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

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=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: