PROSES POISSON MAKALAH
UNTUK MEMENUHI TUGAS MATAKULIAH Proses Stokastik yang dibina oleh Bapak Dr. Swasono Rahardjo, S.Pd, M.Si
Disusun oleh: Acika Karunila
(140312604858) (140312604858)
Alfonsa Maria Gita M
(130312602889) (130312602889)
Meytika Islamia W
(140312602792) (140312602792)
UNIVERSITAS NEGERI MALANG FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM JURUSAN MATEMATIKA Maret 2017
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BAB I PENDAHULUAN 1.1
Latar Belakang
Secara umum pemodelan yang berkembang saat ini bisa dikategorikan kedalam dua proses yaitu proses poisson dan proses deterministik.Pada proses stokastik banyak fenomena di alam yang bisa dibawa kearah proses poisson. Proses antrian merupakan contoh nyata proses Poisson yang banyak terjadi pada berbagai fasilitas pelayanan saat ini. Proses antrian suatu proses yang berhubungan dengan kedatangan pelanggan pada suatu fasilitas pelayanan, menunggu dalam baris antrean jika belum dapat dilayani, kemudian seorang pelanggan begitu dilayani hanya akan meninggalkan sarana pelayanan tersebut setelah selesai pelayanan. Bentuk distribusi kedatangan ada dua yaitu distribusi jumlah kedatangan dan distribusi waktu antar kedatangan.Sedangkan bentuk distribusi pelayanan juga ada dua yaitu distribusi jumlah pelayanan dan distribusi waktu pelayanan. Pada proses kedatangan waktu antar kedatangan merupakan distribusi identik dan independen(bebas).Beberapa distribusi waktu antar kedatangan bisa berdistribusi eksponensial,general,deterministik atau poisson.Untuk memperoleh distribusi waktu antar kedatangan beberapa kasus antrean yang terjadi saat ini,seperti panggilan telpon,koneksi server internet,dan lalu lintas kendaraan dijalan tol pada arus mudik dan arus balik lebaran serta kedatangan pelanggan dikantor pos pada menjelang tahun baru dan lebaran tidak mudah didapatkan. Hal ini terjadi karena waktu antar kedatangannya yang sangat kecil sekali,sehingga sulit untuk mendapatkan data waktu antar kedatangannya, atau dalam interval waktu kecil,misalkan satu menit jumlah kedatangannya sangat besar. 1.2 Rumusan Masalah 1. Apa yang dimaksud dengan proses Poisson? 2. Bagaimana penerapan proses Poisson dalam kehidupan sehari-hari? 1.3
Tujuan
1. Mengetahui pengertian dari dengan proses Poisson. 2. Mengetahui bagaimana penerapan proses Poisson dalam kehidupan seharihari.
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BAB II PEMBAHASAN
Proses Poisson adalah proses menghitung (counting ( counting process) process) untuk banyaknya kejadian yang terjadi hingga hingga suatu waktu. Contoh: kedatangan nasabah suatu bank, munculnya item cacat pada proses pemeriksaan, masuknya pesan SMS pada handphone anda, dll. 2.1
Proses Menghitung
Definisi 2.1:
Proses stokastik proses) proses) jika
dikatakan proses menghitung (counting ( counting
menyatakan banyaknya kejadian yang terjadi
selama waktu t (S.Osaki, 1992) Contoh: 1.
adalah banyaknya bayi yang lahir selama waktu t. Maka proses menghitung.
2.
adalah banyaknya orang yang datang ke Toserba Grya dalam
waktu [0,t]. Maka
proses menghitung.
Proses menghitung i. ii. iii. iv.
memenuhi sifat:
adalah bilangan bulat
Jika
, maka
Untuk
menyatakan banyaknya kejadian yang
terjadi pada interval waktu (s,t]
Proses menghitung disebut proses dengan kenaikan bebas (independent ) jika banyaknya kejadian yang terjadi pada interval waktu increments terpisah adalah saling bebas. Artinya, banyaknya kejadian yang terjadi pada waktu t , (yaitu N (t )), )), bebas dari banyaknya kejadian yang terjadi pada waktu antara t dan dan t + s, s, (yaitu N (yaitu N (t + s) s)- N N (t )). )).
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terjadi pada interval waktu tertentu hanya tergantung pada panjang dari interval tersebut, tidak bergantung pada letak interval tersebut.
Artinya, banyaknya kejadian pada interval waktu
(yaitu
mempunyai distribusi yang sama dengan
banyaknya kejadian pada interval waktu
(yaitu
untuk semua
Definisi 2.2: Fungsi
dikatakan
jika
Contoh:
Untuk interval waktu yang kecil
tidak ada kejadian pada interval waktu yang
kecil h>0
peluang ada kejadian pada interval waktu
yang kecil h>0
2.2
Proses Poisson
Definisi 2.3: (S.Osaki, 1992)
Suatu proses menghitung
dikatakan proses Poisson
dengan laju (parameter) λ > 0 jika memenuhi: i.
ii. Proses
mempunyai
kenaikan
bebas
stasioner
(stationary
independent increments)
iii. iv.
Dari definisi ini, untuk
berlaku,
Menyatakan peluang bahwa ada k kejadian yang terjadi pada
interval (0,t]. Hukum Peluang Total:
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Karena proses Poisson stasioner, maka
Untuk sebarang
I ndepe ndependent i ncr ements
Proses mencacah (counting (counting process) process) mempunyai sifat independent increments jika increments jika untuk sembarang 0 s t u v, X(t) – X(t) – X(s) X(s) saling bebas dengan X(v) – X(v) – X(u) X(u) Jumlah kejadian yang terjadi pada selang waktu yang tidak tumpah tindih adalah peubah acah yang saling bebas
Definisi 2.4: (S.Osaki, 1992)
Suatu proses menghitung
dikatakan proses Poisson
dengan laju (parameter) λ > 0 jika memenuhi: i. N(0) = 0
ii. Proses mempunyai kenaikan bebas (independent increments) iii. Peluang ada k kejadian dalam interval waktu t: Pk (t )
P N (t s ) N (s ) k
s, t 0. N (s t ) N (s ) Maka
( t ) k
e
t
k !
, k 0,1, ...
POI ( t ).
] t ,
E [ N t
] t ,
Var[ N t
]
E[ N t t
Rate (laju dari proses)
= rata-rata banyaknya kejadian yang terjadi per waktu t stationar y in crements crements
Suatu counting process process mempunyai stationary increments increments jika untuk sembarang s sembarang s < < t , sebaran bagi X (t ) – X X ( s) s) tergantung hanya pada selang waktu t – – s: s: X (t – s) s)
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a) Definisi 2.3 ⟶ Definisi 2.4 Sifat i,ii jelas Selanjutnya, tulis
Untuk
Dari bentuk
diperoleh:
Dengan syarat awal
Untuk
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Dari sini diperoleh:
Atau ditulis
(PDB Linier)
Untuk
Dengan syarat
, diperoleh:
Dengan induksi matematik diperoleh
P k (t )
( t ) k
e
t
k !
Hal ini menunjukkan
(sifat (iii) Definisi 2)
b)
Definisi 2.4 ⟶ Definisi 2.3 Sifat (i) jelas
Dari sifat (iii) definisi 2,
mempunyai distribusi yang
sama dengan N(t)
Artinya, punya kenaikan stasioner (sifat (ii) definisi 1). Selanjutnya, dari sifat (iii) definisi 2,
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(memenuhi sifat (iv) definisi 1)
Ekspektasi dan Variansi dari
2.2.1
Contoh soal
1. Pelanggan tiba di toko mengikuti proses Poisson dengan laju 2 orang per jam selama jam kerja dari pukul 10.00 (t=0) sampai pukul 18.00 a. Tentukan peluang bahwa k pelanggan (k=0,1,2) datang pada pukul 13.00 – 13.00 – 15.00 15.00 b. Tentukan mean dan variansi dari kedatangan pelanggan selama jam kerja Jawab:
a. Waktu 13.00-15.00 → t=2
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2. Panggilan telepon mengikuti proses Poisson dengan laju 10/jam a. Tentukan peluang bahwa ada 8 panggilan telepon terjadi pada satu jam pertama b. Tentukan peluang terdapat 3 panggilan telepon pada setengah jam pertama dan 6 panggilan telepon pada setengah jam kedua Jawab:
λ =10 =10 a. b.
2.3
Waktu Antar Kedatangan
Berdasarkan proses menghitung
,
menyatakan
banyaknya kejadian sampai waktu . Perhatikan bahwa kejadian-kejadian tersebut dapat terjadi kapan saja dalam interval pertama terjadi pada saat
, disini
Kejadian kedua terjadi pada saat
. Misalkan kejadian
dan
, maka
untuk
dan
.
untuk
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Definisi 2.5:
Berdasarkan proses menghitung waktu dari kejadian pertama. Untuk antara kejadian ke
, misalkan
, misalkan
dan kejadian ke
. Maka
adalah
adalah waktu
disebut
basrisan waktu antar kedatangan atau waktu antar kejadian.
2.4
Distribusi Waktu Kedatangan Teorema :
Waktu antar kedatangan
,
dari suatu proses Poisson adalah
saling bebas dan berdistribusi eksponensial dengan parameter . Bukti:
Akan ditunjukkan Catat bahwa,
terjadi jika tidak ada kejadian dari proses Poisson
yang terjadi pada interval
. Ini identik dengan
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Maka
Dengan induksi matematika, kita dapatkan, tiap waktu antar kedatangan adalah saling bebas dan berdistribusi eksponensial dengan parameter (terbukti)
.
λ
λ
2.4.1 Contoh soal
1. Kerusakan terjadi di sepanjang kabel di bawah laut, dengan jumlah kerusakan yang mengikuti proses Poisson dengan laju
per mil. Berapa peluang bahwa tidak terdapat kerusakan pada 2 mil pertama sepanjang kabel tersebut? Penyelesaian
Diketahui:
per mil
Ditanya : Jawab:
Perhatikan bahwa
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Realisasi dari waktu antar kedatangan dan waktu tunggu Untuk proses Poisson,
Perhatikan waktu tunggu
n
Sn
X k ,
S 0
0.
k 1
Karena
P ( S n
Hubungan antara S n
t N (t ) n
t
T)
0
x
n 1
e
( n 1) !
x
dx
, maka
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Teorema:
Diketahui bahwa
, n waktu kedatangan
memiliki
distribusi yang sama dengan statistik terurut sesuai dengan n variabel acak bebas berdistribusi seragam pada interval (0,t) Bukti: Kita akan menghitung fungsi kepadatan bersyarat dari
Diketahui bahwa Misal
dan misal
cukup kecil sehingga
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Yang memenuhi pembuktian.
Proporsisi
Jika
menyatakan banyaknya kejadian tipe-i yang terjadi pada
waktu , acak
maka maka
dengan
dan
mean
adalah bebas Poisson variabel
dan
,
dimana
Bukti
Kitaakan menghitung distribusi bersama dari
berdasarkan pada
dan
:
=n,
Sekarang berdasarkan sebarang kejadian yang terjadi pada interval . Jika terjadi pada waktu
, maka probability nya dalah type-I
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Pembuktian selesai
2.7
The M/G/1 Busy Period
Berdasarkan sistem antrian, diketahui sebagai M/G/1, yang mana pelanggannya tiba pada kesesuaian dengan proses Poisson dengan laju
. Saat Kedatangan menuju pelayanan jika servernya
bebas atau dengan bersama antrian. Kesuksesan waktu pelayanan adalah bebas dan berdistribusi identik berdasarkan G, dan juga bebas pada proses kedatangan. Ketika kedatangan pada server yang bebes, kita dapat sebutkan bahwa waktu atau periode sibuk telah dimulai. Hal itu akan berakhir saat tidak ada pelanggan yang panjang pada periode sibuk. Anggap bahwa periode sibuk akan dimulai pada beberapa saat waktu, yang mana kita tetapkan pada waktu 0. Misal
menyatakan
waktu sampai penjumlahan dari pelanggan yang tiba. (maka, contoh
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sekarang proses kedatangan bebas dari waktu pelayanan dan
dinaman
adalah kelipatan n pikiran dari G dengan dirinya
sendiri. Dengan penjumlahan, kita dapat membentuk teorema 2.3.1
yang diberikan
kedatanagn di
, urutan waktu kedatangan
berdistribusi berdasarkan urutan nilai dari himpunan seragam
bebas
variabel acak. Karena menggunakan fakta bahwa
(2.3.3) dan (2.3.2) yaitu
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Bukti:
Pembuktian dengan Induksi Matematika pada n
Ketika Ketika
, kita harus menghitung
adalah seragam (0,t). Tetapi
Maka asumsi lemma ketika n diganti dengan n-1 dan sekarang berdasarkan kasus n. Karena hasil adalah jelas untuk misal bahwa
,
. Untuk memanfaatkan hipotesis induksi akan
dihitung ruas kiri persamaan pada lemma dengan pengkondisian pada nilai-nilai
dan menggunakan fakta
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Lemma 3
Misal
menyatakan urutan nilai dari himpunan
bebas seragam (0,t) variabel acak, dan misal
, bebas
dan berdistribusi identik nonnegatif variabel acak yang bebas pula
dari
. Maka
Bukti:
Untuk menghitung probabilitynya kita akan menggunakan
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(iv)
Proses Poisson homogen mempunyai parameter λ
Proses Poisson nonhomogen nonhomogen mempunyai parameter λ(t), λ(t) disebut fungsi intensitas Yakni,
Maka kita punyai,
2.9 Compoun Variabel Acak Poisson dan Prosesnya Variabel Compoun d
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Dengan
menurunkan
fungsi
tersebut
, , dimana
Ketika
didapatkan
memiliki distribusi
adalah fungsi distribusi diskret representasi dari
liniear dari Variabel acak Poisson yang bebas. Misalkan
kombinasi
adalah variabel acak
diskrit sehingga
Jika dimisalkan maka
mendefinisikan bilangan dari
dapat di tulis sebagai
varian dari
.
yang sama ke
. Sehingga dapat dicari mean dan
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. . . Dan seterusnya
Corollary 2
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dimana
adalah Proses Poisson, dan
adalah keluarga dari variabel bebas dan identitas
distribusi variabel acak yang bebas dari proses
jika
adalah compound Proses Poisson maka
adalah compound variabel acak Poisson.
. Jadi,
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Jawab:
a. b.
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Maka dapat diketahui bahwa lajunya adalah Pertanyaannya:
, dengan
a. Hitung peluang sebuah tulisan selesai dalam 5 jam kedepan? b. Berapa peluang 3 blog selesai sekaligus dalam 5 jam? Jawab:
menit
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6)
Seorang mahasiswi UII sedang menjadi selebritis medsos. Dia memiliki 3 akun media sosial yang masing-masing memiliki jumlah follower yang selalu bertambah setiap waktu. Penambahan jumlah
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