a) Here, i+1 = 1+0.08 = 1.08
n | Project Balance of A | Project Balance of D |
0 | -3000 | -6,000 |
1 | -3000(1.08)1+750 = -2490 | -6000(1.08)1-600 = -7080 |
2 |
-2490(1.08)1 + 750 = -1939 |
-7080(1.08)1-600 = -8246 |
3 | -1939(1.08)1+750 = -1344 | -8246(1.08)1+5000 = -3906 |
4 |
-1344(1.08)1+700 = -752 |
-3906(1.08)1+4000 = 218 |
5 |
-702(1.08)1+700 = -58 |
218(1.08)1+4000 = 4235 |
6 | -58(1.08)1+700 = 637 | 4235(1.08)1+3000 = 7574 |
7 | 637(1.08)1+350 = 1038 | 7574(1.08)1+4000 = 12,180 |
8 | 1038(1.08)1+350 = 1471 |
b) Future worth of project A:
FW = -3000(F/P,8%,8) +750(P/A,8%,3)(F/P,8%,8) + 700(P/A,8%,3)(F/P,8%,8) + 350(F/A,8%,2)
= -3000 * 1.851 +750 * 2.577 * 1.851 + 700 * 2.577 * 1.851 +350 * 2.080
= -5553 + 3577.5 + 3339 + 728
FW of A = 2091.5 = $2092
Future worth of Project D:
FW = -6,000(F/P,8%,7) -600(P/A,8%,2)(F/P,8%,7) +5000(P/A,8%,1)(F/P,8%,7)+4000(P/A,8%,2)(F/P,8%,7) +3000(P/A,8%,1)(F/P,8%,7) +4000(P/A,8%,1)(F/P,8%,7)
= -6000*1.714 -600*1.783*1.714 + 5000*0.926*1.714 + 4000 *1.783*1.714 + 3000*0.926*1.714 +4000*0.926*1.714
= -10,284 - 1833.6 + 7935.8 + 12224.2 + 4761.4 + 6348.6
FW of D = 19,152.4 = $19,152
c) PW of project B:
-7,300 -3000(P/A,8%,1)-2500(P/A,8%,1)-2000(P/A,8%,4)-2800(P/A,8%,2)
= -7,300 - 3000*0.926 - 2500 *0.926 - 2000 * 3.312 - 2800 *1.783
= -7300 - 2778 - 2315 - 6624 - 4992.4 = - $24,009.
PW of project C:
= -6000 - 2500(P/A,8%,7) -2800(P/A,8%,1)
= -6000 - 2500 * 5.206 -2800*0.926
= -6000 - 13015 - 2592.8
= -21,607.8 = -$21,608
So, as both the PW are negative, it implies that both are costs. So, the lower cost is preferred. So, project C will be the better choice as $21,608 < $24,009.
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