Contoh Studi Kasus Decision Tree

Solusi Quiz Senin,10 Mei 2010

Sistem Pendukung Keputusan

Berdasarkan data-data berikut ini buat Decision tree-nya

Solusi: Node Awal

[1]

Hitung Entropy awal Jumlah Instance Total = 7 Jumlah Instance Yes = 4 Jumlah Instance No = 3

EntropyS ) = − PYes log 2 PYes − P No log 2 PNo

4

= − log 2

4

3

− log 2

3

7 7 7 7 = −0.57 log 2 0.57 − 0.43 log 2 0.43

= −0.57 (− 0.811) − 0.43(− 1.218 ) = 0.462 + 0.524 = 0.986 [2]

Hitung Entropy dan Infromation Gain per Atribut untuk menentukan node awal Atribut <=30

Age Yes No Yes No Yes No

31..40 >40

0 2 2 0 2 1

Entropy ( Age <= 30) = − PYes log 2 PYes − P No log 2 PNo

0

= − log 2

0

2

− log 2

2

2 2 2 2 = −0 log 2 0 − 1log 2 1

=0

Entropy ( Age = 31..40) = − PYes log 2 PYes − P No log 2 PNo

2

= − log 2

2

0

− log 2

0

2 2 2 2 = −1log 2 1 − 0 log 2 0

=0

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Entropy ( Age > 40) = − PYes log 2 PYes − P No log 2 PNo

2

= − log 2

2

1

− log 2

1

3 3 3 3 = −0.67 log 2 0.67 − 0.33 log 2 0.33

= −0.67(− 0.578) − 0.33(− 1.6) = 0.387 + 0.528 = 0.915

∑

InformatonGain( Age ) = Entropy(S ) −

v∈{>= 30 , 31..40 , > 40}

= 0.986 −

S <=30 S

Entropy ( Age <= 30 ) −

S 31..40 S

S v S

Entropy(S v )

Entropy( Age = 31..40) −

S > 40 S

Entropy( Age > 40)

⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ = 0.986 − ⎜ × 0 ⎟ − ⎜ × 0 ⎟ − ⎜ × 0.915 ⎟ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ = 0.986 − 0.392 = 0.594 Atribut High

income Yes 1 No 2 Yes 1 No 0 Yes 2 No 1

Medium Low

Entropy ( Income = High ) = − PYes log 2 PYes − P No log 2 PNo

1

= − log 2

1

2

− log 2

2

3 3 3 3 = −0.33 log 2 0.33 − 0.67 log 2 0.67

= −0.33(− 1.6) − 0.67(− 0.578) = 0.528 + 0.387 = 0.915 Entropy ( Income = Medium) = − PYes log 2 PYes − P No log 2 PNo 1

1

0

= − log 2 − log 2

0

1 1 1 1 = −1log 2 1 − 0 log 2 0

=0 Entropy ( Income = Low) = − PYes log 2 PYes − P No log 2 PNo

2

= − log 2

2

1

− log 2

1

3 3 3 3 = −0.67 log 2 0.67 − 0.33 log 2 0.33

= −0.67(− 0.578) − 0.33(− 1.6) = 0.387 + 0.528 = 0.915 IF-UTAMA

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S v

∑

InformatonGain( Income) = Entropy(S ) −

v∈{ High , Medium , Low}

= 0.986 −

S High S

Entropy( High ) −

S Medium S

S

Entropy( S v )

Entropy( Medium) −

S Low S

Entropy( Low)

⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ = 0.986 − ⎜ × 0.915 ⎟ − ⎜ × 0 ⎟ − ⎜ × 0.915 ⎟ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ = 0.986 − 0.392 − 0 − 0.392 = 0.202 Atribut Yes

student Yes 2 No 1 Yes 2 No 2

No

Entropy(Student = No ) = − PYes log 2 PYes − P No log 2 PNo

2

= − log 2

2

2

− log 2

2

4 4 4 4 = −0.5 log 2 0.5 − 0.5 log 2 0.5

= −0.5(− 1) − 0.5(− 1) = 0.5 + 0.5 =1 Entropy (Student = Yes) = − PYes log 2 PYes − P No log 2 PNo 2

= − log 2

2

1

− log 2

1

3 3 3 3 = −0.67 log 2 0.67 − 0.33 log 2 0.33

= −0.67(− 0.578) − 0.33(− 1.6) = 0.387 + 0.528 = 0.915

∑

InformatonGain(Student ) = Entropy(S ) −

v∈{Yes , No}

= 0.986 −

S Yes S

Entropy (Student = Yes) −

S No S

S v S

Entropy(S v )

Entropy(Student = No )

⎛ 3 ⎞ ⎛ 4 ⎞ = 0.986 − ⎜ × 0.915 ⎟ − ⎜ × 1⎟ ⎝ 7 ⎠ ⎝ 7 ⎠ = 0.986 − 0.392 − 0.571 = 0.023 Atribut Fair

Excelent

IF-UTAMA

Leasing_rating Yes 1 No 3 Yes 2 No 1

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Entropy (Leasing_rating = Fair ) = − PYes log 2 PYes − P No log 2 PNo

1

= − log 2

1

3

− log 2

3

4 4 4 4 = −0.25 log 2 0.25 − 0.75 log 2 0.75

= −0.25(− 2) − 0.75(− 0.415) = 0.5 + 0.311 = 0.811 Entropy (Leasing_rating = Excelent) = − PYes log 2 PYes − P No log 2 PNo 2

= − log 2

2

1

− log 2

1

3 3 3 3 = −0.67 log 2 0.67 − 0.33 log 2 0.33

= −0.67(− 0.578) − 0.33(− 1.6) = 0.387 + 0.528 = 0.915 InformatonGain(Leasing_rating ) = Entropy(S ) −

S v

∑ v∈{Fair , Excelent }

= 0.986 −

S Fair S

Entropy (Leasing_rating = Fair ) −

S Excelent S

S

Entropy(S v )

Entropy(Leasing_rating = Excelent)

⎛ 4 ⎞ ⎛ 3 ⎞ = 0.986 − ⎜ × 0.811⎟ − ⎜ × 0.915 ⎟ ⎝ 7 ⎠ ⎝ 7 ⎠ = 0.986 − 0.463 − 0.392 = 0.131 Atribut Age Income Student Leasing_Rate

Information Gain 0.594 0.202 0.023 0.131

Karena Atribut Age memiliki Nilai Information Gain tertinggi maka Atribut tersebut dijadikan node awal, sehingga decision tree-nya menjadi

[3]

Hitung Entropy dan Infromation Gain per Atribut untuk menentukan node cabang dari edge >40

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Jumlah Instance untuk Atribut Age > 40 = 3 Jumlah Instance Yes = 2 Jumlah Instance No = 1 Atribut High

income Yes 0 No 0 Yes 1 No 0 Yes 1 No 1

Medium Low

Entropy ( Income = High ) = − PYes log 2 PYes − P No log 2 PNo

0

= − log 2

0

0

− log 2

0

0 0 0 0 = −0 log 2 0 − 0 log 2 0

=0 Entropy ( Income = Medium) = − PYes log 2 PYes − P No log 2 PNo

1

1

0

= − log 2 − log 2

0

1 1 1 1 = −1log 2 1 − 0 log 2 0

=0

Entropy ( Income = Low) = − PYes log 2 PYes − P No log 2 PNo

1

= − log 2

1

1

− log 2

1

2 2 2 2 = −0.5 log 2 0.5 − 0.5 log 2 0.5

= −0.5(− 1) − 0.5(− 1) = 0.5 + 0.5 =1 InformatonGain( Income) = Entropy(S ) −

= 0.986 −

S High S Age > 40

Entropy ( High ) −

S Medium S Age > 40

∑

S v

v∈{ High , Medium , Low} S Age > 40

Entropy( Medium) −

Entropy(S v ) S Low S Age > 40

Entropy( Low)

⎛ 0 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ = 0.986 − ⎜ × 0 ⎟ − ⎜ × 0 ⎟ − ⎜ × 1⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ = 0.986 − 0 − 0 − 0.667 = 0.319 Atribut Yes

No

IF-UTAMA

student Yes 1 No 1 Yes 1 No 0

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Entropy(Student = No ) = − PYes log 2 PYes − P No log 2 PNo

1

= − log 2

1

1

− log 2

1

2 2 2 2 = −0.5 log 2 0.5 − 0.5 log 2 0.5

= −0.5(− 1) − 0.5(− 1) = 0.5 + 0.5 =1 Entropy (Student = Yes) = − PYes log 2 PYes − P No log 2 PNo 1

1

0

= − log 2 − log 2

0

1 1 1 1 = −1log 2 1 − 0 log 2 0

= −1(0) − 0 =0

∑

InformatonGain(Student ) = Entropy(S ) −

v∈{Yes, No}

= 0.986 −

S Yes S Age > 40

Entropy (Student = Yes) −

S v S Age > 40

S No S Age > 40

Entropy(S v )

Entropy(Student = No)

⎛ 2 ⎞ ⎛ 1 ⎞ = 0.986 − ⎜ × 1⎟ − ⎜ × 0 ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ = 0.986 − 0.667 − 0 = 0.319 Atribut Fair

Leasing_rating Yes 2 No 0 Yes 0 No 1

Excelent

Entropy (Leasing_rating = Fair ) = − PYes log 2 PYes − P No log 2 PNo

2

= − log 2

2

1

− log 2

1

3 3 3 3 = −0.67 log 2 0.67 − 0.33 log 2 0.33

= −0.67(− 0.578) − 0.33(− 1.6) = 0.387 + 0.528 = 0.915 Entropy (Leasing_rating = Excelent) = − PYes log 2 PYes − P No log 2 PNo 0

= − log 2

0

1

− log 2

1

1 1 1 1 = −0 log 2 0 − 1log 2 1

= 0 − 1(0) =0

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S v

∑

InformatonGain(Leasing_rating ) = Entropy(S ) −

v∈{Fair , Excelent }

= 0.986 −

S Fair S Age > 40

Entropy (Leasing_rating = Fair ) −

S Excelent S Age > 40

S Age > 40

Entropy(S v )

Entropy(Leasing_rating = Excelent)

⎛ 2 ⎞ ⎛ 1 ⎞ = 0.986 − ⎜ × 0.915 ⎟ − ⎜ × 0 ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ = 0.986 − 0.61 − 0 = 0.376 Atribut Income Student Leasing_Rate

Information Gain 0.319 0.319 0.376

Karena Atribut Leasing_Rate memiliki Nilai Information Gain tertinggi maka Atribut tersebut dijadikan node cabang untuk edge >40, sehingga decision tree-nya menjadi

[4]

Decision tree yang dihasilkan adalah

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