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Contoh Minimumkan Z = 4 X1 + X2 Kendala 3 X1 + X2 = 3 4 X1 + 3 X2 ≥ 6 X1 + 2 X2 ≤ 4 X1, X2 ≥ 0 Tahap I minimumkan R = R1 + R2 Dg batasan 3 X1 + X2 +R1.

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Presentasi berjudul: "Contoh Minimumkan Z = 4 X1 + X2 Kendala 3 X1 + X2 = 3 4 X1 + 3 X2 ≥ 6 X1 + 2 X2 ≤ 4 X1, X2 ≥ 0 Tahap I minimumkan R = R1 + R2 Dg batasan 3 X1 + X2 +R1."— Transcript presentasi:

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2 Contoh Minimumkan Z = 4 X1 + X2 Kendala 3 X1 + X2 = 3 4 X1 + 3 X2 ≥ 6 X1 + 2 X2 ≤ 4 X1, X2 ≥ 0 Tahap I minimumkan R = R1 + R2 Dg batasan 3 X1 + X2 +R1 = 3 4 X1 + 3 X2 – S1 + R2 = 6 X1 + 2 X2 + S2 = 4 X1, X2, S1, R1, R2, S2 ≥ 0 Substitusikan R1 dan R2 keluar dari fungsi tujuan: R = R1 + R2 = (3 – 3 X1 – X2) + (6 – 4 X1 – 3 X2 + S1) = -7 X1 – 4 X2 + S1 + 9

3 BasisX1X2S1R1R2S2NK r R R S Dengan demikian tabel awal menjadi BasisX1X2S1R1R2S2NK r05/3-7/3002 X111/ R205/3-4/3102 S205/30-1/3013

4 BasisX1X2S1R1R2S2NK r05/3-7/3002 X111/ R205/3-4/3102 S205/30-1/3013 BasisX1X2S1R1R2S2NK r X1101/53/5-1/503/5 X201-3/5-4/53/506/5 S

5 BasisX1X2S1R1R2S2NK r X1101/53/5-1/503/5 X201-3/5-4/53/506/5 S X1 + 1/5 S1 = 3/5 X2 – 3/5 S1 = 6/5 S1 + S2 = 1 Minimumkan Z = 4 X1 + X2 Kendala X1 + 1/5 S1 = 3/5 X2 – 3/5 S1 = 6/5 S1 + S2 = 1 X1, X2, S1, S2 ≥ 0 Z = 4 X1 + X2 4 (3/5- 1/5 S1) + ( 6/5 + 3/5 S1) = -1/5 S1 + 18/5 Z + 1/5 S1 = 18/5

6 BasisX1X2S1S2NK Z001/5018/5 X1101/503/5 X201-3/506/5 S Z = 4 X1 + X2 4 (3/5- 1/5 S1) + ( 6/5 + 3/5 S1) = -1/5 S1 + 18/5 Z + 1/5 S1 = 18/5 Dg kendala X1 + 1/5 S1 = 3/5 X2 – 3/5 S1 = 6/5 S1 + S2 = 1 X1, X2, S1, S2 ≥ 0 Jadi tabel awal untuk Tahap II menjadi

7 BasisX1X2S1S2NK Z000-1/517/5 X1100-1/52/5 X20103/59/5 S100111


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