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Jawaban Quis Online

Untuk C1                                                                                                        Untuk C2

r1.1 = 3: 4 = 0.75                                                                                r1.2 = 3 : 4 = 0.75

r1.2 = 3: 4 = 0.75                                                                                r2.2 = 3 : 4 = 0.75

r1.3 =  2 : 4 = 0.5                                                                                r3.2 =  4 : 4 = 1

r1.4 = 1 : 4= 0.25                                                                                r4.2 =  4 : 4 = 1

r1.5 = 1 : 4= 0.25                                                                                r5.2 = 3 : 4 = 0,75

r1.6 = 2 : 4 = 0.5                                                                                 r6.2 =  2 : 4 = 0.5

r1.7 = 4 : 4 = 1                                                                                    r7.2 =  2 : 4 = 0,5

Untuk C3                                                                                                        Untuk C4

r1.3 = 3 : 4 = 0.75                                                                               r1.4 = 2 : 2 = 1

r2.3 = 2 : 4 = 0.5                                                                                 r2.4 = 2 : 2 =  1

r3.3 = 2 : 4 = 0.5                                                                                 r3.4 = 1 : 2 = 0.5

r4.3 = 2 : 4 = 0.5                                                                                 r4.4 = 1 : 2 = 0.5

r5.3 = 4 : 4 = 1                                                                                    r5.4 = 2 : 2 = 1

r6.3 = 4 : 4 = 1                                                                                    r6.4 = 2 : 2 = 1

r7.3 = 3 : 4 = 0.75                                                                                r7.4 = 2 : 2 = 1

Sehingga Matrix

V1= 4 (0.75) + 4 (0.75) + 5 (0.75) + 3(1)

    = 3 + 3 + 3.75 + 3

    =  12.75

V2= 4 (0.75) + 4 (0.75) + 5 (0.5) + 3(1)

      =  11.5                                                                        

V3 =  4 (0.5) + 4 (1) + 5 (0.5) + 3 (0.5)

     =   2 + 4 + 2.5 +1.5

     = 10

v4= 4 (0.25) + 4 (1) + 5 (0.5) + 3(0.5)

   =  1 + 4 + 2.5 + 1.5

   =  9

v5 =  4 (0.25) + 4 (0.75) + 5 (1) + 3(1)

    =   1 + 3 + 5 + 3

    =    12

v6 =  4 (0.5) + 4 (0.5) + 5 (1) + 3(1)

     =   2 + 2 + 5 + 3

     =   12

v7 =  4 (1) + 4 (0.5) + 5 (0.75) + 3(1)

     =  4 + 2 + 3.75 + 3

     =   12.75

Tugas

Contoh Kasus Menggunakan Linear Programming

1.        Suatu perusahaan akan memproduksi 2 jenis produk yaitu : lemari dan kursi . untuk memproduksi 2 produk tersebut dibutuhkan 2 kegiatan yaitu : proses perakitan dan pengecatan. Untuk produksi 1 unit lemari diperlukan waktu 8 jam perakitan dan 5 jam pengecatan. Sedangkan untuk 1 unit kursi diperlukan 7 jam perakitan dan 12 jam pengecatan. Jika masing-masing produk adalah Rp. 200.000 untuk lemari dan Rp. 100.000 untuk kursi.

Tentukan solusi optimal agar mendapatkan maksimal....

Penyelesaian :

Produk	Kegiatan		Laba
	Perakitan	Pengecatan
Lemari	8	5	200
Kursi	7	12	100
Waktu	56	60

a.       Fungsi Tujuan :

           Z = 200x + 100y

b.      Fungsi kendala :

           8x + 7y   <= 56

           5x + 12y <=60

          Jika x=0                                                          jika y=0

          8x + 7y <=56                                                  8x + 7y <=56             

          8(0) + 7y <=56                                                8x + 7 (0) <=56

         7y =56                                                             8x =56

          y =56/7                                                            x =56/8

          = 8                                                                   =7

         Koordinat (0,8)                                               Koordinat (7,0)          

        Jika x=0                                                           Jika y=0

        5x+12y<=60                                                   5x+12y<=60

        5(0) + 12y <=60                                              5x + 12(0)  <=60

        12y=60                                                            5x=60

        y= 60/12                                                          x= 60/5

         =5                                                                    =12

        Koordinat (0,5)                                               Koordinat (12,0)

   c.       Maka grafik yang terbentuk :

 

·         A(0,5)

·         ?

·         (7,0) Mencari nilai B dengan melakukan eliminasi :

8x + 7y = 56         x12                         96x +84y = 672

5x +12 y = 60       x7                           35x + 84y = 420  

                                                                        61x = 252

                                                                        x=252/61

                                                                           =4,13     dibulatkan jadi 4

x=4

8x +7y = 56

8 (4) + 7y = 56

32 + 7y + 56

7y = 56 – 32

y= 24/7

  = 3,42 dibulatkan jadi 3

•    A (0,5)             A (0,5)

Z = 200x + 100y

    = 200 (0) + 100 (5)

    = 500

•     B (4,3)             B (4,3)

Z = 200x + 100y

   = 200 (4) + 100 (3)

   = 800 + 300

   = 1.100

•   C (7,0)             C (7,0)

Z = 200x + 100y

    = 200 (7) + 100 (0)

    = 1.400

Dengan demikian dapat ditarik kesimpulan bahwa nilai optimum ada pada titik C (7,0) senilai Rp. 1.400....