September 30, 2017 | Author: Bunga Safhira Wirata | Category: N/A
Statistical Quality Control Montgomery’s 6th edition Solutions for Chapter 06 Jan Rohlén
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Question 6.04 Sample No.
Ri
Xi
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
9 7 5 7 6 2 8 6 5 6 8 7 7 6 9 5 4 8 6 4
10 7.75 7.5 9 9.75 10.75 10.75 6.5 9 13.5 12.5 9.75 13.25 10.5 11 12.5 9.75 10.75 8.75 13.25
Table 1: Table 6E.4
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Statistical Quality Control
Chapter 06
(a) P
P
Ri R= = 6.25 m Sample Size
4
X= A2 0.729
D3 0
Xi = 10.325 m D4 2.282
U CL = X + A2 R = 10.325 + 0.729 × 6.25 = 14.88 X−R Chart CL = X = 10.325 LCL = X − A2 R = 10.325 − 0.729 × 6.25 = 5.77
U CL = D4 × R = 2.282 × 6.25 = 14.26 R Chart CL = R = 6.25 LCL = D4 × R = 0 × 6.25 = 0
The process is in-control.
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Chapter 06
(b) Specs. (350V ± 5V ) σ=
The real σ can be calculated: c = C p
R d2
=
6.25 2.059
= 3.0355
U SL − LSL 3550 − 3450 = = 5.491 6×σ 6 × 3.0355
The minimum capability index for existing processes is 1.33 (i.e., 43 ). Obviously, this process is higher (5.49 1.33). (c)
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Question 6.15 (a) X=
1000 72 = 20 s = = 1.44 50 50
Sample Size
4
X −S
A3 1.628
B3 0
B4 2.266
U CL = X + A3 R = 20 + 1.628 × 1.44 = 22.34 Chart CL = X = 10.325 LCL = X − A3 R = 205 − 1.628 × 1.44 = 17.66 S
U CL = B4 s = 2.266 × 1.44 = 3.26 Chart CL = s = 1.44 LCL = B s = 0 × 1.44 = 0 3
(b) Natural Tolerance Limits First of all, we need to calculate the real σ: σb = (
s C4
=
1.44 0.9213
= 1.563
U N T L = X + 3σ = 20 + 3 × 1.563 = 24.69 LN T L = X − 3σ = 20 − 3 × 1.563 = 15.31
(c) Specs Limits: 19 ± 4 U SL − LSL 23 − 15 = = 0.85 6σ 6 × 1.563 =⇒ the process is not capable!
c = C p
Cp < 1.33 (d)
Pbrework = P (S > U SL) = 1 − P (X ≤ U SL) = 1 − P ( 4 LaTeX Typesetting by : Amirkiarash Kiani
U SL − µ )= σb
Jan Rohlén
Statistical Quality Control
= 1 − φ(
Chapter 06
23 − 20 ) = 1 − φ(1.919) = 0.0275 = 2.75% 1.563
15 − 20 ) = φ(−3.199) 1.563 = 1 − 0.99931 = 0.00069 ∼ = 0.069% Total: Pb = 2.75% + 0.065 ∼ = 2.8%. Pbscrap = P (X < LSL) = φ(
(e) µ = 19
PbRework = 1−φ(
23 − 19 ) = 1−φ(2.56) = 1−0.9947 = 0.0053 = 0.53% 1.563
PbScrap = 1 − φ(
15 − 19 ) = φ(−2.563) = 0.0053 = 53% 1.563
Total: Pb = 0.53% + 0.53% = 1.06% Centring decreased the Rework but increased the Scrap. Cost analysis should be done.
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Question 6.21 Spec. Limits 100 ± 10, X = 104, R = 9.3 and n=5 d2 of (n=5) = 2.236 R 9.3 = = 3.998 d2 2.326 Natural Tolerance is: 6 × σ = 6 × 3.998 = 23.99 The Tolerance limit is wider than specification limit (2 × 10 = 20). Therefore, even adjusting the centre of the process (i.e., change the mean to 100) wouldn’t meet the specification: σ=
Cbp =
U SL − LSL 110 − 90 = = 0.83 6σ 6 × 2.998
Question 6.34 X ∼ N (µx , σx ) Y ∼ N (µy , σy )
)
Independent
each sample n=5. (a) We have mx = 20 Samples and my = 10 20 X
10 X
Rxi = 18.608
i=1
Ryi = 6.978
i=1
Estimated σx and σy ? P
Rxi 18.608 = = 0.9304 20 20 P Ryi 6.978 Ry = = = 0.6978 20 10 0.9304 0.6978 = 0.4 σby = = 0.3 σbx = 2.326 2.326 Rx =
(b) Axis must fit into the hole. Condition: P (X − Y < 0.09) = 0.006 What difference µ0 = µx − µy should be specified? 6 LaTeX Typesetting by : Amirkiarash Kiani
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D = X − y (if x and y are normally distributed then D would be normally distributed too). D ∼ N (µD , σD ) q √ σb = σbx2 + σby2 = 0.42 + 0.32 = 0.5 D − µ0 0.09 − µ0 P (D < 0.09) = P < σ0 σ0 0.09 − µ0 φ σ0
!
!
= 0.0006
0.09 − µ0 φ(−x) = 1 − φ(x) = φ − σ0
!
= 1 − 0.0006 = 0.994
By checking the table, the answer is 2.51. −
0.09 − µ0 = 2.51 =⇒ µ0 = 0.09 + 2.51σD = 1.345 σ0
Question 6.41 Type II Error =⇒ alarm by at least 3rd plot point =⇒ 1 − P {no alram by 3rd sample} |
σx =
{z
}
β3
√σ n
!
U CL − µ LCL − µ φ −φ σx σx !
104 − 98 96 − 98 √ √ φ −φ 8/ 5 8/ 5
!
!
= 0.66578 = β
1 − β 3 = 0.7049
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Question 6.42 ARL1 =
1 1 = = 2.992 1−β 1 − 0.66578
Question 6.45 No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Weight 16.11 16.08 16.12 16.1 16.1 16.11 16.12 16.09 16.12 16.1 16.09 16.07 16.13 16.12 16.1 16.08 16.13 16.15 16.12 16.1 16.08 16.07 16.11 16.13 16.1
P24
MR =
MRi 0.03 0.04 0.02 0 0.01 0.01 0.03 0.03 0.02 0.01 0.02 0.06 0.01 0.02 0.02 0.05 0.02 0.03 0.02 0.02 0.01 0.04 0.02 0.03
M Ri = 0.02375 24
i=1
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Chapter 06
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d2 = 1.128,
σb =
Chapter 06
MR 0.02375 = = 0.021055 d2 1.128
U CL = D4 M R = 3.267 × 0.02375 = 0.07759 CL = M R = 0.02375 LCL = D3 M R = 0 LSL − µ φ σx
!
16 − 16.1052 =φ 0.021055
!
∼ = 3 × 10−7 ∼ = 3 × 10−5 %
U CL = X + 3 Md2R = 16.1052 + 3 × 0.02375 = 16.1683 CL = X = 16.1052 LCL = X − 3 M R = 16.1052 + 3 × 0.02375 = 16.04204 d2
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Chapter 06
Yes, it is normally distributed. Question 6.49 x chart { U CL = 710, CL = 700 and LCL = 690 y chart { U CL = 18.08, CL = 0 and LCL = 7.979 (a) µ = x = 700,
σbx =
s 7.979 = = 8.661 C4 0.9213
(b) Specs 705 ± 15 !
U SL − µ LSL − µ = Pb = P {x < LSL}+p{x > U SL} = φ + 1−φ σbx σbx 690 − 700
=φ
8.661
720 − 700
+ 1−φ
8.661
φ(−1.15) + 1 − φ(2.31) = 0.1339 ' 13.3% (c) !
U CL − µ LCL − µ P = P {X < LCL}+P {X > U CL} = φ + 1−φ σ σ
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!
Jan Rohlén
Statistical Quality Control !
690 − 700 710 − 700 √ √ =φ +1−φ 8.661/ 4 8.661/ 4
Chapter 06 !
= 0.0208
(d) Pb = P {X − LCL} + P {X > U CL} LCL − µ
φ 690 − 693
!
σnew
new
U CL − µnew +1−φ σnew
710 − 693 q √ φ +1−φ 12/ 4 12 + (4)
!
!
= φ(−0.51)+1−φ(2.83) = 0.3108
(e)
ARL1 =
1 1 1 1 = = = 1−β 1 − P {notdetecting} P {detect} 0.3108
In average; after 3 points, we can expect alarm.
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