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ANALISIS PENGGANTIAN (REPLACEMENT ANALYSIS)



Analisis Penggantian? Digunakan kembali setelah diperbaiki?? Mengganti dengan aset baru??



Mesin Rusak



Memerlukan analisis ekonomi teknik agar dapat diperoleh informasi yang dibutuhkan untuk membuat keputusan logis yang selanjutnya dapat memperbaiki efisiensi operasi serta posisi persaingan perusahaan



Penyebab Utama Penggantian Aset Kerusakan Fisik Keperluan Perubahan Teknologi



Pendanaan



Umur Aset?



Umur Aset Umur Ekonomi



Umur Kepemilikan



Umur Fisik



Masa Manfaat



Menentukan Umur Ekonomis Umur ekonomi (economic life) adalah periode waktu (tahun) yang menghasilkan equivalent uniform annual cost (EUAC) minimum dari kepemilikan dan pengoperasian sebuah aset. •Menghitung Nilai Capital recovery cost (CR) di setiap tahunnya 𝐶𝑅𝑡 = 𝑃(𝐴 𝑃 , 𝑖%, 𝑡 − 𝑆(𝐴 𝐹 , 𝑖%, 𝑡 𝐶𝑅𝑡 = 𝑃(𝐴 𝑃 , 𝑖%, 𝑡 − 𝑀𝑉(𝐴 𝐹 , 𝑖%, 𝑡



•Menghitung Nilai AW (Annual Worth) di setiap tahunnya 𝐴𝑊𝑡 = 𝐶𝑅𝑡 + 𝐴𝑂𝐶



• Umur ekonomis adalah nilai AW tertinggi



Contoh 1-Menentukan UMUR EKONOMIS Helcrow, Inc. expect to replace a downtime tracking system currently installed on CNC machines. The Challenger system has a first cost of $ 70.000. An estimated annual operating cost of $20.000. A maximum useful life of 5 years, and a $ 10.000 salvage value anytime it is replaced. At an interest rate of 10% per year, determine is economic service life and corresponding AW value. # Year 1: AW = CR +AOC



# Year 3:



# Year 5:



AW = CR + AOC



AW = CR + AOC



= -70000 (A/P, 10%, 1) + 10000 (A/F, 10%,1) -20000



= -70000 (A/P, 10%, 3) + 10000 (A/F, 10%,3) -20000



= -67000 – 20000



= -25,127– 20000



=-87000



=-45,571



# Year 2:



# Year 4:



AW = CR + AOC



AW = CR + AOC



= -70000 (A/P, 10%, 2) + 10000 (A/F, 10%,2) -20000



= -70000 (A/P, 10%, 4) + 10000 (A/F, 10%,4) -20000



= -35,571 – 20000



= -19,928– 20000



=-55,571



=-39,928



= -70000 (A/P, 10%, 5) + 10000 (A/F, 10%,5) -20000 = -16,828– 20000 =-36,828



Year



MV



AOC



0



-70,000



1 2 3 4 5



10,000 10,000 10,000 10,000 10,000



MV: Market Value AOC: Annual of Cost CR: Capital Recovery AW: Annual Worth



- 20,000 - 20,000 - 20,000 - 20,000 - 20,000



CR -67,000.00 -35,571 -25,127 -19,928 -16,828



total AW -87,000 -55,571 -45,127 -39,928 -36,828



Contoh 2 –Hubungan UMUR EKONOMIS dan Market Value An asset with first cost of $ 250.000 is expected to have a maximum useful life of 10 years and market value that decreases $ 25.000 each year. The annual operating cost is expected to be constant at $ 25.000 per year for 5 year and to increase at substantial 25% per year thereafter. The interest rate is a low 4% per year, because the company , Public Service Corp,. is majority- owned by a municipality and regarded as a semiprivate corporation that enjoys public project interest rate on this loans. A. Verify that the ESL is 5 years. Is the ESL sensitive to the changing market value and AOC estimates? B. The engineer doing a replacement analysis determines that this asset should have an ESL of 10 years when it is pitted against any challenger. If the estimated AOC series has proved to be correct, determine the minimum market value that will make ESL equal 10 years.



Nilai CR dan AW diperoleh dengan rumus:



CR = -P (A/P, 4%, n) + MV (A/F, 4%,n) AW = CR + AOC nilai ESL ini sangat sensitive terhadap perubahan sekecil apapun terhadap Market Value (MV) maupun terhadap Annual Operating Cost (AOC).



Year 0 1 2 3 4 5 6 7 8 9 10



MV - 250,000 225,000 200,000 175,000 150,000 125,000 100,000 75,000 50,000 25,000 -



AOC - 25,000 - 25,000 - 25,000 -25,000 - 25,000 - 31,250 - 31,250 - 31,250 -31,250 -31,250



CR* -35,000 -34,510 -34,026 -33,549 -33,078 -32,614 -32,157 -31,706 -31,261 -30,823



total AW** -60,000 -59,510 -59,026 -58,549 -58,078 -63,864 -63,407 -62,956 -62,511 -62,073



B. Agar ESL 10 tahun maka AW harus lebih kecil dari AW saat 5 tahun yakni $ 58,078. AW ≤ CR + AOC



-58078 ≤ CR - 31250 CR ≤ -58078 + 31250 CR ≤ -26828 CR ≤ - 250000 (A/P, 4%, 10) + MV (A/F, 4%, 10) -26828 ≤ -250000 (0.1232) + MV (0.0833) -26828 ≤ -30800 + MV (0.0833) MV (0.0 833) ≥ 3978 MV ≥ 47,683



Contoh 3-Defender Vs Challanger During a 3-year period Shanna, a project manager with Sherlockhome Agriculture, performed replacement studies on drying equipment. She tabulate the ESL and AW value each year. (a) What decision should be made each year? (b) From the data, describe what changes took place in the defender and challenger over the 3 years. maximum life, years Defender Challenger Defender challenger 1 Defender challenger 2



AW, ESL, years $/year First Year, 2018 3 3 -10.000 10 5 -15.000 Second Year, 2019 2 1 -14.000 10 5 -15.000 Third Year, 2020 1 1 -14.000 5 3 -9.000



Contoh 4-Up-Grading Defender vs Challanger Five years ago, the Alim Rugi Ltd purchased several for $350.000 each. Last year a replacement study was performed with the decision to retain the vehicles for 2 more years. However, this year the situation has changed in that each transport vehicles is estimated to have a value of only $ 8000 now. If they are kept in service, upgrading at cost of $50.000 will make them useful for up to 2 more years. Operating cost is expected to be $10.000 the first year and $15.000 the second year. Alternatively, the company can purchase a new vehicle with an ESL of 7 years, no salvage value and an equivalent annual cost of $-55,540 per year. The MARR is 10% per year. If the budget to upgrade the current vehicles is available this year, use these estimates to determine (a) When the company should replaced the upgraded vehicles (b) The minimum future salvage value of a new vehicles necessary to indicate that purchasing now is economically advantageous to upgrading Year



AOC



P 0



CR



total AW



-50,000



1 8,000



-10,000



-47,000



-57,000



2 -



-15,000



-28,810



-41,190



Year



P



AOC



CR



total AW



0 1 2 3 4 5 6 7



-350,000 300,000 250,000 200,000 150,000 100,000 50,000 -



AW -57000 CR CR CR 1150 1150 S



≤ CR + AOC ≤ CR - 55540 ≤ 57000 - 55540 ≤ 1150 ≤ - 350000 (A/P, 10%, 1) + S (A/F, 10%, 1) ≤ -350000 (1.1) + S (1) ≤ -385000 + S (1) ≥ 386150



-55,540 -55,540 -55,540 -55,540 -55,540 -55,540 -55,540



-85,000 -82,619 -80,317 -78,094 -75,949 -73,882 -71,892



-140,540 -138,159 -135,857 -133,634 -131,489 -129,422 -127,432



Nilai minimum agar menghasilkan formula yang mengharuskan pergantian peralatan pada tahun ini adalah jika nilai AW challenger ≤ AW defender



Tidak Mungkin Karena Nilai S > P