ABSTRACT
The amount of adjustment now is based on the difference between Y2 and µ0 which is
(7.8)
Now the operator adjusts the machine such that the current mean µ2 is reduced by k2(Y2 – µ0), given in Eq. (7.8). That is, the amount of adjustment is k2(Y2 – µ0) and the new mean is
Yi Xi Vi+ µi Wi Vi+ += =
Y1 µ1 W1 V1+ + µ0 d W1 V1+ + += =
Y1 µ1-d W1 V1+ +=
µ2 µ1 k1 d W1 V1+ +( )–= µ0 d k1 d W1 V1+ +( )–+= µ0 d 1 k1-( ) k1 W1 V1+( )–+=
Y2 X2 V2+ µ2 W2 V2+ += = µ0 d 1 k1-( ) k1 W1 V1+( )– W2 V2+( )+ +=
Y2 µ0-d 1 k1-( ) k1 W1 V1+( )– W2 V2+( )+=
µ3 µ2 k2 Y2 µ0-( )–= µ2 k2 d 1 k1-( ) k1 W1 V1+( )– W2 V2+( )+{ }–= µ0 d 1 k1-( ) k1 W1 V1+( )– k2 d 1 k1-( ) k1 W1 V1+( )– W2 V2+( )+{ }–+= µ0 d 1 k1-( ) 1 k2-( ) k1 W1 V1+( ) 1 k2-( ) k2-W2 V2+( ) (7.9)–+=
This setting procedure is illustrated in Figure 7.1. If we proceed in a similar manner, the mean of the (n + 1)st observation (the
reading after the nth adjustment) is
(7.10)
where Now the problem is to determine the optimal values of k1, k2, ..., kn such that
the mean after the nth adjustment is equal to µ0 as soon as possible. A probabilistic expression of this requirement results in the following:
(7.11)
and
(7.12)
Now let us examine these requirements.
