This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
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.
 Fall '12
 ZHOU,Xiang

Click to edit the document details