Statistic Process Control Week 3 Ananda Sabil Hussein, SE, MCom

Latar Belakang Pertengahan tahun 80 an pangsa pasar pager Motorola di rebut oleh produk-produk Jepang seperti halnya NEC, TOSHIBA dan Hitachi. Motorola melakukan perubahan radikal dengan memperbaiki mutu, pengembangan produk dan penurunan biaya yang berbasis statistik

Statistical Process Control Teknik statistik yang secara luas digunakan untuk memastikan bahwa proses yang sedang berjalan telah memenuhi standar.

Start Produce Good Provide Service Stop Process Yes No Assign. Causes? Take Sample Inspect Sample Find Out Why Create Control Chart

Variasi Alami dan Khusus Variasi alami adalah sumber-sumber variasi dalam proses yang secara statistik berada dalam batas kendali Variasi Khusus/dapat dihilangkan yaitu variasi yang muncul disebabkan karena peralatan yang tidak sesuai, karyawan yang lelah atau kurang terlatih serta bahan baku baru.

Diagram Pengendalian

17 = UCL 15 = LCL 16 = Mean Sample number ||||||||||||

Konsep Rata-rata dan Jarak Rata-rata

Menentukan Batas Diagram Rata-rata Batas Kendali Atas (UCL) = Batas Kendali Bawah (LCL) = = rata-rata dari sampel = = Standar deviasi = 2 (95.5%) 3(99.7%) = Standar deviasi rata-rata sampel

Cara Lain Batas Kendali Atas = Batas Kendali Bawah Dimana : = rentangan rata-rata sampel = Nilai batas kendali = rata-rata dari sampel rata-rata

Batas Bagan Rentangan

Bagan Rata-rata(a) These sampling distributions result in the charts below (Sampling mean is shifting upward but range is consistent) R-chart (R-chart does not detect change in mean) UCL LCL x-chart (x-chart detects shift in central tendency) UCL LCL

R-chart (R-chart detects increase in dispersion) UCLLCL Figure S6.5 (b) These sampling distributions result in the charts below (Sampling mean is constant but dispersion is increasing) x-chart (x-chart does not detect the increase in dispersion) UCLLCL Bagan Jarak

Bagan Kendali Atribut Mengukur persentase penolakan dalam sebuah sampel, bagan-p Menghitung jumlah penolakan, bagan-c

 For variables that are categorical  Good/bad, yes/no, acceptable/unacceptable  Measurement is typically counting defectives  Charts may measure  Percent defective (p-chart)  Number of defects (c-chart) Control Charts for Attributes

Control Limits for p-Charts Population will be a binomial distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics UCL p = p + z  p ^ LCL p = p - z  p ^ wherep=mean fraction defective in the sample z=number of standard deviations  p =standard deviation of the sampling distribution n=sample size ^ p(1 - p) n p =p =p =p = ^

Contoh Soal JamRata2JamRata2JamRata

Ditanyakan : Batas kendali proses 9 boks yang mencakup 99.7% Jawab : UCLx = = LCLx ==

Setting Control Limits Process average x = ounces Average range R =.25 Sample size n = 5

Setting Control Limits UCL x = x + A 2 R = (.577)(.25) = = ounces Process average x = ounces Average range R =.25 Sample size n = 5 From Table S6.1

Setting Control Limits UCL x = x + A 2 R = (.577)(.25) = = ounces LCL x = x - A 2 R = = ounces Process average x = ounces Average range R =.25 Sample size n = 5 UCL = Mean = LCL =

Contoh Soal SampleNumberFractionSampleNumberFraction Numberof ErrorsDefectiveNumberof ErrorsDefective Total = 80 (.04)(1 -.04) 100  p = =.02 ^ p = =.04 80(100)(20)

– – – – – – – – – – – – Sample number Fraction defective |||||||||| p-Chart for Data Entry UCL p = p + z  p = (.02) =.10 ^ LCL p = p - z  p = (.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04

– – – – – – – – – – – – Sample number Fraction defective |||||||||| UCL p = p + z  p = (.02) =.10 ^ LCL p = p - z  p = (.02) = 0 ^ UCL p = 0.10 LCL p = 0.00 p = 0.04 p-Chart for Data Entry Possible assignable causes present

Control Limits for c-Charts Population will be a Poisson distribution, but applying the Central Limit Theorem allows us to assume a normal distribution for the sample statistics wherec=mean number defective in the sample UCL c = c + 3 c LCL c = c - 3 c

c-Chart for Cab Company c = 54 complaints/9 days = 6 complaints/day |1|1 |2|2 |3|3 |4|4 |5|5 |6|6 |7|7 |8|8 |9|9 Day Number defective – – – 8 8 – 6 6 – 4 – 2 – 0 0 – UCL c = c + 3 c = = LCL c = c - 3 c = = 0 UCL c = LCL c = 0 c = 6