TUGAS SISTEM PENUNJANG KEPUTUSAN (TIN 713) NON NUMERIC MULTI PERSONS MULTI CRITERIA DECISION MAKING (ME-MCDM) DOSEN : Prof. Dr. Ir. Marimin, M.Sc. Oleh : EVANILA SILVIA F361190121 PROGRAM PASCASARJANA INSTITUT PERTANIAN BOGOR 2020 NON NUMERIC MULTI PERSONS MULTI CRITERIA DECISION MAKING (ME-MCDM) Contoh Kasus : Manajer suatu perusahaan dihadapkan pada usulan beberapa alternatif kajian dan beberapa kriteria. Alternatif ada 3, yaitu : Alternatif 1 : Persediaan bahan baku dan JIT Alternatif 2 : Peningkatan mutu melalui TQC Alternatif 3 : Efisiensi teknologi melalui grup teknologi Kriteria ada 6 yaitu : Kriteria 1 : Urgensi dengan rencana pembangunan perusahaan Kriteria 2 : Biaya yang diperlukan Kriteria 3 : Komprehensifitas proposal Kriteria 4 : Bonafitas pelaksana Kriteria 5 : Operasionalisasi Kriteria 6 : Efektifitas Skala penilaian ada 7, yaitu : P : Perfect ST : Sangat Tinggi T : Tinggi S : Sedang R : Rendah SR : Sangat Rendah PR : Paling Rendah Tingkat Kepentingan Kriteria (berdasarkan adjustment expert), misal : Kriteria 1 : P Kriteria 2 : ST Kriteria 3 : ST Kriteria 4 : S Kriteria 5 : R Kriteria 6 : R =7 =6 =5 =4 =3 =2 =1 Tabel 1. Hasil Penilaian Expert Kriteria Penilaian Pakar Expert 1 Expert 2 Expert 3 Expert 4 Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Alt 1 P T R ST P T Alt 2 ST P ST ST ST ST Alt 3 P ST ST ST ST ST Alt 1 ST T R T S T Alt 2 ST P S R SR ST Alt 3 P T ST P S R Alt 1 P T S R ST ST Alt 2 ST ST ST ST ST T Alt 3 ST SR P S ST SR Alt 1 ST T P S ST SR Alt 2 R T S P SR T Alt 3 P ST ST ST ST ST I. PENGAMBILAN KEPUTUSAN ME-MCDM (AGREGASI KRITERIA – PAKAR) PROSES AGREGASI KRITERIA Menentukan Negasi Tingkat Kepentingan Kriteria Rumus : 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 Dimana : k = indeks q = jumlah skala Tingkat Kepentingan Kriteria Kriteria 1 (P), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊7 ) = 𝑊7−7+1 = 𝑊1 = 𝑃𝑅 Kriteria 2 (ST), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊6 ) = 𝑊7−6+1 = 𝑊2 = 𝑆𝑅 Kriteria 3 (ST), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊6 ) = 𝑊7−6+1 = 𝑊2 = 𝑆𝑅 Kriteria 4 (S), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊4 ) = 𝑊7−4+1 = 𝑊4 = 𝑆 Kriteria 5 (R), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊3 ) = 𝑊7−3+1 = 𝑊5 = 𝑇 Kriteria 6 (R), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊3 ) = 𝑊7−3+1 = 𝑊5 = 𝑇 Sehingga negasi bobot kriteria adalah PR, SR, SR, S, T, T Alternatif 1 (Persediaan Bahan Baku dan JIT) Tabel 2. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 1 P T R ST P T Expert 2 Alt 1 ST T R T S T Expert 3 Alt 1 P T S R ST ST Expert 4 Alt 1 ST T P S ST SR Rumus : 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] Pakar 1 pada alternatif 1 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉11 = min [𝑃𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑅, 𝑆 ˅ 𝑆𝑇, 𝑇 ˅ 𝑃, 𝑇 ˅ 𝑇] 𝑉11 = min [𝑃, 𝑇, 𝑅, 𝑆𝑇, 𝑃, 𝑇] = 𝑅 Pakar 2 pada alternatif 1 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉21 = min [𝑃𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑅, 𝑆 ˅ 𝑇, 𝑇 ˅ 𝑆, 𝑇 ˅ 𝑇] 𝑉21 = min [𝑆𝑇, 𝑇, 𝑅, 𝑇, 𝑇, 𝑇] = 𝑅 Pakar 3 pada alternatif 1 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉31 = min [𝑃𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑆, 𝑆 ˅ 𝑅, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇] 𝑉31 = min [𝑃, 𝑇, 𝑆, 𝑆, 𝑆𝑇, 𝑆𝑇] = 𝑆 Pakar 4 pada alternatif 1 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉41 = min [𝑃𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑃, 𝑆 ˅ 𝑆, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑅] 𝑉41 = min [𝑆𝑇, 𝑇, 𝑃, 𝑆, 𝑆𝑇, 𝑇] = 𝑆 Sehingga hasil agregasi kriteria untuk alternatif 1 adalah : R, R, S, S Alternatif 2 (Peningkatan Mutu melalui TQC) Tabel 3. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 2 ST P ST ST ST ST Expert 2 Alt 2 ST P S R SR ST Expert 3 Alt 2 ST ST ST ST ST T Expert 4 Alt 2 R T S P SR T Rumus : 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] Pakar 1 pada alternatif 2 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉12 = min [𝑃𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇] 𝑉12 = min [𝑆𝑇, 𝑃, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇] = 𝑆𝑇 Pakar 2 pada alternatif 2 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉22 = min [𝑃𝑅 ˅𝑆𝑇, 𝑆𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑆, 𝑆 ˅ 𝑅, 𝑇 ˅ 𝑆𝑅, 𝑇 ˅ 𝑆𝑇] 𝑉22 = min [𝑆𝑇, 𝑃, 𝑆, 𝑆, 𝑇, 𝑆𝑇] = 𝑆 Pakar 3 pada alternatif 2 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉32 = min [𝑃𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑇] 𝑉32 = min [𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑇] = 𝑇 Pakar 4 pada alternatif 2 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ∪ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉42 = min [𝑃𝑅 ˅ 𝑅, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑆, 𝑆 ˅ 𝑃, 𝑇 ˅ 𝑆𝑅, 𝑇 ˅ 𝑇] 𝑉42 = min [𝑅, 𝑇, 𝑆, 𝑃, 𝑇, 𝑇] = 𝑅 Sehingga hasil agregasi kriteria untuk alternatif 2 adalah : ST, S, T, R Alternatif 3 (Efisiensi Teknologi melalui Grup Teknologi) Tabel 4. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 3 P ST ST ST ST ST Expert 2 Alt 3 P T ST P S R Expert 3 Alt 3 ST SR P S ST SR Expert 4 Alt 3 P ST ST ST ST ST Rumus : 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] Pakar 1 pada alternatif 3 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉13 = min [𝑃𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇] 𝑉13 = min [𝑃, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇] = 𝑆𝑇 Pakar 2 pada alternatif 3 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉23 = min [𝑃𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑇, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆 ˅ 𝑃, 𝑇 ˅ 𝑆, 𝑇 ˅ 𝑅] 𝑉23 = min [𝑃, 𝑇, 𝑆𝑇, 𝑃, 𝑇, 𝑇] = 𝑇 Pakar 3 pada alternatif 3 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉33 = min [𝑃𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑆𝑅, 𝑆𝑅 ˅ 𝑃, 𝑆 ˅ 𝑆, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑅] 𝑉33 = min [𝑆𝑇, 𝑆𝑅, 𝑃, 𝑆, 𝑆𝑇, 𝑆𝑅] = 𝑆𝑅 Pakar 4 pada alternatif 3 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉43 = min [𝑃𝑅 ˅ 𝑃, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆𝑅 ˅ 𝑆𝑇, 𝑆 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇, 𝑇 ˅ 𝑆𝑇] 𝑉43 = min [𝑃, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇] = 𝑆𝑇 Sehingga hasil agregasi kriteria untuk alternatif 3 adalah : ST, T, SR, ST PROSES AGREGASI PADA PAKAR Menentukan bobot nilai Rumus : 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ 𝒒−𝟏 )] 𝒓 Expert 1 𝒒−𝟏 )] 𝒓 7−1 𝑄1 = 𝐼𝑛𝑡 [1 + (1 ∗ )] 4 𝑄1 = 𝐼𝑛𝑡[1 + (1,5)] 𝑄1 = 𝐼𝑛𝑡 [2,5] = 3 = 𝑅 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 2 𝒒−𝟏 )] 𝒓 7−1 𝑄2 = 𝐼𝑛𝑡 [1 + (2 ∗ )] 4 𝑄2 = 𝐼𝑛𝑡[1 + (3)] 𝑄2 = 𝐼𝑛𝑡 [4] = 4 = 𝑆 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 3 𝒒−𝟏 )] 𝒓 7−1 𝑄3 = 𝐼𝑛𝑡 [1 + (3 ∗ )] 4 𝑄3 = 𝐼𝑛𝑡[1 + (4,5)] 𝑄3 = 𝐼𝑛𝑡 [5,5] = 6 = 𝑆𝑇 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 4 𝒒−𝟏 )] 𝒓 7−1 𝑄4 = 𝐼𝑛𝑡 [1 + (4 ∗ )] 4 𝑄4 = 𝐼𝑛𝑡[1 + (6)] 𝑄4 = 𝐼𝑛𝑡 [7] = 7 = 𝑃 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Sehingga bobot nilai Q1, Q2, Q3, Q4 = R, S, ST, P Menentukan penilaian alternatif Rumus : 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ˄ 𝒃𝒋 ] Dimana : 𝒃𝒋 : urutan dari nilai terbesar penilaian pakar ke-j Alternatif 1 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = R, R, S, S maka 𝒃𝒋 = S, S, R, R 𝑉1 = max [𝑅 ˄ 𝑆, 𝑆 ˄ 𝑆, 𝑆𝑇 ˄ 𝑅, 𝑃 ˄ 𝑅 ] 𝑉1 = max [𝑅, 𝑆, 𝑅, 𝑅 ] = S Alternatif 2 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, S, T, R maka 𝒃𝒋 = ST, T, S, R 𝑉2 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑇, 𝑆𝑇 ˄ 𝑆, 𝑃 ˄ 𝑅 ] 𝑉2 = max [𝑅, 𝑆, 𝑆, 𝑅 ] = S Alternatif 3 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, T, SR, ST maka 𝒃𝒋 = ST, ST, T, SR 𝑉3 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑇, 𝑃 ˄ 𝑆𝑅 ] 𝑉3 = max [𝑅, 𝑆, 𝑇, 𝑆𝑅 ] = T Jadi hasil agregasi pakar adalah S, S, T Sehingga alternatif yang dipilih adalah alternatif ke-3 (T) yaitu Efisiensi Teknologi melalui Grup Teknologi Rangkuman Metode Agregasi Kriteria Pakar : Negasi bobot kriteria adalah PR SR SR S Hasil agregasi kriteria untuk alternatif 1 adalah R R S S Hasil agregasi kriteria untuk alternatif 2 adalah ST S T R Hasil agregasi kriteria untuk alternatif 3 adalah ST T SR ST Bobot nilai Q1, Q2, Q3, Q4 adalah R S ST P Hasil agregasi pakar adalah S S T Alternatif terpilih adalah alternatif ke-3 yaitu T T T II. PENGAMBILAN KEPUTUSAN ME-MCDM (AGREGASI PAKAR – KRITERIA) PROSES AGREGASI PENILAIAN PAKAR Menentukan bobot nilai , dengan rumus : 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ 𝒒−𝟏 )] 𝒓 Expert 1 𝒒−𝟏 )] 𝒓 7−1 𝑄1 = 𝐼𝑛𝑡 [1 + (1 ∗ )] 4 𝑄1 = 𝐼𝑛𝑡[1 + (1,5)] 𝑄1 = 𝐼𝑛𝑡 [2,5] = 3 = 𝑅 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 2 𝒒−𝟏 )] 𝒓 7−1 𝑄2 = 𝐼𝑛𝑡 [1 + (2 ∗ )] 4 𝑄2 = 𝐼𝑛𝑡[1 + (3)] 𝑄2 = 𝐼𝑛𝑡 [4] = 4 = 𝑆 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 3 𝒒−𝟏 )] 𝒓 7−1 𝑄3 = 𝐼𝑛𝑡 [1 + (3 ∗ )] 4 𝑄3 = 𝐼𝑛𝑡[1 + (4,5)] 𝑄3 = 𝐼𝑛𝑡 [5,5] = 6 = 𝑆𝑇 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Expert 4 𝒒−𝟏 )] 𝒓 7−1 𝑄4 = 𝐼𝑛𝑡 [1 + (4 ∗ )] 4 𝑄4 = 𝐼𝑛𝑡[1 + (6)] 𝑄4 = 𝐼𝑛𝑡 [7] = 7 = 𝑃 𝑸𝒌 = 𝑰𝒏𝒕 [𝟏 + (𝒌 ∗ Sehingga bobot nilai Q1, Q2, Q3, Q4 = R, S, ST, P Penilaian Alternatif 1 (Persediaan Bahan Baku dan JIT) Tabel 5. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 1 P T R ST P T Expert 2 Alt 1 ST T R T S T Expert 3 Alt 1 P T S R ST ST Expert 4 Alt 1 ST T P S ST SR Agregasi pakar Rumus : 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ˄ 𝒃𝒋 ] Dimana : 𝒃𝒋 : urutan dari nilai terbesar penilaian pakar ke-j Kriteria 1 Kriteria 4 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = P, ST, P, ST maka 𝒃𝒋 = P, P, ST, ST 𝑿𝒋 = ST, T, R, S maka 𝒃𝒋 = ST, T, S, R 𝑉1 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑃, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑆𝑇 ] 𝑉1 = max [𝑅, 𝑆, 𝑆𝑇, 𝑆𝑇 ] = ST 𝑉4 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑇, 𝑆𝑇 ˄ 𝑆, 𝑃 ˄ 𝑅 ] 𝑉4 = max [𝑅, 𝑆, 𝑆, 𝑅 ] = S Kriteria 2 Kriteria 5 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = T, T, T, T maka 𝒃𝒋 = T, T, T, T 𝑿𝒋 = P, S, ST, ST maka 𝒃𝒋 = P, ST, ST, S 𝑉2 = max [𝑅 ˄ 𝑇, 𝑆 ˄ 𝑇, 𝑆𝑇 ˄ 𝑇, 𝑃 ˄ 𝑇 ] 𝑉2 = max [𝑅, 𝑆, 𝑇, 𝑇 ] = T 𝑉5 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑆 ] 𝑉5 = max [𝑅, 𝑆, 𝑆𝑇, 𝑆 ] = ST Kriteria 3 Kriteria 6 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = R, R, S, P maka 𝒃𝒋 = P, S, R, R 𝑿𝒋 = T, T, ST, SR maka 𝒃𝒋 = ST, T, T, SR 𝑉3 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑆, 𝑆𝑇 ˄ 𝑅, 𝑃 ˄ 𝑅 ] 𝑉3 = max [𝑅, 𝑆, 𝑅, 𝑅 ] = S 𝑉6 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑇, 𝑆𝑇 ˄ 𝑇, 𝑃 ˄ 𝑆𝑅 ] 𝑉6 = max [𝑅, 𝑆, 𝑇, 𝑆𝑅 ] = T Sehingga hasil agregasi pakar adalah ST, T, S, S, ST, T Penilaian Alternatif 2 (Peningkatan Mutu melalui TQC) Tabel 6. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 2 ST P ST ST ST ST Expert 2 Alt 2 ST P S R SR ST Expert 3 Alt 2 ST ST ST ST ST T Expert 4 Alt 2 R T S P SR T Agregasi pakar Rumus : 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ˄ 𝒃𝒋 ] Dimana : 𝒃𝒋 : urutan dari nilai terbesar penilaian pakar ke-j Kriteria 1 Kriteria 4 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, ST, ST, R maka 𝒃𝒋 = ST, ST, ST, R 𝑿𝒋 = ST, R, ST, P maka 𝒃𝒋 = P, ST, ST, R 𝑉1 = max[𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑅 ] 𝑉1 = max [𝑅, 𝑆, 𝑆𝑇, 𝑅 ] = ST 𝑉4 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑅 ] 𝑉4 = max [𝑅, 𝑆, 𝑆𝑇, 𝑅 ] = ST Kriteria 2 Kriteria 5 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = P, P, ST, T maka 𝒃𝒋 = P, P, ST, T 𝑿𝒋 = ST,SR,ST,SR maka 𝒃𝒋 = ST,ST, SR, SR 𝑉2 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑃, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑇 ] 𝑉2 = max [𝑅, 𝑆, 𝑆𝑇, 𝑇 ] = ST 𝑉5 = max[𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑅, 𝑃 ˄ 𝑆𝑅 ] 𝑉5 = max [𝑅, 𝑆, 𝑆𝑅, 𝑆𝑅 ] = S Kriteria 3 Kriteria 6 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, S, ST, S maka 𝒃𝒋 = ST, ST, S, S 𝑿𝒋 = ST, ST, T, T maka 𝒃𝒋 = ST, ST, T, T 𝑉3 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆, 𝑃 ˄ 𝑆 ] 𝑉3 = max [𝑅, 𝑆, 𝑆, 𝑆 ] = S 𝑉6 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑇, 𝑃 ˄ 𝑇 ] 𝑉6 = max [𝑅, 𝑆, 𝑇, 𝑇 ] = T Sehingga hasil agregasi pakar adalah ST, ST, S, ST, S, T Penilaian Alternatif 3 (Efisiensi Teknologi melalui Grup Teknologi) Tabel 7. Hasil Penilaian Expert ke-j Kriteria Pakar Alternatif Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Expert 1 Alt 3 P ST ST ST ST ST Expert 2 Alt 3 P T ST P S R Expert 3 Alt 3 ST SR P S ST SR Expert 4 Alt 3 P ST ST ST ST ST Agregasi pakar Rumus : 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ˄ 𝒃𝒋 ] Dimana : 𝒃𝒋 : urutan dari nilai terbesar penilaian pakar ke-j Kriteria 1 Kriteria 4 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱 [𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = P, P, ST, P maka 𝒃𝒋 = P, P, P, ST 𝑿𝒋 = ST, P, S, ST maka 𝒃𝒋 = P, ST, ST, S 𝑉1 = max[𝑅 ˄ 𝑃, 𝑆 ˄ 𝑃, 𝑆𝑇 ˄ 𝑃, 𝑃 ˄ 𝑆𝑇 ] 𝑉1 = max [𝑅, 𝑆, 𝑆𝑇, 𝑆𝑇 ] = ST 𝑉4 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑆 ] 𝑉4 = max [𝑅, 𝑆, 𝑆𝑇, 𝑆 ] = ST Kriteria 2 Kriteria 5 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, T, SR, ST maka 𝒃𝒋 = ST, ST, T, SR 𝑿𝒋 = ST,S,ST,ST maka 𝒃𝒋 = ST,ST, ST, S 𝑉2 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑇, 𝑃 ˄ 𝑆𝑅 ] 𝑉2 = max [𝑅, 𝑆, 𝑇, 𝑆𝑅 ] = T 𝑉5 = max[𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑆 ] 𝑉5 = max [𝑅, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇 ] = ST Kriteria 3 Kriteria 6 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑽𝒊 = 𝒇(𝑽𝒊 ) = 𝐦𝐚𝐱[𝑸𝒋 ∩ 𝒃𝒋 ] 𝑿𝒋 = ST, ST, P, ST maka 𝒃𝒋 = P, ST, ST, ST 𝑿𝒋 = ST, R, SR, ST maka 𝒃𝒋 = ST, ST, R, SR 𝑉3 = max [𝑅 ˄ 𝑃, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑆𝑇, 𝑃 ˄ 𝑆𝑇 ] 𝑉3 = max [𝑅, 𝑆, 𝑆𝑇, 𝑆𝑇 ] = ST 𝑉6 = max [𝑅 ˄ 𝑆𝑇, 𝑆 ˄ 𝑆𝑇, 𝑆𝑇 ˄ 𝑅, 𝑃 ˄ 𝑆𝑅 ] 𝑉6 = max [𝑅, 𝑆, 𝑅, 𝑆𝑅 ] = S Sehingga hasil agregasi pakar adalah ST, T, ST, ST, ST, S PROSES AGREGASI KRITERIA Menentukan Negasi Tingkat Kepentingan Kriteria Rumus : 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 Dimana : k = indeks q = jumlah skala Tingkat Kepentingan Kriteria Kriteria 1 (P), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊7 ) = 𝑊7−7+1 = 𝑊1 = 𝑃𝑅 Kriteria 2 (ST), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊6 ) = 𝑊7−6+1 = 𝑊2 = 𝑆𝑅 Kriteria 3 (ST), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊6 ) = 𝑊7−6+1 = 𝑊2 = 𝑆𝑅 Kriteria 4 (S), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊4 ) = 𝑊7−4+1 = 𝑊4 = 𝑆 Kriteria 5 (R), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊3 ) = 𝑊7−3+1 = 𝑊5 = 𝑇 Kriteria 6 (R), maka negasi tingkat kepentingan kriteria adalah 𝑵𝒆𝒈 (𝑾𝒌 ) = 𝑾𝒒−𝒌+𝟏 𝑁𝑒𝑔 (𝑊3 ) = 𝑊7−3+1 = 𝑊5 = 𝑇 Sehingga negasi bobot kriteria adalah PR, SR, SR, S, T, T Penilaian Alternatif Tabel 6. Hasil Penilaian Expert ke-j Kriteria Pakar Kriteria 1 Kriteria 2 Kriteria 3 Kriteria 4 Kriteria 5 Kriteria 6 Negasi Bobot T. Kepentingan Kriteria PR SR SR S T T Hasil Agregasi Expert (Alt 1) ST T S S ST T Hasil Agregasi Expert (Alt 2) ST ST S ST S T Hasil Agregasi Expert (Alt 3) ST T ST ST ST S Rumus : 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] Alternatif 1 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉11 = min [𝑃𝑅 ˅ 𝑺𝑻 , 𝑆𝑅 ˅ 𝑻, 𝑆𝑅 ˅ 𝑺, 𝑆 ˅ 𝑺, 𝑇 ˅ 𝑺𝑻, 𝑇 ˅ 𝑻] 𝑉11 = min [𝑆𝑇, 𝑇, 𝑆, 𝑆, 𝑆𝑇, 𝑇] = 𝑆 Alternatif 2 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉21 = min [𝑃𝑅 ˅ 𝑺𝑻, 𝑆𝑅 ˅ 𝑺𝑻, 𝑆𝑅 ˅ 𝑺, 𝑆 ˅ 𝑺𝑻, 𝑇 ˅ 𝑺, 𝑇 ˅ 𝑻] 𝑉21 = min [𝑆𝑇, 𝑆𝑇, 𝑆, 𝑆𝑇, 𝑇, 𝑇] = 𝑆 Alternatif 3 𝑽𝒊𝒋 = 𝐦𝐢𝐧 [𝑵𝒆𝒈 (𝑾𝒂𝒌 ) ˅ 𝑽𝒊𝒋 (𝒂𝒌 )] 𝑉31 = min [𝑃𝑅 ˅ 𝑺𝑻, 𝑆𝑅 ˅ 𝑻, 𝑆𝑅 ˅ 𝑺𝑻, 𝑆 ˅ 𝑺𝑻, 𝑇 ˅ 𝑺𝑻, 𝑇 ˅ 𝑺] 𝑉31 = min [𝑆𝑇, 𝑇, 𝑆𝑇, 𝑆𝑇, 𝑆𝑇, 𝑇] = 𝑇 Jadi hasil agregasi kriteria untuk alternatif adalah : S, S, T Sehingga alternatif terpilih adalah alternative ke-3 (T) yaitu Efisiensi Teknologi melalui Grup Teknologi Rangkuman Metode Agregasi Pakar – Kriteria : Bobot nilai Q1, Q2, Q3, Q4 adalah R S ST P Hasil agregasi pakar untuk alternatif 1 adalah ST T S S ST T Hasil agregasi pakar untuk alternatif 2 adalah ST ST S ST S T Hasil agregasi pakar untuk alternatif 3 adalah ST T ST ST ST S Negasi bobot kriteria adalah PR SR SR S T T Hasil agregasi kriteria untuk alternatif adalah S S T Alternatif terpilih adalah alternatif ke-3 yaitu T
