QUIZ

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 :

Nama KK            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

Pembobotan dari kriteria diatas dapat dilihat dibawah ini :
C1 : Jumlah Tanggungan (Attribut Keuntungan)
1-2 : 1
3-4 : 2
5-6 : 3

C2 : Pendapatan Kepala Keluarga (Attribut Biaya)
2.000.000    : 1
2.400.000    : 2
2.800.000    : 3
3.200.000    : 4
3.600.000    : 5

C3 : Luas Bangunan Rumah (Attribut Biaya)
50-70    : 1
71-90          : 2
91-110        : 3
111-130      : 4
131-150      : 5

C4 : Memiliki KK (Attribut Keuntungan)
Ada     : 2
Tidak Ada    : 

Penyelesaiannya :

Nama KK            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

Tabel Kecocokan :

Tahap 1:

Tahap 2:

Y₂₁    =  w₁r₂₁       = 5* 0,5477                   = 2,7385

Y₃₁    = w₁r₃₁        = 5* 0,3651                   = 1,8255

Y₄₁    = w₁r₄₁        = 5* 0,3651                   = 1,8255

Y₅₁    = w₁r₅₁        = 5* 0.5477                   = 2,7385

Y₁₂    = w₂r₁₂        = 4* 0,1601                   = 0,6404

Y₂₂    = w₂r₂₂        = 4 0,4804                     = 1,9232

Y₃₂    = w₂r₃₂        = 4* 0­­­­,6405                   = 2,563

Y₄₂    = w₂r₄₂        = 4* 0,3202                   = 1 ,2808

Y₅₂    = w₂r₅₂        = 4* 0,4804                   = 1,9232

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                   = 1,6164

Y₁₄    = w₄r₁₄        = 3* 0,2425                   = 0,7275

Y₂₄    = w₄r₂₄        = 3* 0,4850                   = 1,445

Y₃₄    = w₄r₃₄        = 3* 0,4850                   = 1,445

Y₄₄    = w₄r₄₄        = 3* 0,4850                   = 1 ,445

Y₅₄    = w₄r₅₄        = 3* 0,4850                   = 1,445

Tahap 3:

Y₁⁺       = Max   1.8255    2,7385        1,8255       1,8255       2,7385        = 2,7385

Y₂^-        = Min   0,6404     1,9232        2,562          1,2802       1,9232        = 0,6404

Y₃^-        = Min   1,6164     0,5388        0,5388       1,0776       1,6164        = 0,5388

Y₄⁺       = Max   0,7275    1,445      1,445     1,445                  1,445          = 1,,445

A^{+         =}     2,2738    0,6404        0,5388        1445

 

Y₁^-        = Min   1,8255     2,7385        1,8255       1,8255       2,7385        = 1,8255

Y₂⁺       = Max   0,6404    1,9232          2,562                  1,2808       1,9232        = 2,562

Y₃⁺       = Max   1,6164    0,5388        0,5388       1,0776       1,6164        = 1,6164

Y₄^-        = Min   0,,7275              1,445                   1,445         1,445          = 1,445

A^-          =     1,8255   2,562          1,6164        0,7275

Tahap 4:

Tahap 5:

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Tampilan Tugas Kriptografi

           TAMPILAN TUGAS PEMOGRAMAN KRIPTOGRAFI

Tampilan Kriptografi
Berikut designnya dan cara penyelesaiannya :          
                          

1.tampilan dari menu utama

 berikut designnya:

                      

Dan berikut cara penyelesaian from-from yang akan

  di tampilkan dimenu utama:

Public Class Form1

Private Sub PEMOGRAMANKRIPTOGRAFIToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles PEMOGRAMANKRIPTOGRAFIToolStripMenuItem.Click

        Form2.MdiParent = Me

        Form2.Show()

    End Sub

    Private Sub PEMOGRAMANKRIPTOGRAFIToolStripMenuItem1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles PEMOGRAMANKRIPTOGRAFIToolStripMenuItem1.Click

        Form3.MdiParent = Me

        Form3.Show()

    End Sub

    Private Sub KRIPTOGRAFIToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles KRIPTOGRAFIToolStripMenuItem.Click

        Form4.MdiParent = Me

        Form4.Show()

    End Sub

    Private Sub KRIPTOGRAFIVEGENERIToolStripMenuItem_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles KRIPTOGRAFIVEGENERIToolStripMenuItem.Click

        Form5.MdiParent = Me

        Form5.Show()

    End Sub

    Private Sub MenuStrip1_ItemClicked(ByVal sender As System.Object, ByVal e As System.Windows.Forms.ToolStripItemClickedEventArgs) Handles MenuStrip1.ItemClicked

    End Sub

    Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

    End Sub

End Class

Hasilnya:

                       

2.tampilan  dari pemograman kriptografi Caesar Chiper

  berikut designnya:

                  

Dan cara penyelesaian kriptografi Caesar Chiper:

Public Class Form2

    Private Sub enkripsi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles enkripsi.Click

        Dim x As String = ""

        Dim xkalimat As String = ""

        For i = 1 To Len(plaintext.Text)

            x = Mid(plaintext.Text, i, i)

            x = Chr(Asc(x) + 3)

            xkalimat = xkalimat + x

        Next

        chipertext.Text = xkalimat

    End Sub

    Private Sub deskripsi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles deskripsi.Click

        Dim x As String = ""

        Dim xkalimat As String = ""

        For i = 1 To Len(plaintext.Text)

            x = Mid(plaintext.Text, i, i)

            x = Chr(Asc(x) + 3)

            xkalimat = xkalimat + x

        Next

        plaintext.Text = xkalimat

    End Sub

    Private Sub keluar_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles keluar.Click

        Dim widia As Integer

        widia = MsgBox("anda mau keluar ?", MsgBoxStyle.YesNo)

        If widia = MsgBoxResult.Yes Then

            Me.Close()

        End If

    End Sub

    Private Sub hapus_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles hapus.Click

        plaintext.Text = ""

        chipertext.Text = ""

    End Sub

    Private Sub kkkriptogrfi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles kkkriptogrfi.Click

        Me.Hide()

        kkkriptogrfi.Show()

    End Sub

    Private Sub Form2_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

    End Sub

End Class

Hasilnya:

   

                  

3.tampilan  dari pemograman kriptografi Vernam Chiper

  berikut designnya:

                    

Dan berikut ini adalah penyelesaian kriptografi Vernam Chiper:

Public Class Form3

    Private Sub Form3_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

        Plaintext.Text = ""

        kunci.Text = ""

        chipertext.Text = "'"

    End Sub

    Private Sub enkripsi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles enkripsi.Click

        Dim j As Integer

        Dim jum As Integer

        Dim skey As String

        Dim nkata As Integer

        Dim nkunci As Integer

        Dim skata As String

        Dim splain As String = ""

        Dim nEnc As Integer

        j = 0

        skata = Plaintext.Text

        jum = Len(skata)

        skey = kunci.Text

        For i = 1 To jum

            If j = Len(skey) Then

                j = 1

            Else

                j = j + 1

            End If

            nkata = Asc(Mid(skata, i, 1)) - 65

            nkunci = Asc(Mid(skey, j, 1)) - 65

            nEnc = ((nkata + nkunci) Mod 26)

            splain = splain & Chr((nEnc))

        Next i

        chipertext.Text = splain

    End Sub

    Private Sub kunci_KeyPress(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyPressEventArgs) Handles kunci.KeyPress

        e.KeyChar = UCase(e.KeyChar)

        Dim tombol As Integer = Asc(e.KeyChar)

        If Not (((tombol >= 65) And (tombol <= 90)) Or (tombol = 8)) Then

            e.Handled = True

        End If

    End Sub

    Private Sub Plaintext_KeyPress(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyPressEventArgs) Handles Plaintext.KeyPress

        e.KeyChar = UCase(e.KeyChar)

        Dim tombol As Integer = Asc(e.KeyChar)

        If Not (((tombol >= 65) And (tombol <= 90)) Or (tombol = 8)) Then

            e.Handled = True

        End If

    End Sub

End Class

Hasilnya menggunakan enkripsi:

                  

 4.tampilan  dari pemograman kriptografi Gronsfeld

   berikut designnya:

                   

Dan berikut ini adalah penyelesaian kriptografi Gronsfeld:

Public Class Form4

    Private Sub Form4_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

        plaintext.Text = ""

        chipertext.Text = ""

    End Sub

    Private Sub enkripsi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles enkripsi.Click

        Dim j As Integer

        Dim jum As Integer

        Dim skey As String

        Dim nkata As Integer

        Dim nkunci As Integer

        Dim skata As String

        Dim splain As String

        Dim nEnc As Integer

        j = 0

        jum = Len(Text)

        splain = ""

        skey = kunci.Text

        skata = Text

        For i = 1 To jum

            If j = Len(skey) Then

                j = 1

            Else

                j = j + 1

            End If

            nkata = Asc(Mid(skata, i, 1)) - 65

            nkunci = Asc(Mid(skey, j, 1))

            nEnc = (nkata + nkunci) Mod 26

            splain = splain & Chr((nEnc) + 65)

        Next i

        chipertext.Text = splain

    End Sub

End Class

Berikut hasilnya:

                 

5.tampilan  dari pemograman kriptografi Vigenere  

  berikut designnya:

                  

Dan berikut ini adalah penyelesaian kriptografi Vigenere:

Public Class Form5

 Private Sub Form5_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load

        plaintext.Text = ""

        kunci.Text = ""

        chipertext.Text = ""

    End Sub

    Private Sub enkripsi_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles enkripsi.Click

        Dim j As Integer

        Dim jum As Integer

        Dim skey As String

        Dim nkata As Integer

        Dim nkunci As Integer

        Dim skata As String

        Dim splain As String

        Dim nEnc As Integer

        j = 0

        jum = Len(Text)

        splain = ""

        skey = kunci.Text

        skata = Text

        For i = 1 To jum

            If j = Len(skey) Then

                j = 1

            Else

                j = j + 1

            End If

            nkata = Asc(Mid(skata, i, 1))

            nkunci = Asc(Mid(skey, j, 1))

            nEnc = ((nkata + nkunci) Mod 256)

            splain = splain & Chr((nEnc))

        Next i

        chipertext.Text = splain

    End Sub

End Class

Berikut hasinya:

                     

Demikianlah Tampilan Tugas Kriptografi beserta designnya: