Soal QUIZ
Suatu
kelurahan mendapatkan Bantuan Langsung Tunai dari pemerintah untuk masing masing kepala
keluarga dengan syarat ketentuan sebagai berikut :
C1 : Jumlah Tanggungan
C2 : Pendapatan Kepala Keluarga
C3 : Luas Bangunan Rumah
C4 : Memiliki KK
Pilihlah 5 alternatif KK yang akan mendapatkan bantuan dari beberapa KK berikut ini :
C1 : Jumlah Tanggungan
C2 : Pendapatan Kepala Keluarga
C3 : Luas Bangunan Rumah
C4 : Memiliki KK
Pilihlah 5 alternatif KK yang akan mendapatkan bantuan dari beberapa KK berikut ini :
|
NKK
|
C1
|
C2
|
C3
|
C4
|
|
Aldyan
|
4
|
2.350.000
|
100M²
|
Tidak ada
|
|
Hendro
|
5
|
3.050.000
|
50M²
|
Ada
|
|
Joko
|
3
|
3.350.000
|
70M²
|
Ada
|
|
Doni
|
4
|
2.550.000
|
90M²
|
Ada
|
|
Dono
|
6
|
2.850.000
|
120M²
|
Ada
|
|
Kasino
|
3
|
2.650.000
|
80M²
|
Ada
|
|
Susanto
|
2
|
3.350.000
|
150M²
|
Tidak ada
|
Bobot W = [5,4,3,4]
|
NKK
|
C1
|
C2
|
C3
|
C4
|
|
Aldyan
|
2
|
1
|
3
|
1
|
|
Hendro
|
3
|
3
|
1
|
2
|
|
Joko
|
2
|
4
|
1
|
2
|
|
Doni
|
2
|
2
|
2
|
2
|
|
Dono
|
3
|
3
|
4
|
2
|
Bobot W=
[5,4,3,4]
Penyelesaian
1.X1 = √2²+3²+2²+2²+2²
= 5.4772
r11=
2/5.4772 =0.3651 r41= 2/5.4772
=0.3651
r21=
3/5.4772 =0.5477 r51=
3/5.4772 =0.5477
r31=
2/5.4772 =0.3651
X2 =√1²+3²+4²+2²+3² = 6.7823
r12=
2/6.7823 =0.1474 r41= 2/6.7823
=0.2948
r22=
3/6.7823 =0.4423 r51=
3/6.7823 =0.423
r32=
4/6.7823 =0.5897
X3 =√3²+1²+1²+2²+4² = 5.5677
r13=
3/5.5677 =0.5388 r43= 2/5.5677
=0.3592
r23=
1/5.5677 =0.1786 r53=
4/5.5677 =0.7184
r33=
1/5.5677 =0.1786
X4=√1²+2²+2²+2²+2² = 4.1231
r14=
1/4.1231 =0.2425 r44= 2/4.1231
=0.4850
r24=
2/4.1231 =0.4850 r45=
2/4.1231 =0.4850
r34=
2/4.1231 =0.4850
0.3651 0.1474
0.5388 0.2425
0.5477 0.4423
0.1796 0.4850
Matrik R 0.3651
0.5897 0.1796 0.4850
0.3651 0.2984
0.3592 0.4850
0.5477 0.4423
0.7184 04850
2. y11=w1 r11=
5*0.3651 = 1.8255
y21=w1 r21= 5*0.5477 = 2.7385
y31=w1 r31= 5*0.3651 = 1.8255
y41=w1 r41= 5*0.3651 = 1.8255
y51=w1 r51= 5*0.5477 = 2.7385
y12=w2 r12= 4*0.1474 = 0.5896
y22=w2 r22= 4*0.4423 = 1.7692
y32=w2 r32= 4*0.5877 = 2.3588
y42=w2 r42= 4*0.2984 = 1.1936
y52=w2 r52= 4*0.4423 = 1.7692
y13=w3r13 = 3*0.5388 = 1.6164
y23=w3r23 = 3*0.1796 = 0.5388
y33=w3r33 = 3*0.1796 = 0.5388
y43 =w3r43 = 3*0.3592 = 1.0776
y53 = w3r53 = 3*0.7184 = 2.1552
y14 = w4r14 = 4*0.2425 = 0.97
y24 = w4r24 = 4*0.4850 = 1.94
y34 = w4r34 = 4*0.4850 = 1.94
y44 = w4r44 = 4*0.4850 = 1.94
y54 = w4r54 = 4*0.4850 = 1.94
1.8255 0.5896
1.6164 0.97
2.7385 1.7692
0.5388 1.94
Matrik y 1.8225 2.3588
0.5388 1.94
1.8255 1.1936
1.0776 1.94
2.7385
1.7692 2.1552 1.94
3. y1⁺ = max
{1.8255 ; 2.7385 ; 1.8255 ; 1.8255 ; 2.7385} = 2.7385
y2 ̄
= min {0.5896 ; 1.7692 ; 2.3588 ; 1.1936 ; 1.7692} = 0.5896
y3 ̄
= min {1.6164 ; 0.5388 ; 0.5388 ; 1.0776 ; 2.1552} = 0.5388
y4⁺ = max {0.97 ; 1.94 ; 1.94 ; 1.94 ;
1.94} = 1.94
A⁺
={ 2.7385 ; 0.5896 ; 0.5388 ; 1.94 }
y1 ̄ = min {1.8255 ; 2.7385 ;1.8255 ; 1.82556
; 2.7385} = 1.8255
y2⁺ = max {0.5896 ; 1.7692 ; 2.3588 ;
1.1936 ; 1.7692} = 2.3588
y3 ⁺= max {1.6164 ; 0.5388 ; 0.5388 ;
1.0776 ; 2.1552} = 2.1552
y4 ̄
= max {0.97 ; 1.94 ; 1.94 ; 1.94 ; 1.94} = 0.97
A ̄
={ 1,8255 ; 2.3588 ; 2.1552; 0.97}
4. D1⁺= √(1.8225 - 2.7385)²+(0.5896 – 0.5896)² + (1.6164 – 0.5388)²+ (0.97 - 1.94)²
= 1.7149
D2 ̄ = √(2.7385 - 2.7385)²+(1.7672 – 0.5896)² + (0.5388 – 0.5388)²+ (1.94- 1.94)²
= 1.1796
D3 ̄ = √(1.8255 - 2.7385)²+(2.3588 – 0.5896)² + (0.5388 – 0.5388)²+ (1.94- 1.94)²
= 1.9908
D4⁺= √(2.7385- 2.7385)²+(1.7692 – 0.5896)² + (2.1552 – 0.5388)²+ (1.94 - 1.94)²
= 2.0010
D1 ̄=
√(1.8225 – 1.8225)²+(0.5896 – 2.3588)² + (1.6164 – 2.1552)²+ (0.97 – 0.97)²
= 1.8494
D2⁺ = √(2.7385
– 1.8225)²+(1.7672
– 2.3588)²
+ (0.5388 – 2.1552)²+ (1.94- 0.97)²
= 2.1772
D3⁺= √(1.8255 – 1.8225)²+(2.3588 – 2.3588)² + (0.5388 – 2.1552)²+ (1.94- 0.97)²
= 1.8850
D4 ̄=
√(2.7385- 1.8225)²+(1.7692 – 2.3588)² + (2.1552 – 2.1552)²+ (1.94 – 0.97)²
= 1.4585
5. V1 = 1.8479/1.9479+1.7149 = 0.5186
V2 = 2.1772/2.1172+1.1796 = 0.6485
V3 = 1.8850/1.8850+1.9908 = 0.4863
V4 = 1.4585/1.4585+2.0010 =
0.4215
Tidak ada komentar:
Posting Komentar