i* = IRR i*? IRR more sign changes Passes: unique i* = IRR

Transcription

1 Decision Rules Single Alternative Based on Sign Changes of Cash Flow: Simple Investment i* = IRR Accept if i* > MARR Single Project start with zero, one sign change Non-Simple Investment i*? IRR Net Investment Test: Based on Signs of PB (all need to be negative) more sign changes Passes: unique i* = IRR Fails: Compute an unique True i* using external rate Accept if TRUE i* > MARR Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 1 / 17

2 Decision Rules Mutually Exclusive Alternatives Based on Sign Changes of Incremental Cash Flow Accept High Investment if i* B - A > MARR Mutually Exclusive Alternatives Incremental Analysis Simple Investment start with negative, one sign change Non-Simple Investment i* B - A = IRR i* B - A? IRR Accept Low Investment otherwise Net Investment Test: Based on Signs of PB (all need to be negative) more sign changes Passes: unique i* B - A = IRR Fails: Compute an unique True i* B - A using external rate Accept Low Investment Otherwise Accept High Investment if TRUE i* > MARR Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 2 / 17

3 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows Example The Flyer Aircraft Company has an opportunity to supply a large airplane to Interair. Interair will pay $19 million when the contract is signed and $10 million one year later. Flyer estimates its cost to be $50 million each for the years two and three when the aircraft is being produced. Interair will take delivery of the airplane during year 4, and agrees to pay $20 million at the end of that year and $60 million at the end of year 5. Compute the rate of return on this project i*=? Years Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 3 / 17

4 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows 1 Two sign changes, maximum of two i values. 2 PW = (P/F, i%, 1) 50(P/F, i%, 2) 50(P/F, i%, 3) + 20(P/F, i%, 4) + 60(P/F, i%, 5) = i 50 (1 + i) 2 50 (1 + i) (1 + i) (1 + i) 5 i% PW i1 = (10) = 10.1% i2 = ( 1.2) (10) = 47% 0.5 ( 1.2) Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 4 / 17

5 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows Need for an External Interest Rate Multiple i exists, this means that this investment is a mixed investment, and surely fails the net investment test. Therefore, we need to use an external rate to calculate an unique True IRR. Assume an external interest rate of 6%: For r=6% Year Cash Flow 0 $19 is invested for 2 years at external investment at 6% F = 19(F/P, 6%, 2) = $ $10 is invested for 1 years at external investment at 6% F = 10(F/P, 6%, 1) = $ $ = -18 (return to internal investment) 3 $-50 4 $20 5 $60 Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 5 / 17

6 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows For r=6% Year Cash Flow $-18 3 $-50 4 $20 5 $60 Now, there is one sign change, so we can solve for the internal value i. PW = 18(P/F, i%, 2) 50(P/F, i%, 3) + 20(P/F, i%, 4) + 60(P/F, i%, 5) = 18 (1 + i) 2 50 (1 + i) (1 + i) (1 + i) 5 i% PW Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 6 / 17

7 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows For r=6% Year Cash Flow $-18 3 $-50 4 $20 5 $60 i% PW i = (2) = 8.4% Assuming an external interest rate of 6%, ROR is found to be 8.4%. Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 7 / 17

8 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows Resolving Multiple Rate of Return with Initial Positive Cash Flows 1 Alter cash flow through the use of an external rate to reduce the number of sign changes to one and to eliminate the initial positive cash flows 2 Solve the ROR equation Formally, Set up the relation: PB t = PB t 1 (1 + i) + A t where t = 1, 2,...,n, n is the total years in the project, A t is the net cash{ flow in year t r if PBt 1 > 0 i = i otherwise. where r is the external investment rate. Set PB n = 0 and solve for i. Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 8 / 17

9 Resolution of Multiple Rates of Return The Special Case with Initial Positive Net Cash Flows For the last example, we have r = 6%, n = 5 and the following cash flow stream: For r=6% Year Cash Flow 0 $19 1 $10 2 $-50 3 $-50 4 $20 5 $60 PB 0 = 19, PB 1 = 19(F/P, 6%, 1) + 10 = 30.14, PB 2 = 30.14(F/P, 6%, 1) 50 = 18, PB 3 = 18(F/P, i %, 1) 50, PB 4 = [ 18(F/P, i %, 1) 50](F/P, i, 1) + 20 PB 5 = ( [ 18(F/P, i %, 1) 50](F/P, i, 1) + 20 ) (F/P, i, 1) = 18(1 + i ) 3 50(1 + i ) (1 + i ) + 60 i = 8.4% will satisfy the equation. (according to trial and error) Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 9 / 17

10 Incremental Analysis for Mutually Exclusive Alternatives Extras ROR (High Investment) < ROR (Low Investment) PW of Receipts Y X PW of Disbursement Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 10 / 17

11 Incremental Analysis for Mutually Exclusive Alternatives Extras ROR (High Investment) < ROR (Low Investment) ROR (Incremental) < ROR (High Investment) PW of Receipts Y X PW of Disbursement Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 11 / 17

12 Incremental Analysis for Mutually Exclusive Alternatives Extras ROR (High Investment) > ROR (Low Investment) PW of Receipts Y X PW of Disbursement Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 12 / 17

13 Incremental Analysis for Mutually Exclusive Alternatives Extras ROR (High Investment) > ROR (Low Investment) ROR (Incremental) > ROR (High Investment) PW of Receipts Y X PW of Disbursement Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 13 / 17

14 Incremental Analysis for Mutually Exclusive Alternatives Incremental Rate of Return Evaluation Using AW Incremental ROR using AW Comparing alternatives by ROR analysis method always leads to the same selection as PW, AW, FW analysis method, whether ROR is determined using PW-based, AW-based or FW-based. All methods must consider equal service principle For AW-based ROR method there are two ways to perform the evaluation for i for the incremental cash flow series: 1 The incremental cash flows must be evaluated over the LCM of lives (results in i ) 2 Set the difference between AW relations for each alternative to zero 0 = AW B AW A (resulting i = i ) This way requires less computational effort Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 14 / 17

15 Incremental Analysis for Mutually Exclusive Alternatives Incremental Rate of Return Evaluation Using AW Example Our company has the option of supplying an equipment from two different suppliers - a U.S. firm (A) and an Asian firm (B). Estimates of this equipment from vendors A and B are given below. Decide which vendor should be selected using AW-based incremental ROR method and MARR of 15%. (external investment rate is 12.65%) A B Initial cost, $ -8,000-13,000 Annual costs, $ -3,500-1,600 Salvage Value, $ 0 2,000 Life, years 10 5 Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 15 / 17

16 Incremental Analysis for Mutually Exclusive Alternatives Incremental Rate of Return Evaluation Using AW Example Year Cash Flow Cash Flow Incremental Cash Flow A B (B-A) 0-8,000-13,000-5, ,500-1,600 1, ,000-11,000-13, ,500-1,600 1, ,000 2,000 1 Using the incremental cash flows: it must be evaluated over 10 years 2 Set the difference between AW relations for each alternative to zero using 5 and 10 years 0 = AW B AW A Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 16 / 17

17 Incremental Analysis for Mutually Exclusive Alternatives Incremental Rate of Return Evaluation Using AW Example 1 i = 12.65% is found from the following equation using LCM years and incremental cash flows: 0 = 5, 000(A/P, i, 10) + 1, , 000(A/F, i, 10) 11, 000(P/F, i, 5)(A/P, i, 10) 2 Same i = 12.65% can be also found from: 0 = AW B AW A = 13, 000(A/P, i, 5) 1, , 000(A/F, i, 5) [ 8, 000(A/P, i, 10) 3, 500] Since i = 12.65% < 15%, the additional $5,000 of investment is not justified, and alternative A is chosen. Dr.Serhan Duran (METU) IE 347 Week 8 Industrial Engineering Dept. 17 / 17