91Ó°ÊÓ

First: Locate and connect desired ARLs on the in-control and out-of-control scales; Second: Solve for\(n\)equation

\({k^\prime } = \frac{{\Delta /2}}{{\sigma /\sqrt n }}\)

and round\(n\)up to the nearest integer.

Third: Use second line to connect the point on the\({K^\prime }\)scale with the point on the in-control ARL scale. Note where this second line crosses\({h^\prime }\)scale. Then\(h\)is

\(h = {h^\prime }\frac{\sigma }{{\sqrt n }}\)

Locate 250 on the in-control ARL scale, and\(4.8\)on the out-of-control scale and extending to\({k^\prime }\)scale on the Kemp Nonogram gives\(k = 0.7\).

Solve for\(n\)equation

\({k^\prime } = \frac{{\Delta /2}}{{\sigma /\sqrt n }} = \frac{{\sigma /2}}{{\sigma /\sqrt n }} = \frac{{\sqrt n }}{2}\)

because of shift of 1 standard deviation in the process mean, which yields

\(n = 1.96 \approx 2\)

Connect connect\(0.7\)on the\({K^\prime }\)scale to 250 on the in-control ARL scale and extending to\({h^\prime }\)you get\({h^\prime } = 2.85\)(Cusum procedure). Find\(h\)from

\(h = {h^\prime }\frac{\sigma }{{\sqrt n }}. = 2.85 \cdot \frac{1}{{\sqrt n }} \cdot \sigma = 2.015 \cdot \sigma \)

The out-of-control signal results when one of\({d_i},e_i^\prime s\)is larger than\(h\).

\(h = 2.015 \cdot \sigma .\)