UCL A R UCL A R UCL D R LCL D R...Massachusetts Institute of Technology and the Singapore-MIT...
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Massachusetts Institute of Technology and the Singapore-MIT Alliance Programme 2.830 Control of Manufacturing Processes Solution to Assignment 5 Problem 1: Montgomery 5-12 Problem 5-12 X bar Chart Problem 5-12 R bar Chart 0 5 10 15 20 25 30 35 40 X 0 5 10 15 20 25 30 35 40 0 10 20 30 40 50 60 70 80 90 R 100 110 120 130 140 150 160 170 180 190 xbarbar = 130.87 A2 = 0.577 UCL = 154.4318 LCL = 107.3082 bar Rbar = 40.835 D3 = 0 D4 = 2.115 UCL = 86.366 LCL = 0 bar Run Number Run Number %this is for the X bar chart A2=0.577; %Table VI , n=5 UCLx=xbarbar+A2*Rbar; LCLx=xbarbar-A2*Rbar; %this is for the R chart D3=0.0; %Table VI , n=5 D4=2.115; %Table VI , n=5 UCLr=Rbar*D4; LCLr=Rbar*D3; Problem 2: Montgomery 5-17 First we get the correction factors from Table VI, with n=5: A 2 = 0.577 D 3 =0 D 4 =2.115 a) for Xbar chart, Xbarbar=26.5 5 . 662 9 UCL = x + R A = + * 577 . 0 = 708 . 26 2 25 25 5 . 662 9 UCL = x − R A = − * 577 . 0 = 292 . 26 2 25 25 for R chart, Rbar=0.36 9 UCL = R D = * 115 . 2 = 761 . 0 4 25 9 LCL = R D = * 0 = 0 . 0 3 25 b) We have to get the fraction of parts between the control limits. Note that we are assuming control limits of +-3sigma, this can also be obtained from sigma=Rbar/d 2 =0.36/2.326=0.155;
Transcript of UCL A R UCL A R UCL D R LCL D R...Massachusetts Institute of Technology and the Singapore-MIT...
Massachusetts Institute of Technology and the Singapore-MIT Alliance Programme 2.830 Control of Manufacturing Processes
Solution to Assignment 5
Problem 1: Montgomery 5-12
Problem 5-12 Xbar Chart Problem 5-12 Rbar Chart
0 5 10 15 20 25 30 35 40
X
0 5 10 15 20 25 30 35 40 0
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R
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xbarbar = 130.87
A2 = 0.577
UCL = 154.4318
LCL = 107.3082
bar
Rbar = 40.835
D3 = 0 D4 = 2.115
UCL = 86.366
LCL = 0
bar
Run Number Run Number
%this is for the X bar chart A2=0.577; %Table VI , n=5UCLx=xbarbar+A2*Rbar;LCLx=xbarbar-A2*Rbar;
%this is for the R chart D3=0.0; %Table VI , n=5D4=2.115; %Table VI , n=5UCLr=Rbar*D4;LCLr=Rbar*D3;
Problem 2: Montgomery 5-17
First we get the correction factors from Table VI, with n=5:A2= 0.577 D3=0 D4=2.115
a) for Xbar chart, Xbarbar=26.5
5.662 9UCL = x + R A = + * 577.0 = 708.26 2 25 25 5.662 9UCL = x − R A = − * 577.0 = 292.26 2 25 25
for R chart, Rbar=0.36 9UCL = R D = * 115.2 = 761.0 4 25
9LCL = R D = * 0 = 0.0 3 25 b) We have to get the fraction of parts between the control limits. Note that we are assuming control limits of +-3sigma, this can also be obtained from sigma=Rbar/d2=0.36/2.326=0.155;
1
1
99506.0 ) 00499.0= − ) (1+ −=
− 5.26Φ−
+
− 9.26Φ−
(1 99995.0
9.251
So our percentages are: 5.26
(1 99936.0 99936.0
155.
4.26 9.25 155.
19.26 4.26
155.
155.
c)
Problem 3:
Problem 4:
00128.0)=Φ− Φ−
This is acceptably close to the book answer, the difference is the result of slight round-off error.
−) (1+−=
−
+
−
Problem 5:
