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### Konsep Nilai Uang Terhadap Waktu

• 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
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