INDU 6331 Assignment 6 Advanced Quality Control

The data in table represent individual observations on molecular weight taken hourly from a chemical process. The target value of molecular weight is 1050 and the process standard deviation is thought to be about σ= 25. a) Set up a tabular cusum for the mean of this process. Design the cusum to quickly detect a shif of about 1.0σ in the process mean. σ= 25 μ0= 1.050 Magnitude of shift= 1,0(25)= 25 μ1= 1.050 + 1,0(25)= 1.075 ¿1.075− 1.050/¿ 2 =12.5 ¿ μ1− μ0 /¿ 2 =¿ K=¿ k= K/σ= 12,5/25= 0,5 h= 5 (recommended 4 or 5 in book) H= h*σ= 5(25)= 125 μ0+K= 1.050+12.5= 1.062,5 +¿ 0, xi−( μ0+K)+Ci−1 ¿ +¿=max ¿ Ci ¿ μ0-K= 1.050-12.5= 1.037,5 INDU 6331 Assignment 6 Advanced Quality Control −¿ 0,( μ0−K)−xi+Ci−1 ¿ +¿=max ¿ Ci ¿ −¿=0 +¿=C0 ¿ C0 ¿ Control limits tabular CUSUM chart UCL= H= 125 CL= 0 LCL= -H= -125 Considering a decision interval H = 5σ= 125, tabular CUSUM shows that there exists a shift of the data. The shift is detected in the upper-side CUSUM at observation 10 C ¿ +¿=171 ¿ ¿ . We can conclude that the process is out-of-control in the 10 observation, and considering N+ equal to 3 in that observation, it can be assumed that the shift occurred between observation 7 and observation 8. This analysis can be confirmed in the tabular CUSUM control chart below.