log 0log Berdasarkan data-data berikut ini buat Decision tree-nya Solusi: Node Awal

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 − PNo log 2 PNo 4 4 3 3 = − log 2 − log 2 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 31..40 >40

Age Yes No Yes No Yes No

0 2 2 0 2 1

Entropy ( Age <= 30) = − PYes log 2 PYes − PNo log 2 PNo 0 0 2 2 = − log 2 − log 2 2 2 2 2 = −0 log 2 0 − 1log 2 1 =0 Entropy ( Age = 31..40) = − PYes log 2 PYes − PNo log 2 PNo

2 2 0 0 = − log 2 − log 2 2 2 2 2 = −1log 2 1 − 0 log 2 0 =0

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Entropy ( Age > 40) = − PYes log 2 PYes − PNo log 2 PNo 2 2 1 1 = − log 2 − log 2 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}

Sv S

Entropy (S v )

S <=30 S S Entropy ( Age <= 30) − 31..40 Entropy ( Age = 31..40) − > 40 Entropy ( Age > 40) S S S ⎛2 ⎞ ⎛2 ⎞ ⎛3 ⎞ = 0.986 − ⎜ × 0 ⎟ − ⎜ × 0 ⎟ − ⎜ × 0.915 ⎟ ⎝7 ⎠ ⎝7 ⎠ ⎝7 ⎠ = 0.986 − 0.392 = 0.594

= 0.986 −

Atribut High Medium Low

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

Entropy (Income = High ) = − PYes log 2 PYes − PNo log 2 PNo 1 1 2 2 = − log 2 − log 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 − PNo log 2 PNo 1 1 0 0 = − log 2 − log 2 1 1 1 1 = −1log 2 1 − 0 log 2 0 =0 Entropy (Income = Low) = − PYes log 2 PYes − PNo log 2 PNo 2 2 1 1 = − log 2 − log 2 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

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InformatonGain(Income) = Entropy (S ) −

Sv

∑

v∈{ High , Medium , Low}

S High

S

Entropy (S v )

S Medium S Entropy (Medium ) − Low Entropy (Low) S S S ⎛3 ⎞ ⎛1 ⎞ ⎛3 ⎞ = 0.986 − ⎜ × 0.915 ⎟ − ⎜ × 0 ⎟ − ⎜ × 0.915 ⎟ ⎝7 ⎠ ⎝7 ⎠ ⎝7 ⎠ = 0.986 − 0.392 − 0 − 0.392 = 0.202

= 0.986 −

Atribut Yes No

Entropy (High ) −

student Yes 2 No 1 Yes 2 No 2

Entropy (Student = No ) = − PYes log 2 PYes − PNo log 2 PNo 2 2 2 2 = − log 2 − log 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 − PNo log 2 PNo 1 2 2 1 = − log 2 − log 2 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}

Sv S

Entropy (S v )

S Yes S Entropy (Student = Yes ) − No Entropy (Student = No ) S S ⎛3 ⎞ ⎛4 ⎞ = 0.986 − ⎜ × 0.915 ⎟ − ⎜ × 1⎟ ⎝7 ⎠ ⎝7 ⎠ = 0.986 − 0.392 − 0.571 = 0.023 = 0.986 −

Atribut Fair Excelent

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Leasing_rating Yes 1 No 3 Yes 2 No 1 Ver/Rev:0/0

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Entropy (Leasing_rating = Fair ) = − PYes log 2 PYes − PNo log 2 PNo 1 1 3 3 = − log 2 − log 2 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 − PNo log 2 PNo 2 2 1 1 = − log 2 − log 2 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 ) −

∑

v∈{Fair , Excelent }

Sv S

Entropy (S v )

S Fair S Entropy (Leasing_rating = Fair ) − Excelent Entropy (Leasing_rating = Excelent ) S S ⎛4 ⎞ ⎛3 ⎞ = 0.986 − ⎜ × 0.811⎟ − ⎜ × 0.915 ⎟ ⎝7 ⎠ ⎝7 ⎠ = 0.986 − 0.463 − 0.392 = 0.131

= 0.986 −

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 Medium Low

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

Entropy (Income = High ) = − PYes log 2 PYes − PNo log 2 PNo 0 0 0 0 = − log 2 − log 2 0 0 0 0 = −0 log 2 0 − 0 log 2 0 =0 Entropy (Income = Medium ) = − PYes log 2 PYes − PNo log 2 PNo 1 1 0 0 = − log 2 − log 2 1 1 1 1 = −1log 2 1 − 0 log 2 0 =0 Entropy (Income = Low) = − PYes log 2 PYes − PNo log 2 PNo 1 1 1 1 = − log 2 − log 2 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 ) −

∑

v∈{ High , Medium , Low}

= 0.986 −

S High S Age> 40

Entropy (High ) −

Sv S Age > 40

Entropy (S v )

S Medium S Entropy (Medium ) − Low Entropy (Low) S Age > 40 S Age > 40

⎛0 ⎞ ⎛1 ⎞ ⎛2 ⎞ = 0.986 − ⎜ × 0 ⎟ − ⎜ × 0 ⎟ − ⎜ × 1⎟ ⎝3 ⎠ ⎝3 ⎠ ⎝3 ⎠ = 0.986 − 0 − 0 − 0.667 = 0.319 Atribut Yes No

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student Yes 1 No 1 Yes 1 No 0

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Entropy (Student = No ) = − PYes log 2 PYes − PNo log 2 PNo 1 1 1 1 = − log 2 − log 2 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 − PNo log 2 PNo 1 1 0 0 = − log 2 − log 2 1 1 1 1 = −1 log 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 ) −

Sv 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 Excelent

Leasing_rating Yes 2 No 0 Yes 0 No 1

Entropy (Leasing_rating = Fair ) = − PYes log 2 PYes − PNo log 2 PNo 2 2 1 1 = − log 2 − log 2 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 − PNo log 2 PNo 0 0 1 1 = − log 2 − log 2 1 1 1 1 = −0 log 2 0 − 1 log 2 1 = 0 − 1(0 ) =0

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InformatonGain(Leasing_rating ) = Entropy (S ) − = 0.986 −

∑

v∈{Fair , Excelent }

Sv S Age > 40

Entropy (S v )

S Fair S Entropy (Leasing_rating = Fair ) − Excelent Entropy (Leasing_rating = Excelent ) S Age> 40 S Age > 40

⎛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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log 0log Berdasarkan data-data berikut ini buat Decision tree-nya Solusi: Node Awal

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