8/9/2019 Problem Set No. 1

http://slidepdf.com/reader/full/problem-set-no-1 1/2

8/9/2019 Problem Set No. 1

http://slidepdf.com/reader/full/problem-set-no-1 2/2

1. ,@#51- 8 111#56 = .,562

!. ro"th 7ate Anal%sis

i. nternal: 07OA F b 01(7OAFb

1.   b=1−11126

39510=0.718

2. 0.1-6F.1<801(.1-6F.1< = .1,@6

ii. Sustainable: 07OE F b 01 ( 7OEFb

1. 0.,562F.1<801(.,562F.1< = .,61iii. nterpretation:

1. nternal: Apple an $ro" a ma+imum o! 1,.@6G "ithout usin$ an% *ind o! E?N.

2. Sustainable: Apple an $ro" a ma+imum o! ,6.1G "ithout e+ternal e4uit%

!inanin$ and at its urrent debt(e4uit% ratio.

5.   C 0(1+  rm )

m

→3500(1+ .104 )7∗4

=$ 6987.73

;.   FV =C [ (1+r )T −1

r   ]→5000[ (1+.08)10−1

.08   ]=$72432.81

.

a. Option A:  PV =  C 

r−g=

  5000

.08−.02=$83,333.33→

83333.33

(1+.08)5 =$56715.26

 b. Option B: PV =C [ 1−(1+r )T 

r   ]→10000[ 1−(  1

(1+0.1)5)

.10   ]=$37907.87

. "ould hoose Option A beause it has a hi$her Present 3alue.

<. See attahed

@..07 (120000 )

1−(1+ .07 )−10=$17085.30

1-..12 (1,200,000 )

1−(1+.12)− x  =$ 150,000

a. + = 2<.6-, %ears H 2< %ears 6.5 months

11. D

12. See Attahed

1,. See Attahed

16. See Attahed

15. &oth the P and NP3 onsider the initial in'estment o! the proIet# the present o! 'alue o! all ash

!lo"s# as "ell as the interest rate. The pro!itabilit% inde+# ho"e'er# is a ratio that ompares the ost to

the P3 o! !uture ash !lo"s# "hereas NP3 simpl% sho"s the net return on the proIet in present 'alue

terms. P does not re!let the sale o! the in'estment# but is a better indiator o! the better proIet i! both

are mutuall% e+lusi'e. Jo"e'er# in most ases NP3 is a better tool beause it uses all o! the ash !lo"s

o! the proIet and disounts them properl%.