An example on the PW, FW, AW, IRR, and B/C



An example on the PW, FW, AW, IRR, and B/C

Prepared by: Parisay,   Date: Summer of 2003 

An expansion to a business is under study.  Four feasible alternatives are found with different cash flow during 10 years of the useful life of such expansion.   The initial investment and estimated revenue for all four alternatives (I, J, K, and L) are listed in the following table.  All numbers are in thousands.

1)  Assuming that MARR=12%, rank these alternatives using a) present worth method, b) future worth method, c) annual worth method.  Analyze the result obtained using different methods of ranking.

2)  Assuming that MARR=5%, rank these alternatives using present worth method.  Compare this ranking with the one in part (1).

3)  Find IRR of each alternative.  Rank alternatives using IRR method (MARR=5%). Compare this ranking with the one in part (2).

4)  Rank the alternatives using Benefit-Cost Ratio method (MARR=5%).  Compare this ranking with the one in part (3). 

Cash flow for four alternatives I, J, K, and L

|  |year |      I |     J |    K |      L |

|  |0 |-1100 |-1000 |-1000 |-900 |

|  |1 |180 |0 |260 |0 |

|  |2 |175 |0 |260 |0 |

|  |3 |170 |0 |260 |0 |

|  |4 |165 |0 |260 |0 |

|  |5 |160 |0 |260 |0 |

|  |6 |155 |370 |0 |0 |

|  |7 |150 |370 |0 |0 |

|  |8 |145 |370 |0 |0 |

|  |9 |140 |370 |0 |0 |

|  |10 |135 |370 |0 |1375 |

|  |  |  |  |  |  |

|total,  i=0 |  |475 |850 |300 |475 |

 

 

 

Solution:

Question 1:  i = 12%

PW(I) = -1100 +180(P|A, 12%, 10) -5(P|G, 12%, 10) = -184.27

PW(J) = -1000 +370(P|A, 12%, 5)(P|F, 12%, 5) = -243

PW(K) = -1000 +260(P|A, 12%, 5) = -62.7

PW(L) = -900 +1375(P|F, 12%, 10) = -457.25

 

FW(I) = PW(I)*(F|P, 12%, 10) = -572.34

FW(J) = -1000(F|P, 12%, 10), 370(F|A, 12%, 5) = -755.39

FW(K) = PW(K)*(F|P, 12%, 10) = -194.75

FW(L) = -900(F|P, 12%, 10) + 1375 = -1420.4

 

AW(I) =  -1100(A|P, 12%, 10) +180 -5(A|G, 12%, 10) = -32.63

AW(J) = PW(J)*(A|P, 12%, 10) = -43.01

AW(K) = PW(K)*(A|P, 12%, 10) = -11.1

AW(L) =  -900(A|P, 12%, 10) +1375(A|F, 12%, 10) = -80.93

CONCLUSION:

Question 2:  i = 5%

PW(I) = -1100 +180(P|A, 5%, 10) -5(P|G, 5%, 10) = 131.7

PW(J) = -1000 +370(P|A, 5%, 5)(P|F, 5%, 5) = 254.96

PW(K) = -1000 +260(P|A, 5%, 5) = 125.54

PW(L) = -900 +1375(P|F, 5%, 10) = -55.89

CONCLUSION:

 

Question 3: 

For alternative I:

If:  i1 = 12%     Then: PW(I)1 = -184.27;   If i2 = 5%        Then: PW(I)2 = 131.7

PW(I) 1 - PW(I)2                PW(I) 1 – 0

---------------------  =   -------------------            conclusion:  i = 7.92%= IRR(I)

       i1    -  i2                      i1    -  i     

 [pic]

For alternative L:

If:  i1 = 12%     Then: PW(L)1 = -457.25;   If i2 = 5%       Then: PW(L)2 = -55.89

PW(L) 1 - PW(L)2                         i1    -  i2                      

  ---------------------  =   -------------------            conclusion:  i = 4.02% = IRR(L)

   PW(L) 1 – 0                          i1    -  i      

[pic] 

For alternative J:

-243 -254.96               -243 -0

-----------------    =     -------------          conclusion:  i = 8.58% = IRR(J)

12 -5                               12 -  i     

 

For alternative K:

-62.7 -125.54               -62.7 -0

-----------------    =     -------------          conclusion:  i = 9.67% = IRR(K)

12 -5                               12 -  i     

 CONCLUSION:

 

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