Lecture6

# Cost alternatives the rst alternative in the ordered

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Unformatted text preview: t acceptable alternative in the ordered list. Cost alternatives: the ﬁrst alternative in the ordered list (the one with the least capital investment). Step 3. Evaluate the difference between the next alternative and the current best alternative. If the incremental cash ﬂow is acceptable, choose the next alternative as the current best alternative. Otherwise, keep the current best alternative and drop the next alternative from further consideration. Step 4. Repeat Step 3 until the last alternative is considered. Select the current best alternative as the best one. SEEM5740/ECLT5930 18 Incremental analysis by ERR If the IRR of the incremental cash ﬂow stream does not exist or is not unique, then what should we do? We can apply the ERR method to the incremental cash ﬂow. Accept the incremental cash ﬂow if ERR ≥ MARR. In ERR analysis we usually set = MARR (the result is consistent with PW analysis from Chapter 5) SEEM5740/ECLT5930 19 Case 2: Unequal useful lives Two types of assumptions: Repeatability: We can replace an alternative when it ends by the same one. Under repeatability, the study period is either equal to a common multiple of the lives of the alternatives or indeﬁnitely long. Coterminated: A ﬁnite and identical study period for all alternatives. SEEM5740/ECLT5930 20 Repeatable projects Example 6.5: MARR=10%. Investment Annual revenues Useful life (years) MV at end of useful life A 3,500 1,255 4 0 B 5,000 1,480 6 0 Let’s suppose the repeatability assumption holds. A study period of 12 years (the least common multiple (LCM) of 4 and 6) where Alternative A repeats itself 3 times and Alternative B repeats itself twice during the entire period. SEEM5740/ECLT5930 21 1255 0 4 3500 3500 8 12 12 3500 3 project A over 12 years SEEM5740/ECLT5930 22 1480 0 6 5000 12 12 5000 2 project B over 12 years SEEM5740/ECLT5930 23 PW and AW analysis: Example 6.5 PW (10%)A = −3500 − 3500[(P/F , 10%, 4) + (P/F , 10%, 8)] + 1255(P/A, 10%, 12) = 1028. PW (10%)B = −5000−5000(P/F , 10%, 6)+1480(P/A, 10%, 12) = 2262. Select B. Let’s ﬁnd the AW(MARR) of each alternative using the PW(MARR): AW(10%)A =102...
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## This note was uploaded on 03/23/2013 for the course SEEM 5740 taught by Professor Zhou,xiang during the Fall '12 term at CUHK.

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