PTI

1. Azmi MuthiAzzahra
41614010015
Latihan 1.
Diketahui :
P = $120.000
n = 8
i = 10 %
G = $3.000
Ditanyakan :
Apakah mesin tersebut layak secara ekonomis?
Penyelesaian:
Syarat layak secara ekonomis :
NPV = benefits > cost
PW of benefits :
= 29.000 (P/A,10%,8) – 3.000 (P/G,10%,8)
0 1 2 3 4 5 6 7 8
29.000
P = 120.000
F = 12.000
G =
3.000
A = 11.00026.000 23.000 20.000 17.000 14.000 11.000
8.000

2. = 29.000 (5,3349) – 3.000 (16,0287)
= 154.712 – 48.086
= 106.635
Karena benefits < cost maka mesin tersebut tidak layak secara ekonomis. Mesin tersebut
akan layak secara ekonomis apabila tingkat bunga nya di perkecil.
Latihan 2.
Diketahui :
Ada 3 alternatif yang dapat dilakukan dalam pembangunan gedung ini, yakni:
1) Membangun gedung 2 lantai dengan luas 2000 m2
2) Membangun gedung 4 lantai dengan luas 3500 m2
3) Membangun gedung 8 lantai dengan luas 8000 m2
Proyeksi biaya dan pendapatan dari gedung tersebut selama 20 tahun adalah :
Alternatif A
Gd. 2 lantai
Alternatif B
Gd. 4 lantai
Alternatif C
Gd. 8 lantai
Biaya konstruksi (diluar harga
tanah)
$ 400,000 $ 800,000 $ 2,100,000
Harga sewa per m
2
/ th $ 25 $ 30 $ 32
Nilai gedung dan tanah pada akhir
tahun ke-20
$ 200,000 $ 300,000 $ 400,000
Ditanyakan :
a) Alternatif mana yang sebaiknya dipilih investor tersebut?
b) Berapa minimal harga sewa harus diturunkan dari alternatif A dan C, agar investor
tersebut memilih alternatif B?
c) Berapa maksimal biaya pembangunan gedung dari alternatif yang dipilih pada
pertanyaan a), agar investor konsisten / tetap pada pilihannya.

3. Penyelesaian :
Gedung 2 Lantai
NPV= 50.000 (P/A,10%,20) + 200.000 (P/F,10%,20) – 400.000
= 50.000 (8,5136) + 200.000 (0,1486) – 400.000
= 425.680 + 29.720 – 400.000
= 55.400
Gedung4 Lantai
400.000
F = 200.000
P = 100.000
A = 50.000
0 1 5 10 15 20
800.000
F = 300.000
P = 100.000
A = 105.000
0 1 5 10 15 20

4. NPV= 105.000 (P/A,10%,20) + 300.000 (P/F,10%,20) – 800.000
= 105.000 (8,5136) + 300.000 (0,1486) – 800.000
= 893.928 + 44.580 – 800.000
= 138.508
Gedung8 Lantai
NPV= 256.000 (P/A,10%,20) + 400.000 (P/F,10%,20) – 2.100.000
= 256.000 (8,5136) + 400.000 (0,1486) – 2.100.000
= 2.179.481 + 59.440 – 2.100.000
= 138.921
a) Alternatifyang sebaiknya dipiliholehinvestortersebutadalah alternative C Gedung8 lantai
karena NPV nya terbesar
b) Jika biaya sewa alternative C diturunkan menjadi $30 maka NPV menjadi
NPV= 240.000 (P/A, 10%, 20) + 400.000 (P/F,10%, 20) – 2.100.000
= 248.000 (8,5136) + 400.000 (0,1486) – 2.100.000
= 2.111.372 + 59.440 – 2.100.000 = 70.812
Jadi nilai minimumharga sewa per m2
pertahun untuk alternative c adalah $31. Alternatifa
tidak perluditurunkan karena NPVsudah lebihkecil dibandingkandengan alternative b.
2.100.000
F = 400.000
P = 100.000
A = 256.000
0 1 5 10 15 20

5. c) Jika NPV C = NPV B maka biaya kontruksi maksimum nya adalah :
PW of benefitsC – NPVB = 256.000 (P/A,10%,20) + 400.000 (P/F,10%,20) – 138.508
= 256.000 (8,5136) + 400.000 (0,1486) – 138.508
= 2.179.481 + 59.440 – 138.508
= 2.100.413
Latihan 3
a)
Q1 = F1 = P (1 + i )n
= 50 ( 1 + 0,12 )2
= 50 (1,12)2
= 50 (1,2544) = 62,72
0 1 2
Q1
50

6. Q2 = P= A (
(1+𝑖) 𝑛−1
𝑖 (1+𝑖) 𝑛
)
Q2 = 50 (
(1+0,12)5−1
0,12(1+0,12)4
)
= 50 (
(1,12)5−1
0,12(1,12)4
)
= 50 (
1,762−1
0,12 (1,762)
)
= 50 (
0,762
0,211
) = 180,56
Q = Q1 + Q2 - 50
= 62,72 + 152,4 – 50 = 165,12
b)
P1 = P = A (P/A,i %,n)
= 20 (P/A,15%, 3)
= 20 (2,2832) = 45,664
P2 = P = F (P/F,i%,n)
= 10 (P/F,15%, 3)
= 10 (0,6575) = 6,575
P = P1 + P2 – 30
= 45,664 + 6,575 – 30 = 22,24
Q2
50
A = 50
2 3 64 5
P1
20
A = 20
0 1 2 3
P2
10 10
0 0

7. c)
P1 = A (P/A, i%,n)
= 300 (P/A 12%, 4)
= 300 (3,037) = 911,1
P2 = G (P/G,i%,n)
= 100 (P/G,i%,n)
= 100 (4,127) = 41,27
P = P1 – P2
= 911,1 – 41,27 = 869,83
d)
P1 = P = A (P/A,i%,n)
= 50 (P/A,10%, 9)
= 50 (5,7590) = 287,95
P
1
0 0 0
A = 50
P1
0 1 42 3
300
0
300 300
P2
0 1 42 3
0
100
0
200

8. P2 = P = F (P/F,i%,n)
= 70 (P/F,10%, 5)
= 70 (0,6209) = 43,463
P3 = P = F (P/F,i%,n)
= 70 (P/F,10%,7)
= 70 (0,5132) = 35,924
P4 = P = F (P/F,i%,n)
= 70 (P/F,10%, 90)
= 70 (0,4241) = 29,687
P = P1 + P2 + P3 + P4
= 287,95 + 43,463 + 35,924 + 29,687 = 397,024
P2
0 0 0 0
70
P3
0 0 0 0 0 0
70
P4
0 0 0 0 0 0
70
0 0

9. a) Q = P = F (P/F,i%,n)
= 200 (P/F,10%, 4)
= 200 (0,6830) = 136,6
b) R = F = A (F/A,i%,n) - 100
= 100 (F/A,15%, 4) - 100
= 100 (4,993) - 100 = 399,3
c)
S1 = P = A (P/A,i%,n)
= 50 (P/A,12%,4)
= 50 (3,037) = 151,85
S1
0 1 42 3
50
0
50 50

10. S2 = P = G (P/G,i%,n)
= 50 (P/G,12%, 4)
= 50 (4,127) = 206,35
S = S1 + S2 = 151,85 + 206,35 = 358,2
d)
1) T1’ = P = A (P/A,i%,n)
= 30 (P/A,12%,n)
= 30 (3,605) = 108,15
T1’’ = P = G (P/G,i%,n)
= 30 (P/G,i%,5)
= 30 (6,397) = 191,91
T1 = T1’ + T1’’ = 108,15 + 191,91 = 300,06
S2
0 1 42 3
0
50
0
100
0
T1’
30 30 30
T2
30
T3 T4
T5
T1’’
0
30
60
T2
90
T3 T4 T5

11. 2)
T2’ = P = A (P/A,i%,n)
= 60 (P/A,12%,4)
= 60 (3,037) = 182,22
T2’’ = P = G (P/G,i%,n)
= 30 (P/G,12%, 4)
= 30 (4,127) = 123,81
T2 = T2’ + T2’’ – 30 = 182,22 + 123,81 – 30 = 276,03
T2
30
60
T3
90
T4
120
T5
T2’
30
60
T3 T4 T5
60 60
T2’’
30
T3 T4 T5
60
30
0

12. 3)
T3’ = P = A (P/A,i%,n)
= 60 (P/A,12%,3)
= 60 (2,402) = 144,12
T3’’ = P = G (P/G,i%,n)
= 30 (P/G,12%, 3)
= 30 (2,221) = 66,63
T3 = T3’ + T3’’ – 60 = 144,12 + 66,63 – 60 = 150,75
60
T3 T4 T5
90 120
60
T3’’ T4 T5
30
60
60
T3’ T4 T5
60 60

13. 4) T4 = P = F (P/F,i%,n) - 90
= 120 (P/F,12%, 2) – 90
= 120 (0,797) – 90 = 95,64 – 90 = 5,64
T5 – 120 = 0
T5 = 120
a)
B1 = P1 = F (P/F,i%,n)
= 100 (P/F,10%, 1)
= 100 (0,9091) = 90,91
B2 = P2 = F (P/F,i%,n)
= 100 (P/F,10%, 3)
= 100 (0,7513) = 75,13
B2
100
B1
100
0
1
T4 T5
90 120

14. B3 = P3 = F (P/F,i%,n)
= 100 (P/F,10%, 5)
= 100 (0,6209) = 62,09
B = 90,91 + 75,13 + 62,09 = 228,13
b) P = A (P/A,i%,n)
P = A . X
X =
𝑃
𝐴
=
634
200
= 3,17
Dari table menunjukkanbahwa(P/A,10%,4) = 3,17
c) F = A (F/A,i%,n)
= 10 (F/A,10%,5)
= 10 (6,105) = 61,05
d)
P1 = A (P/A,i%,n)
= x (P/A,10%,4)
= x (3,170) = 3,170x
P2 = G (P/G,i%,n)
= x (P/G,10%, 4)
= x (4,378) = 4,378x
B3
100
P1
xxxx
P2
3x
2x
x
0

15. P = P1 + P2 = 3,170x + 4,378x = 7,548x