% interest payable quarterly is for sale for $8000. If the bond matures in 5 years, what is the ROR (a) per quarter (b) per year Solution: (a) I = 10,000(0.06) 4 = $150 per quarter The rate of return equation is: 0 = -8000 + 150(P/A,i,20) + 10,000(P/F,i,20) By trial and error or Excel, i = 2.8% per quarter (b) Nominal i per year = 2.8(4) = 11.2% per year Effective i per year = (1 + 0.11.2/4) 4 – 1 = 11.7% per year

5 For a net cash flow sequence which has the signs shown below, the maximum number of i* values that may satisfy the rate of return equation is (A) 1(B) 2(C) 3(D) 4 Signs for net cash flow sequence: + + - + + + + + - Answer: 3

6 On the basis of Descarte’s rule, the number of possible i* values for the cash flow shown below is: Year 123456 Net Cash Flow+200-300-10,000+15,000-50+800 Answer = 4

7 Write the general rate of return equation. Answer: 0 = -P ± A(P/A, i, n) ± F(P/F, i, n) Determine the rate of return amount for a $10,000 bond with 6% interest per year that is for sale at $6,000 and matures in 10 years with a rate of return of 10% per year. Rate of Return Problems I = 10000(.06)/1 = 600 6.1446.3855 0 = -6000 + 600(P/A, 10%, 10) + 10000(P/F, 10%, 10) = 1541.76 Answer: PW = 1541.76

8 An investment of $30,000 resulted in profits of $10,000 per year for 6 years and $20,000 in year 7. Determine the rate of return amount if the rate of return is 10% per year Rate of Return Problems Answer: 0 = -30000 + 10000(P/A, 10%, 6) + 20000(P/F, 10%, 7) = $ 23,817.00 For a net cash flow sequence which has the signs shown below, the maximum number of i* values that may satisfy the rate of return equation is: Signs for net cash flow sequence: - + + + + + + + - For a net cash flow sequence which has the signs shown below, the maximum number of i* values that may satisfy the rate of return equation is: Signs for net cash flow sequence: - + + + + + + + - For a net cash flow sequence which has the signs shown below, the maximum number of i* values that may satisfy the rate of return equation is: Signs for net cash flow sequence -++++++++- Answer: 2


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