tugas5ppmnadprils

TI 3201 – Pengendalian dan Penjaminan Mutu Tugas 5

Nadia Fadhilah Riza (13406069) Prilla Sista Lily Jane (13406080)

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EXERCISE 5-6

The net weight (oz) of a dry bleach product is to be monitored by 𝒙 and R control charts using a sample size of n=5. Data for 20 preliminary sample as follows

a. Set up 𝒙 and R control charts using those data. Does the process exhibit statistical control?



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Berdasarkan peta R dan peta xbar diatas, dapat dilihat bahwa semua data proses in control.

b. Estimate the process mean and standard deviation

process mean = CL xbar chart = 16.27

process standard deviation = σ =R

d2=

0.475

2.326= 0.2042

c. Does fill weight seem to follow a normal distribution



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d. If the specifications are at 16.2 ± 0.5, what conclusions would you draw about process capability?

C p =USL − LSL

6σ=

16.7 − 15.7

6(0.2042)= 0.8161

Karena C p = 0.8161 < 1, maka proses menghasilkan banyak produk yang tidak memenuhi spesifikasi (cacat).

e. What fraction of containers produced by this process is likely to be below the lower specification limit of 15.7 oz?

p = P x < 15.7 = Φ 15.7 − 16.268

0.2042 = Φ −2.78 = 0.0027

Berdasarkan perhitungan diatas, maka diketahui bahwa fraksi produk yang dibawah spesifikasi 15.7 oz sebesar 0.27% [2700 parts per million (ppm)] produk yang diproduksi

EXERCISE 5-12

The table shown presents 20 subgroups of five measurements on the critical dimension of a part produced by a machining process.

a. Set up 𝒙 and R control charts on this process. Verify that the process is in statistical control.



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Berdasarkan peta R dan peta xbar diatas, dapat dilihat bahwa semua data proses in control.

b. Following the establishment of control charts in part (a) above. 10 new samples were collected. Plot the 𝒙 and R control values on the control chart you established in part (a) and draw conclusions.



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Akan diplot nilai R dan xbar pada bagian (b) ke peta kontrol xbar dan R pada bagian (a). Dengan menggunakan UCL dan LCL R dan xbar pada bagian (a), maka dihasilkan peta kontrol sebagai berikut:



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Berdasarkan peta kontrol xbar dan R diatas, dapat diketahui bahwa penambahan sampel yang dilakukan menunjukkan terdapatnya pengerjaan proses yang tidak sesuai dengan spesifikasi proses yang diinginkan. Hal ini ditunjukkan dengan kesepuluh data sampel baru yang semuanya out of control. Namun, belum dapat diketahui apakah data out of control ini merupakan false alarm atau memang kesalahan proses produksi.

c. Suppose that the assignable cause responsible for the action signals generated in part (b) has been identified and adjustments made to the process to correct its performance. Plot the 𝒙 and R control values from the following new subgroups, which were taken following the adjustment, against the control chart limits established in part (a). What are your conclusions?

Akan diplot nilai R dan xbar pada bagian (c) ke peta kontrol xbar dan R pada bagian (a). Dengan menggunakan UCL dan LCL R dan xbar pada bagian (a), maka dihasilkan peta kontrol sebagai berikut:



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Berdasarkan peta control xbar dan R diatas, dapat diketahui bahwa data out sampel baru yang kesepuluhnya out of control pada bagian (b), merupakan kesalahan proses produksi yang dilakukan. Hal ini dibuktikan dengan peta control xbar pada bagian (c) dengan hanya 1 data saja yang out of control. Berarti terjadi perbaikan proses produksi yang dilakukan, yang menyebabkan proses produksi yang tidak memenuhi spesifikasi menjadi berkurang jumlahnya.

EXERCISE 5-22

Samples of n = 5 are taken from a process every hour. The 𝒙 and R control values for the particular quality characteristic are determined. After 25 samples have been collected, we calculate 𝒙 = 20 and 𝑹 = 4.56.

a. What are the three sigma control limits for 𝒙 and 𝑹 ?

Untuk peta R:

CL = R

UCL = R + 3σ = R + 3d3

R

d2= 4.56 + 3 0.864

4.56

2.326 = 9.64

LCL = R − 3σ = R − 3d3

R

d2= 4.56 − 3 0.864

4.56

2.326 = −0.5215

Untul peta xbar

CL = x

UCL = x +3

d2 nR = 20 +

3

(2.326) 54.56 = 22.63

LCL = x −3

d2 nR = 20 −

3

(2.326) 54.56 = 17.37



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b. Both charts exhibit control. Estimate the process standard deviation.

σ =R

d2=

4.56

2.326 = 1.96

c. Assume that the process output is normally distributed. If the specifications are 19 ± 5, what are your conclusions regarding the process capability?

C p =USL − LSL

6σ=

24 − 14

6(1.96)= 0.8501

Karena C p = 0.8501 < 1, maka proses menghasilkan banyak produk yang tidak memenuhi spesifikasi (cacat).

d. If the process mean shifts to 24, what is the probability of not detecting this shift on the first subsequent sample?

UCL = x +3

d2 nR = 24 +

3

(2.326) 54.56 = 26.63

LCL = x −3

d2 nR = 24 −

3

(2.326) 54.56 = 21.37

𝛽 = 𝑃 𝐿𝐶𝐿 ≤ 𝑥 ≤ 𝑈𝐶𝐿 = 𝛷 26.63 − 24

1.96 − 𝛷

21.37 − 24

1.96 = 𝛷 1.34 − 𝛷 −1.34

= 0.9099 − 0.0901 = 0.8198

Maka, probabilitas tidak dapat dideteksinya perubahan pada subsequent sample untuk pertama kalinya sebesar 0.8198

EXERCISE 5-44

Control charts for 𝒙 and 𝑺 have been maintained on a process and have exhibited statistical control. The sample size is n = 6. The control chart parameters are follows:

a. Estimate the mean and standard deviation of the process.

process mean = CL xbar chart = 16.27

process standard deviation = σ =S

c4=

1.738

0.9515= 1.83

b. Estimate the natural tolerance limits for the process.

Natural tolerance limit for process = three sigma above and below the mean

UCL = 708.2



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CL = 706

LCL = 703.8

c. Assume that the process output is well modeled by a normal distribution. Is specifications are 703 and 709, estimate the fraction noncorming.

𝑝 = 𝑃 𝑥 < 703 + 𝑃 𝑥 > 709 = 𝛷 703 − 706

1.83 + 1 − 𝛷

709 − 706

1.83

= 𝛷 −1.64 + 1 − 𝛷 1.64 = 0.0505 + 1 − 0.9505 = 0.1

Berdasarkan perhitungan diatas, maka diketahui bahwa fraksi produk yang tidak memenuhi spesifikasi sebesar 0.1% [100000 parts per million (ppm)]

d. Suppose the process mean shifts to 702 while the standard deviation remains constant. What is the probability of an out-of-control signal occurring on the first sample following the shifts?

UCL = x + A3S = 702 + 1.266 1.738 = 704.2

LCL = x − A3S = 702 − 1.266 1.738 = 699.8

𝛽 = 𝑃 𝐿𝐶𝐿 ≤ 𝑥 ≤ 𝑈𝐶𝐿 = 𝛷 704.2 − 702

1.83 − 𝛷

699.8 − 702

1.83 = 𝛷 1.20 − 𝛷 −1.20

= 0.8849 − 0.1357 = 0.7492 Maka, probabilitas tidak dapat dideteksinya perubahan pada subsequent sample untuk pertama kalinya sebesar 0.7492

e. For the shift in part (d), what is the probability of detecting the shift by at least the third subsequent sample?

β2 1 − β = 0.7492 2 1 − 0.7492 = 0.14

Maka, probabilitas tidak dapat dideteksinya perubahan pada subsequent sample untuk

ketiga kalinya sebesar 0.14

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