Nongkrong Bersama: 2014

Jumat, 13 Juni 2014

Kuis SPK susulan

No	Nama	C1	C2	C3	C4
1	Joko	3	3	3	2
2	Widodo	3	3	2	2
3	Simamora	4	4	2	1
4	Susilawati	1	4	3	1
5	Dian	2	3	4	2
6	Roma	2	2	4	2
7	Hendro	3	2	3	2

Bobot W=(3,4,5,4)

Penyeselesaian :

Matrik:

3 3 3 2

3 3 2 2

4 4 2 1

X= 1 4 3 1

2 3 4 2

2 2 4 2

3 2 3 2

Identikdengan:

r11 r12 r13 r14

r21 r22 r23 r24

r31 r32 r33 r34

x = r41 r42 r43 r44

r51 r52 r53 r54

r61 r62 r63 r64

r71 r72 r73 r74

UntukC1 :

r11 = 3/4 = 0,75

r21 = 3/4 = 0,75

r31 = 4/4 = 1

r41 = 1/4 = 0.25

r51 = 2/4 = 0.5

r61 = 2/4 = 0.5

r71 = 3/4 = 0,75

UntukC2 :

r12 = 3/4 = 0,75

r22 = 3/4 = 0,75

r32 = 4/4 = 1

r42 = 4/4 = 1

r52 = 3/4 = 0.75

r62 = 2/4 = 0.5

r72 = 2/4 = 0.5

UntukC3 :

r13 = 3/2 = 1.5

r23 = 2/2 = 1

r33 = 2/2 = 1

r43 = 3/2 = 1.5

r53 = 4/2 =2

r63 =4/2 =2

r73 =3/2 = 1.5

UntukC4 :

r14 = 2/2 = 1

r24 = 2/2 = 1

r34 = 1/2 = 0.5

r44 = 1/2 = 0.5

r54 = 2/2 = 1

r64 = 2/2 = 1

r74 = 2/2 = 1

SehinggaMatri X yaitu :

0.75 0.75 1.5 1

0.75 0.75 1 1

1 0.75 1 0.5

0.25 1 1.5 0.5

0.5 1 2 1

0.5 0.5 2 1

0.75 0.5 1.5 1

Preferensi

V1 =3( 0.75) + 4(0.75) + 5(1.5) + 4(1) = 16.75

V2 = 3(0.75) + 4(0.75) + 5(1) + 4(1) = 14.25

V3 = 3(1) + 4(0.75) + 5(1) + 4(0.5) = 13

V4 = 3(0.25) + 4(1) + 5(1.5) + 4(0.5) = 14.25

V5 = 3(0.5) + 4(1) + 5(2) + 4(1) = 19.5

V6 = 3(0.5) + 4(0.5) + 5(2) + 4(1) = 17.5

V7 = 3(0.75) + 4(0.5) + 5(1.5) + 4(1) = 15.75

Minggu, 20 April 2014

Latihan SPK Kelas TI-P1102

Penyelesaian :

X = Lemari

Y = Kursi

PRODUK	PERAKITAN	PENGECATAN	KEUNTUNGAN
Lemari (X)	8	5	200
Kursi (Y)	7	12	100
Batasan	56	60

Fungsi Tujuan :

Z = 200 X + 100 Y

a. Fungsi kedua

(1) 8 X + 7 Y = 56

(2) 5 X + 12 Y = 60

b. Menentukan Titik Potong Dari Persamaan 1

Jika X = 0

8 X + 7 Y = 56 Titik Potong (0,8) , (7,0)

8 (0) + 7 Y= 56

X =

= 8

Jika Y = 0

8 X + 7 Y = 56

8 X + 7 (0)=56

X=

= 7

Titik Potong Dari Persamaan 2

Jika X = 0

5 X +12 Y = 60 Titik Potong (0,5) , (12,0)

5 (0) + 12 Y=60

X =

= 5

Jika Y = 0

5X + 12 Y = 60

5 X + 12 (0)=60

X=

= 12

c. Menyelesaikan Persamaan Eliminasi

X 5

X 8

8 X + 7 Y = 56 40 X + 35 Y = 280

5 X + 12 Y = 60 40 X + 96 Y = 480

0 + 61 =-200

Y =

Titik Potong (0 , 3,3) , (4,08 , 0)

Y = 3,3

5 X + 12 Y = 60

5 X + 12 (3,3) = 60

5 X + 39,6 = 60

5 X = 60 – 39,6

5 X = 20,4

X =

X = 4,08

d. Penentuan solusi

· (0,8)

Z = 200 X + 100 Y

= 200(0)+ 100(8)

= 0 + 800

= 800

· (7,0)

Z = 200 X + 100 Y

= 200(7)+ 100(0)

= 1400+ 0

= 1400

· (0 , 3,3)

Z = 200 X + 100 Y

= 200(0)+ 100(3,3)

= 0+ 330

= 330

· (7,0)

Z = 200 X + 100 Y

= 200(4,08)+ 100(0)

= 816 + 0

= 816

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