INDU 6331 Assignment 5 Solutions

9.1. 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 shift 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 K = /μଵ − μ଴/ 2 = /1.075 − 1.050/ 2 = 12.5 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 C୧ ା = max[0, x୧ − (μ଴ + K) + C୧ିଵ ା ] μ0-K= 1.050-12.5= 1.037,5 C୧ ା = max[0, (μ଴ − K) − x୧ + C୧ିଵ ି ] C଴ ା = C଴ ି = 0 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 (Cଵ଴ ା = 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. 9.25. Rework exercise 9.1. using an EWMA control chart with λ=0,1 and L= 2,7. Compare results to those obtained with the cusum. σ= 25 μ0= 1.050 λ= 0,1 L= 2,7