UCL A R UCL A R UCL D R LCL D R ... Massachusetts Institute of Technology and the Singapore-MIT Alliance
UCL A R UCL A R UCL D R LCL D R ... Massachusetts Institute of Technology and the Singapore-MIT Alliance
UCL A R UCL A R UCL D R LCL D R ... Massachusetts Institute of Technology and the Singapore-MIT Alliance

UCL A R UCL A R UCL D R LCL D R ... Massachusetts Institute of Technology and the Singapore-MIT Alliance

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Text 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

    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

    ba r

    Rbar = 40.835

    D3 = 0 D4 = 2.115

    UCL = 86.366

    LCL = 0

    ba r

    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:

    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+ − =   

    − 5.26 Φ−   

     

    +

      

    − 9.26 Φ−   

     (1 99995.0 9.251

    So our percentages are: 5.26

    (1 99936.0 99936.0

    155.

    4.26 9.25 155.

    1 9.26 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: