Tugas Analisis Kuantitatif
M. Rachmat Gunawan NIM: 23207045 Program MCIO ITB
Operation Research PT. Philips Ralin PT Philips Ralin memproduksi 3 jenis/model radio, yaitu model A, B, dan C yang masing-masing memberikan keuntungan sebagai berikut: • • •
Model A: $16/set Model B: $30/set Model C: $50/set
Menurut informasi dari bagian penjualan, keperluan per minggu untuk masing-masing model adalah: • • •
Model A: 20 set Model B: 120 set Model C: 60 set
Proses pembuatan radio melalui proses-proses pembuatan komponen, perakitan, dan pengepakan dengan masing-masing modelnya sebagai berikut: •
•
•
Model A: o Pembuatan komponen: 3 jam/set o Perakitan : 3.5 jam/set o Pengepakan : 5 jam/set Model B: o Pembuatan komponen : 5 jam/set o Perakitan : 5 jam/set o Pengepakan : 8 jam/set Model C: o Pembuatan komponen : 1 jam/set o Perakitan : 1.5 jam/set o Pengepakan : 3 jam/set
Untuk minggu yang akan datang, perusahaan memiliki waktu sebanyak: • • •
Pembuatan komponen : 1440 jam Perakitan : 1920 jam Pengepakan : 576 jam
Pertanyaan: 1. Formulasikan persoalan di atas sebagai persoalan programa linier 2. Buatlah tabel simplex untuk iterasi awalnya saja.
Jawaban
a = banyaknya produksi model A b = banyaknya produksi model B c = banyaknya produksi model C fungsi tujuan : memaksimalkan z = 16a + 30b + 50c berdasarkan : 3a + 4b + c ≤ 1440 3,5a + 5b + 1,5c ≤ 1920 5a + 8b + 3c ≤ 576 a ≥ 20 b ≥ 120 c ≥ 60 Konversi ke bentuk standard Z – 16a – 30b – 50c + MR4 + MR5 + MR6 3a + 3.5b + 5c + S1 = 1440 5a + 5b + 8c + S2 = 1920 a + 1.5b + 3c + S3 = 576 a – S4 + R4 = 20 Æ R4 = 20 - a + S4 b – S5 + R5 = 120 Æ R5 = 120 – b + S5 c – S6 + R6 = 60 Æ R6 = 60 – c + S6 a, b, c, S1, S2, S3, S4, S5, S6, R4, R5, R6 > 0
=0
fungsi tujuan berubah menjadi : z = 16a + 30b + 50c – M(20 –a + S4) – M(120 – b + S5) – M(60 – c + S6) Æ z – (16+M)a – (30+M)b – (50+M)c + MS4 + MS5 + MS6 = -200M Tabel Simplex Iterasi awal: Basis z a b c S1 S2 S3 S4 S5 S6 R4 R5 R6 Solusi z 1 (-16-M) (-30-M) (-50-M) 0 0 0 M M M 0 0 0 -200M S1 0 3 4 11 0 0 0 0 0 0 0 0 1440 S2 0 3,5 5 1,5 0 1 0 0 0 0 0 0 0 1920 S3 0 5 8 30 0 1 0 0 0 0 0 0 576 S4 0 1 0 0 0 0 0 -1 0 0 1 0 0 20 S5 0 0 1 0 0 0 0 0 -1 0 0 1 0 120 S6 0 0 0 1 0 0 0 0 0 -1 0 0 1 60
Money Bank Money Bank sedang dalam proses untuk merumuskan sebuah kebijakan pinjaman yang dengan jumlah total uang sebesar $12 juta. Sebagai bank yang memberikan pelayanan lengkap, bank tersebut berkewajiban untuk memberikan pinjaman kepada berbagai
nasabah. Tabel berikut ini memberikan jenis-2 pinjaman, suku bunga yang dikenakan oleh bank tersebut, dan probabilitas pinjaman macet sebagai mana diestimasi dari pengalaman masa lalu. Jenis pinjaman Suku bunga Probabilitas pinjaman macet Pribadi 0.14 0.1 Mobil 0.13 0.07 Rumah 0.12 0.03 Pertanian 0.125 0.05 Komersial 0.1 0.02 Pinjaman macet diasumsikan tidak dapat diperoleh kembali dan karena itu tidak menghasilkan pendapatan bunga. Persaingan dengan bank lainnya di wilayah tersebut mengharuskan bank untuk mengalokasikan setidaknya 40% dari total dana untuk pinjaman pertanian dan komersial. Untuk membantu industri perumahan di wilayah itu, pinjaman perumahan harus setidaknya sama dengan 50% dari pinjaman pribadi, mobil, dan perumahan. Bank juga memiliki kebijakan tertulis yang menyatakan bahwa rasio keseluruhan pinjaman macet atas semua pinjaman tidak boleh lebih besar dari 0.04. Selesaikan persoalan di atas dengan metode simpleks. Jawaban 40% * 12 juta ≤ X4 + X5 Î X4 + X5 ≥ 4.8 X3 ≥ 50% * (X1 + X2 + X3) Î -0.5X3 + X1 + X2 ≤ 0 X1 + X2 + X3 + X4 + X5 ≤ 12 0.1 X1 + 0.07 X2 + 0.03 X3 + 0.05 X4 + 0.02 X5 ≤ 0.04 (X1 + X2 + X3 + X4 + X5) Î 0.06 X1 + 0.03 X2 – 0.01 X3 + 0.01 X4 – 0.02 X5 ≤ 0 Maksimalkan pendapatan bunga: Z = 0.14 X1 + 0.13 X2 + 0.12 X3 + 0.125 X4 + 0.1 X5 Dirubah ke bentuk standard X4 + X5 –S1 + R1= 4.8 X1 + X2 – 0.5 X3 + S2 = 0 X1 + X2 + X3 + X4 + X5 + S3 = 12 0.06 X1 + 0.03 X2 – 0.01 X3 + 0.01 X4 – 0.02 X5 + S4 = 0 Z - 0.14 X1 - 0.13 X2 - 0.12 X3 - 0.125 X4 - 0.1 X5 + M.R1= 0 Menggunakan Solver dari MS Exel diperoleh: X1 = $ 340.000,00 X2 = $ 6.860.000,00 X3 = $ 4.800.000,00
Motor Perusahaan otomotif yang memproduksi sepeda motor sedang membuat sepeda motor tipe baru yaitu tipe Y dan tipe Z. Proses produksi dilakukan di dua statiun kerja yaitu perakitan dan finishing. Waktu yang tersedia di bagian perakitan 330 jam/minggu, sedangkan di bagian finishing 180 jam/minggu. Setiap unit sepeda motor tipe Y memerlukan waktu 6 jam perakitan dan 3 jam finishing sedangkan setiap unit motor tipe Z memerlukan waktu 7 jam perakitan dan 4 jam finishing. Biaya per unit tipe Y adalah 9 juta dengan harga jual 11 juta. Sedangkan biaya per unit tipe Z adalah 10 juta dengan harga jual 13 juta. Jumlah permintaan maksimum tipe Y adalah 60 unit/minggu sedangkan permintaan maksimum tipe Z adalah 50 unit/minggu. Berapa jumlah masing-2 tipe sepeda motor yang harus diproduksi dan berapa keuntungannya. Selesaikan dengan: a. solusi grafis b. metode simpleks
Jawaban X1 = motor tipe Y X2 = motor tipe Z Perakitan: Finishing :
memaksimalkan : laba = harga jual – biaya produksi
Menggunakan Chart Titik-titik kritisnya ada di: X1 = 20, X2 = 30 ==> Z = 130 juta X1 = 0, X2 = 45 ==> Z = 135 juta X1 = 55, X2 = 0 ==> Z = 110 juta
X2 100
X1 < 60 50 6X1 + 7X2 < 330
A X1 < 50
B ( 20,30 )
X1 0 Z = 2X1 + 3X2
50
C
3X1 + 4X2 < 180
Menggunakan Simplex perubahan persamaan ke bentuk standar : memaksimumkan z -2x1 – 3x2 = 0 berdasarkan : 6x1 + 7x2 + S1 = 330 3x1 + 4x2 + S2 = 180 x1 + S3 = 60 x2 + S4 = 50 x1, x2, S1, S2, S3, S4 ≥ 0 Table simplex: Iterasi Basis Z X1 X2 S1 S2 S3 S4 Solusi 0 Z 1 -2 -3 0 0 0 0 0
100
1
S1 S2 S3 S4 Z S1 X2 S3 S4
0 0 0 0 1 0 0 0 0
6 7 3 4 1 0 0 1 0,25 0 0,75 0 0,75 1 1 0 -0,75 0
1 0 0 0 0 1 0 0 0
0 0 1 0 0 1 0 0 0,75 0 -1,75 0 0,25 0 0 1 -0,25 0
0 0 0 1 0 0 0 0 1
330 180 60 50 135 15 45 60 5
Pada fungsi tujuan sudah tidak ada lagi nilai negatif untuk variabel non-basis sehingga hasil tersebut sudah optimal. solusi maksimum untuk z adalah 135 dengan jumlah produksi tipe Y (x1) sebesar 0 unit dan tipe Z (x2) sebanyak 45 unit; sama seperti hasil menggunakan solusi grafis.
Konveksi Perusahaan konveksi pekaian "Lestari" memproduksi 3 jenis pakaian, yaitu pakaian anak2, pria, dan wanita. Untuk 1 lusin pakaian anak2 diperlukan 2 rol kain berbagai corak dan warna serta 4 orang tenaga kerja. Untuk 1 lusin pakaian pria dan 1 lusin pakaian wanita diperlukan masing2 4 dan 2 rol kain berbagai corak dan warna dengan jumlah tenaga kerja masing2 2 dan 6 orang. Kain yang digunakan setiap harinya adalah 20 rol. Tenaga kerja yang ada mempunyai keahlian yang sama, dan jumlahnya 16 orang. Aturan perusahaan mengharuskan seluruh tenaga kerja digunakan, artinya tidak boleh ada tenaga kerja yang menganggur. Ongkos membuat masing-2 jenis pakaian besarnya Rp 150.000/lusin pakaian anak2, Rp 300.000/lusin pakaian pria, dan 450.000/lusin pakaian wanita. Jika masing-2 jenis pakaian itu dijual dengan harga 250.000/lusin pakaian anak-2, Rp. 540.000/lusin untuk pakaian pria, dan Rp. 530.000/lusin untuk pakaian wanita, berapa keuntungan yang dapat diperoleh perusahaan tersebut? Selesaikan dengan a. Metode Penalty (Teknik M) b. Teknik 2 Fase
Jawaban X1 = lusin pakaian anak-2 X2 = lusin pakaian pria X3 = lusin pakaian wanita Rol Kain : 2X1 + 4X2 + 2X3 <= 20 Tenaga Kerja: 4X1 + 2X2 + X3 = 16 maksimalkan laba = z = 100x1 + 240x2 + 80x3 (dalam ribuan) Metode penalty Normalisasi:
(4)
Tabel Simplex: Z
x1 x2 x3 s1 r1 Solusi Rasio I Z 1 -(100+4M) -(240+2M) -(80+6M) 0 0 -16M S1 0 2 4 2 1 0 20 10 R1 0 4 2 0 1 16 2.67 6 II Z 6 -280 -1280 0 0 80+6M 80 S1 0 4 0 6 -2 88 4.4 20 x3 0 4/6 2/6 1 0 1/6 16/6 8 III Z 120 -480 0 0 7680 80+6M 12627.2 X2 0 4/20 1 0 6/20 -2/20 88/20 x3 0 0 20 0 4 24 12 IV Z 1440 0 0 0 92160 1440M+15680 12480 X2 0 0 12 0 3.6 -24.8 48 4 x1 0 1 0 20 0 4/12 24/12 2 Jawaban: X1 = 2 X2 = 4 Z = 2*100 + 4*480 = 1.160 (dalam ribuan) Kesimpulan: Maksimal laba per hari adalah Rp. 1.160.000 dengan produksi 2 lusin pakaian anak-anak dan 4 lusin pakaian pria.
Metode 2 Fase Normalisasi: Minimalkan r = R2 = 16 – 6X3 – 2X2 – 4X1 Dengan kendala: 2X1 + 4X2 + 2X3 + S1 = 20
4X1 + 2X2 + 6X3 +R2 = 16 Iterasi
Basis S1
R 0
X1 2
X2 4
X3 2
S1 1
R1 0
Solusi 20
R1
0
4
2
6
0
1
16
R
1
4
2
6
0
0
16
0
S1 0
0.666667
0 1
0.666667 0
3.3333 33 0.3333 33 0
X3 1
R
0
1
1 0
0 0
0.3333 3 0.1666 67 -1
14.66667 2.666667 0
Solusi Fisibel, selanjutnya R tidak diikutkan dalam fase 2. 2
/3 X1 + 1/3 X2 + X3 = 2 2/3 Æ 2 X1 + X2 + 3 X3 = 8
2
/3 X1 + 3 1/3 X2 + S1 = 14 2/3 Æ 2X1 + 10 X2 + 3S1 = 44
Maksimalkan Z
= 100X1 + 240X2 + 80X3 = 50 (44 – 10X2 – 3S1) + 240 X2 + 80X3 Æ Æ
= -260 X2 + 80 X3 – 50 S1 + 2200 Z + 260 X2 - 80 X3 + 50 S1 = 2200
Iterasi
Basis
0
S1
0
2
10
0
3
44
4.4
1
X3 Z X2
0 1 0
2 -100 0.2
1 -240 1
3 -80 0
0 0 0.3
8 0 4.4
8 RHS/X3 *
X3
0
1.8
0
3
-0.3
3.6
1.2
Z X2
1 0
-52 0.2
0 1
-80 0
72 0.3
1056 4.4
RHS/X4 22
X3
0
0.6
0
1
-0.1
1.2
2
Z X2 X1 Z
1 0 0 1
-4 0 1 0
0 1 0 0
0 -0.33333 1.666667 6.666667
64 0.333333 -0.16667 63.33333
1152 4 2 1160
RHS/X2
2
3
Z
Kesimpulan: Z = 1160 (dalam ribuan) X1 = 2 X2 = 4
X1
X2
X3
S1
Solusi
Rasio
Statistical Quality Control Gunakan buku "Fundamentals of Quality Control and Improvement" Second Edition, Amitava Mitra, selesaikan kasus-2 berikut:
Chapter 7 Tabel Bantuan
nomor 18 Random samples of size 5 of the length of a connector pin are selected. For each sample, the sample mean and range are calculated (in millimeters), as shown on table bellow. The length specifications are 50+/-3.5 mm. The daily production rate is 2000. The unit cost of scrapt is $1.00, and the unit cost of reword is $0.25. 1. 2. 3. 4.
find the trial control limits for X bar and R-chart Assuming special causes for out-of-control points, find the revised control limits. Find the total daily cost of scrapt and rework If the process average changes to 52 mm, what is the probabability of detecting it on the next sample drawn after the change.
Range R samples samples Range R samples Range R 1 50.3 4 11 52.5 5 21 52.1 2 2 48.4 2 12 47.8 3 22 49.5 3 3 48.5 5 13 49.6 5 23 50.7 4 4 49.1 4 14 50.8 9 24 52.6 5 5 52.6 3 15 48.5 4 25 52.3 6 6 46.2 4 16 49.3 6 26 48.6 5 7 50.8 3 17 53.3 3 27 51 4 8 52.2 4 18 52.1 4 28 52.3 3 9 49.5 5 19 50.2 5 29 50.6 2 10 51.7 4 20 51.9 4 30 51.5 4
Jawaban
a. Xbar dan R-chart
X-Control Chart
R-Control Chart
b. Ada 4 data yang over limit, perhitungan diulang dengan meniadakan data yang over limit (baris 6, 12, 17, dan 14)
X-Control Chart
R-Control Chart
c. Scrap dan Rework
Dari jawaban b diketahui nilai standard deviation nya, sbb.:
sehingga bisa dihitung nilai estimated process
Dengan spesifikasi: 50 ± 3,5 mm didapatkan LSL = 46,5 dan USL = 53,5, dengan Sehingga didapatkan transformasi ke distribusi normal baku (z) sbb.:
(5)
Jadi proporsi untuk produk di bawah LSL (scrap) adalah 0,0078 dan yang di atas USL (rework) adalah (1-0,9452) = 0,0548 Dengan production rate sebesar 2000 per hari, maka didapatkan: Total biaya scrap perhari: (2000)(0,0078)($1) = $15,6 Total biaya rework perhari: (2000)(0,0548)($0,25) = $27,4 d. Rata-2 Xbarnya bergeser menjadi 52. Probabilitas terdeteksi di sample berikutnya adalah:
zLCL =
46.5− 52 53.5 − 52 = − 3.16 zUCL = = 0.862 1.74 1.74
Dari table normal, diperoleh: P(LCL baru) = 0.0008 P(UCL baru) = 1 – 0.8051 = 0.1949 Jadi, probabilitas terdeteksi pergeseran Xbar kearah 52 adalah 0.1949 + 0.0008 = 0.1957 (19.57 %)
nomor 24 The thickness of sheet metal (in milimeters) used for making automobile bodies is a characteristic of interest. Random samples of size 4 are taken. The average and standard deviation are calculated for each sample and are shown in the table below for 20 sample. The specification limits are 9.95+/-0.3 mm. 1. Find the control limits for the X bar and s-chart. If there are out of control points, assume special causes, and revise the limits. 2. Estimate the process mean and the process standard deviation 3. If the thickness of sheet metal exceeds the upper specification limit, it can be reworked. However, if the thickness is less then the lower specification limit, it cannot be used for its inteded purpose and must be scrapped for other users. The cost of rework is $0.25 per linear foot, and the cost of scrap is $0.75 per linear foot. The rolling mills are 100 ft in length. The manufacturer has four (4) such mills and run 80 batches on each mill daily. What is the daily cost of rework? What is the daily cost of scrap? 4. If the manufacturer has the flexibility to change to process mean, should it be moved to 10.00? 5. What alternative course of action should be considered if the product is nonconforming? Sample Sample Average Sample Standard Deviasi 1 10.19 0.15
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
9.8 10.12 10.54 9.86 9.45 10.06 10.13 9.82 10.17 10.18 9.85 9.82 10.18 9.96 9.57 10.14 10.08 9.82 10.15
0.12 0.18 0.19 0.14 0.09 0.16 0.18 0.14 0.13 0.16 0.15 0.06 0.34 0.11 0.09 0.12 0.15 0.09 0.12
Jawaban: XDbar = 9.9945 s-bar = 0.1435 CLX = X = 9.9945 UCLX = XDbar + A s-bar N = 4-> A = 1.5 UCLX = 9.9945 + 1.5*0.1435 = 10.20975 LCLX = XDbar - A s-bar = 9.9945 – 1.5*0.1435 = 9.77925 CLS = 0.1435 UCLS = B4 .s-bar = 2.266 * 0.1435 = 0.325171 LCLS = B3 . s-bar = 0 Chart:
3
5
7
9
11
13
15
17
19
3
5
7
9
11
13
15
17
19
1
10.8 10.6 10.4 10.2 10 9.8 9.6 9.4 9.2 9 8.8
0.4 0.35 0.3 0.25
1
0.2 0.15 0.1 0.05 0
Ada beberapa sampel diluar batas yaitu sampel nomor 4, 6, 14 dan 16, sehingga revised control chartnya menjadi: XDbar = 10.009 s-bar = 0.135 CLX = XDBar = 10.009 UCLX = XDbar + A s-bar N = 4-> A = 1.5 UCLX = 10.009 + 1.5*0.135 = 10.211 LCLX = XDbar - A s-bar = 10.009 – 1.5*0.135 = 9.806 CLS = 0.135 UCLS = B4 .s-bar = 2.266 * 0.135 = 0.306 LCLS = B3 . s-bar = 0
10.3 10.2 10.1 10 9.9 9.8 9.7 9.6 9.5 1
2
3
5
7
8
9
10 11 12 13 15 17 18 19 20
0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 1 2 3 5 7 8 9 10 11 12 13 15 17 18 19 20
Process Standard Deviation = σ = s-bar / c4 = 0.135 / 0.9213 = 0.1465 Process Mean = μ = XDbar = 10.009 USL = 10.25 LSL = 9.65 9.65− 10.009 = − 2.45 0.1465 10.25 − 10.009 = 1.645 zUSL = 0.1465 zLSL =
P (scrap) = 0.0071 P (rework) = 1 – 0.9495 = 0.0505 Produksi per hari = 80 batch x 4 mills x 100 feet /mills = 32.000 feet Cost of rework = 0.25 * 32000 * 0.0505 = $ 404 Cost of scrap = 0.75 * 32000 * 0.0071 = $ 170.4 Kalau Xbar proses bergeser ke 10: 9.65− 10 = − 2.39 0.1465 10.25 − 10 = 1.71 zUSL = 0.1465 zLSL =
P (scrap) = 0.0084 P (rework) = 1 – 0.9564 = 0.0436
Produksi per hari = 80 batch x 4 mills x 100 feet /mills = 32.000 feet Cost of rework = 0.25 * 32000 * 0.0436 = $ 348.8 Cost of scrap = 0.75 * 32000 * 0.0084 = $ 201.6 Kesimpulan: Terjadi penurunan rework dan peningkatan scrap, tapi secara total biaya rework + scrap turun dari 574.8 menjadi 550.4.
nomor 27 The baking time of painted corrugated sheet metal is of interet. Too much time will cause the paint to flake, and too litle time will result in an unacceptable finish. The specification on baking time are 10+/-0.2 min. Random samples of size 6 are selected, and their baking times noted. The sample means and standard deviation are calculated for 20 samples with the following result:
1. Calculate the center line and control limits for the X bar and s-chart 2. Estimate the process mean and standard deviation, assuming the process to be in control. 3. Is the process capable? What proportion of the output is non conforming 4. If the mean of the process can be shifted to 10 min, would you recommend such a change? 5. If the process mean changes to 10.2 min, what is the probability of detecting this change on the first sample taken after the shift? Asume the process variability has not changed. Jawaban CLX = 199.8/20 = 9.99 s-bar = 1.4 / 20 = 0.07 Process Standard Deviation = σ = s-bar / c4 = 0.07 / 0.9515 = 0.074 Process Mean = XDbar = 9.99 UCLX = 9.99 + 1.287*0.07 = 10.08 LCLX = 9.99 – 1.287*0.07 = 9.89 9.08− 9.99 = − 12.29 0.074 10.2 − 9.9 = 4.05 zUSL = 0.074 zLSL =
P(>USL) = 0.0000 P(
Zmin = 4.05 -> CPk = 4.05 / 3 > 1 9.08− 10 = − 12.43 0.074 10.2 − 10 = 2.7 zUSL = 0.074 zLSL =
P(>USL) = 1 – 0.9965 = 0.0035 Kesimpulan: pergeseran rata-rata proses ke 10 min menyebabkan naiknya probabilitas cat yang terlalu matang, dan menurunnya Process Capability Index menjadi 2.7/3 = 0.9. Tidak disarankan untuk merubah rata-ratanya. 9.08− 10.2 = − 15.13 0.074 10.2 − 10.2 =0 zUSL = 0.074
zLSL =
P(
USL) = 0.5 Total Probability = 0.5 Jadi probabilitas terdeteksi pergeseran adalah 0.5 atau 50%.
Chapter 8 nomor 14 Which type of control chart (p, np, c, u, or U) is most appropriate to monitor the following situation? 1. the number of potholes in highway Æ U-chart 2. the proportion of customers that are satisfied with the operation of the local housing authority. Æ p-chart 3. the satisfaction of customers at a restaurant where customers consider the quality of food and attitude of the server to be more important than the decor. Æ c-chart 4. the number of errors in account transaction at a bank where the number of accounts varies from month to month Æ p-chart 5. the number of automobile accident claims filed per month at the insurance dealer, assuming a stable number of insured person Æ np-chart 6. The responsiveness to customer needs of a local library where customers value the longer hours over short line to check out books. Æ c-chart 7. The number of the thefts in a city over a period of time Æ u-chart 8. The proportion of people that favor the construction of condominiums in a certain neighborhood Æ p-chart 9. The number of weld defects in custruction of aircraft Æ np-chart 10. The number of seeds that germinate from a large pack of seeds sold by a nursery, when the number of seeds in a packet may vary Æ p-chart 11. The performance of the braking mechanism of a certain model of automobile. The car is expected to stop within certain distance. Æ c-chart 12. The number of firemen to have on duty to provide acceptable level of service. Each fire usually requires four firemeen. The department has data on the number of calls they receive during the specified period of time Æ np-chart 13. The performance of keypunch operation. if one or more punching errors are made, the card is rendered useless, and a new card has to be punched. Æ np-chart 14. The supervisor of a stenographics pool is interested in controlling the number of typographical errors per page. Data is randomly choosen and obtained daily on the number of errors in 30 pages. Æ np-chart 15. The manager of a plant wants to control the wiring and transistor defects in an electronic component. Wiring defects are considered more serious. Æ c-chart nomor 23 The number of nonconforming door hinges found for samples of size 300 is shown in table bellow. Construct a chart for the number of nonconforming hinges. What control limits would you use for the next period, assuming that special causes for the out-ofcontrol points are identified? Sample Nonconforming Hinges Sample Nonconforming Hinges 1 10 16 8
2 3 4 5 6 7 8 9 10 11 12 13 14 15
14 12 6 11 14 8 9 7 15 10 6 12 7 20
17 18 19 20 21 22 23 24 25 26 27 28 29 30
10 7 12 13 9 7 12 6 8 6 9 10 7 12
Jawaban: Jumlah sample tetap maka digunakan np-chart. P NP NP(1-P) UCL LCL
0.033 9.9 9.5733 NP+SQRT(NP(1-p)) 12.99407 6.805925
25 20 15 10 5 0 1
4
7 10 13 16 19 22 25 28
Revised Control Chart P NP NP(1-P) UCL LCL
0.03127 9.380952 9.087611 NP+SQRT(NP(1-p)) 12.39552 6.366386
14 12 10 8 6 4 2 0 1
3
5
7
9
11
13
15
17
19
21
nomor 30 The number of typographical errors is counted over a certain number of pages for each sample. The data for 25 samples is shown in table bellow. The number of pages used for each samples is not fixed. Construct a control chart for the number of typographical errors per page. Revise the limits, assuming special causes for points that are out of control. Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Page Inspected 10 5 8 6 12 10 8 10 6 8 10 12 6 4 8
Typographical Error 4 3 8 4 14 8 4 5 2 12 7 5 15 3 2
Page Inspected
Sample 16 17 18 19 20 21 22 23 24 25
8 12 10 12 8 6 10 10 8 12
Typographical Error 4 6 8 10 5 8 4 5 4 5
Jawaban: Karena jumlah sample tidak sama, digunakan p-chart. Page Typographical Sample Inspected Error Proporsi UCL 1 10 4 0.4 1.139215
LCL 0.27631
Sample 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Page Typographical Inspected Error Proporsi UCL LCL 5 3 0.6 1.317929 0.097597 8 8 1 1.190141 0.225384 6 4 0.666667 1.264765 0.15076 12 14 1.166667 1.101623 0.313902 10 8 0.8 1.139215 0.27631 8 4 0.5 1.190141 0.225384 10 5 0.5 1.139215 0.27631 6 2 0.333333 1.264765 0.15076 8 12 1.5 1.190141 0.225384 10 7 0.7 1.139215 0.27631 12 5 0.416667 1.101623 0.313902 6 15 2.5 1.264765 0.15076 4 3 0.75 1.389949 0.025576 8 2 0.25 1.190141 0.225384 8 4 0.5 1.190141 0.225384 12 6 0.5 1.101623 0.313902 10 8 0.8 1.139215 0.27631 12 10 0.833333 1.101623 0.313902 8 5 0.625 1.190141 0.225384 6 8 1.333333 1.264765 0.15076 10 4 0.4 1.139215 0.27631 10 5 0.5 1.139215 0.27631 8 4 0.5 1.190141 0.225384 12 5 0.416667 1.101623 0.313902 P P(1-P)
Initial Control Chart
0.707763 0.206835
3 2.5 2 1.5 1 0.5 0 1
3
5
7 9 11 13 15 17 19 21 23 25
Revise Control Chart Page Typographical Sample Inspected Error Proporsi UCL LCL 1 10 4 0.4 1.036929 0.096761 2 5 3 0.6 1.231644 -0.09795 3 8 8 1 1.092414 0.041275 4 6 4 0.666667 1.17372 -0.04003 6 10 8 0.8 1.036929 0.096761 7 8 4 0.5 1.092414 0.041275 8 10 5 0.5 1.036929 0.096761 9 6 2 0.333333 1.17372 -0.04003 11 10 7 0.7 1.036929 0.096761 12 12 5 0.416667 0.995971 0.137719 14 4 3 0.75 1.310112 -0.17642 15 8 2 0.25 1.092414 0.041275 16 8 4 0.5 1.092414 0.041275 17 12 6 0.5 0.995971 0.137719 18 10 8 0.8 1.036929 0.096761 19 12 10 0.833333 0.995971 0.137719 20 8 5 0.625 1.092414 0.041275 22 10 4 0.4 1.036929 0.096761 23 10 5 0.5 1.036929 0.096761 24 8 4 0.5 1.092414 0.041275 25 12 5 0.416667 0.995971 0.137719 P P(1-P)
0.566845 0.245532
1.5 1 0.5 0 1
3
5
7
9
11
13
15
17
19
21
-0.5
nomor 31 The number of imperfections in bond paper produced by a paper mill is observed over a period of several days. The data for 25 samples is shown in table bellow. Construct a control chart for the number of imperfections per square meter. Revise the limits, assuming special causes for points that are out of control. Sample Area Inspected Imperfections Sample Area Inspected Imperfections 1 150 6 16 200 6 2 100 8 17 150 4 3 200 5 18 200 7 4 150 4 19 150 14 5 250 10 20 100 4 6 100 11 21 100 8 7 150 3 22 200 9 8 200 5 23 300 12 9 300 10 24 250 7 10 250 10 25 200 5 11 100 5 12 200 4 13 250 12 14 300 8 15 300 12 Jawaban: Karena diminta control chart per sq meter, maka digunakan u-chart. Initial Control Chart Area Sample Inspected Imperfections %Imperfection UCL 1 150 6 0.04 0.087323 2 100 8 0.08 0.098191
LCL -0.00939 -0.02025
Area Sample Inspected Imperfections %Imperfection UCL LCL 3 200 5 0.025 0.080845 -0.00291 4 150 4 0.026667 0.087323 -0.00939 5 250 10 0.04 0.076424 0.001514 6 100 11 0.11 0.098191 -0.02025 7 150 3 0.02 0.087323 -0.00939 8 200 5 0.025 0.080845 -0.00291 9 300 10 0.033333 0.073161 0.004777 10 250 10 0.04 0.076424 0.001514 11 100 5 0.05 0.098191 -0.02025 12 200 4 0.02 0.080845 -0.00291 13 250 12 0.048 0.076424 0.001514 14 300 8 0.026667 0.073161 0.004777 15 300 12 0.04 0.073161 0.004777 16 200 6 0.03 0.080845 -0.00291 17 150 4 0.026667 0.087323 -0.00939 18 200 7 0.035 0.080845 -0.00291 19 150 14 0.093333 0.087323 -0.00939 20 100 4 0.04 0.098191 -0.02025 21 100 8 0.08 0.098191 -0.02025 22 200 9 0.045 0.080845 -0.00291 23 300 12 0.04 0.073161 0.004777 24 250 7 0.028 0.076424 0.001514 25 200 5 0.025 0.080845 -0.00291 U
0.038969072
0.12 0.1 0.08 0.06 0.04 0.02 0 -0.02 1
3
5
7
-0.04
Revised Control Chart
9 11 13 15 17 19 21 23 25
Area Inspected Imperfections %Imperfection UCL LCL 150 6 0.04 0.081903 -0.0106 100 8 0.08 0.092298 -0.02099 200 5 0.025 0.075706 -0.0044 150 4 0.026667 0.081903 -0.0106 250 10 0.04 0.071478 -0.00017 150 3 0.02 0.081903 -0.0106 200 5 0.025 0.075706 -0.0044 300 10 0.033333 0.068356 0.002948 250 10 0.04 0.071478 -0.00017 100 5 0.05 0.092298 -0.02099 200 4 0.02 0.075706 -0.0044 250 12 0.048 0.071478 -0.00017 300 8 0.026667 0.068356 0.002948 300 12 0.04 0.068356 0.002948 200 6 0.03 0.075706 -0.0044 150 4 0.026667 0.081903 -0.0106 200 7 0.035 0.075706 -0.0044 100 4 0.04 0.092298 -0.02099 100 8 0.08 0.092298 -0.02099 200 9 0.045 0.075706 -0.0044 300 12 0.04 0.068356 0.002948 250 7 0.028 0.071478 -0.00017 200 5 0.025 0.075706 -0.0044
Sample 1 2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 20 21 22 23 24 25
U
0.035652174
0.1 0.08 0.06 0.04 0.02 0 -0.02 -0.04
1
3
5
7
9
11
13 15 17
19 21
23
