FIN1020 TN1
Étude de cas : FIN1020 TN1. Rechercher de 53 000+ Dissertation Gratuites et MémoiresPar loly_frizette • 20 Octobre 2022 • Étude de cas • 739 Mots (3 Pages) • 314 Vues
Travail noté 1
Question 1
a)
[pic 1][pic 2]
[pic 3][pic 4][pic 5][pic 6][pic 7][pic 8][pic 9][pic 10][pic 11][pic 12][pic 13][pic 14][pic 15][pic 16][pic 17][pic 18][pic 19][pic 20]
PMT : ?
i : 7%
b) PV = PMT(1- (1 + i)*-n/i) + PMT(1 - (1 + i)*-n/I) (1 + i)*-n
35 000 ((1 – (1.07)*10/0.07) + 40 000((1 – (1.07)*10/0.07) (1.07)*-10
PV= 245 825.35$ + 142 817.31$
PV = 388 642.66$
Pour retirer 35 000$ pendant 10 ans et 40 000$ pendant 10 ans, mon cousin doit avoir 388 642.66$
c)
PMT = ?
PV = 388 642.66$
I = 7%
N = 40
388 642.66$ = PMT(1 - (1 + 0.07)-40)/0.07
PMT = 6648$
Mon cousin devra verser 6648$ annuellement.
d) 0.5% x 12 = 6% (taux capitalisé mensuellement)
(1 + 6%)*12/12 – 1 = 6.17%
35 000 ((1 – (1.0617)*10/0.0617) + 40 000((1 – (1.0617)*10/0.0617) (1.0617)*-10
255 540.91 + 164 484.91
Avec un taux de rendement de 0.5%, mon cousin devra accumuler 416 025.83$
Question 2
a)
PV : 300 000 – 40 000 = 260 000$ pour la mise de fond
N = 240 périodes
I = ?
(1 + i)2 = (1 + i)12
(1 + 0.08/2)2 = (1 + I)12
(1.04)2/12 – 1 = i
I = 0.06558%
1 + i-n - 1
1 + 0.006558-240
PMT = 2153.73$
b)
21.53.69$ X 240 = 516 885.60$
516 885.76 – 260 000 = 256 885.60$
c)
(1 + i)26 = (1 + i)2
(1 + i)26/26 = (1 + 0.04)2/26
I = (1.04)1/13 – 1
I = 0.3022%
N = 26 x 20 = 520
1 + 0.0322 -520 – 1
PMT = 992.37$
d)
PMT = 2153.69$
N = 240 – 12 = 228
(1 + i)-n - 1
I = 0.006558-228 – 1
PV = 2153.69
Solde restant après 12 mois : 254 418.02$
260 000 – 254 418.02 = 5581.98$
12 x 2153.69 = 25 844.28$
Total des intérêts payés après 12 mois : 25 844.28 – 5581.98 = 20 262.30$
e)
(1 + i)2 = (1 + i)12
(1 + 0.095/2)2 = (1 + i)12
(1.0475)2/12 – 1 = i
I = 0.007764 = 0.7764%
PMT = ?
I = 0.7764%
N = 240 – 12 = 228
1 + 0.007764-228 – 1
PV = 254 418.02$
PMT = 2384.10$
f)
(1 + i)1 = (1 + i)6
(1 + 0.03)1 = (1 + i)6
(1.03)1/6 – 1 = i
I = 0.004939 = 0.4939%
N = 240
1 + i -n -1
1 + 0.004939-240 – 1
PMT = 1851.76$
Question 3
a)
PV = FV (1+ i)-n (1+i)-n
PV = 50 000 (1+012/12)-12x2 (1+0.08/4)-4x2
PV = 50 000 (1.01)-24 (1.02)-6
PV = 50 000 (0.787566)(0.853490)
PV = 33 609 $
b)
FV= PMT (1 + i)n / i * (1 + i)n + PMT (1+i)n -1/i
i = 8%
(1 + I1/m1) m1 = (1 + I2/m2)m2
(1 + 8%/4)4= (1 + i)12
I = 0.006623 ou 0, 6623 %
FV= PMT (1+12%/12)24-1 /0.01 * (1+8%/4)8 + PMT (1+0.00623)24-1/0.00623)
50 000 = PMT 26,973465 X 1,171659 + PMT 25,800672
50 000 = PMT 57.404375
PMT= 871$
c)
(1 + i1/m1) m1= (1 + i2/m2) m2
(1 + 0.12/12)12 = (1 + i) 4
I = 0.03
FV = PMT (1 + i)8 -1 / 0.03 X (1 + 0.08/4 )8 + (1 + 0.08/4)8-1/ 0.08/4
50 000 = PMT 10,418786 + PMT 8,5822969
PMT = 2631$ trimestriellement
Question 4
a)
P0 = ?
N 10 ans ⟹ n =20 périodes
VN= 1000$
IC = 11.5% / 2 = 5.75%
...