Databases Reference
In-Depth Information
2
3
2
3
2.6
1
5.204
3.6
3.6
3.6
3.6
4
5
×
4
5
1
1
2.617
2
2
2
2
1.617
1
1
1
1
1
1
=
2.617
2
2
2
2
1
1
1
1
1
1
1
2.617
2
2
2
2
1
1
2.617
2
2
2
2
Figure 9.14: Replace y by 1.617
The solution for y is y = 17.4/10.76 = 1.617. The improved estimates of U and
V are shown in Fig. 9.14.
We shall do one more change, to illustrate what happens when entries of M
are blank. We shall vary u
31
, calling it z temporarily. The new U and V are
shown in Fig. 9.15. The value of z affects only the entries in the third row.
2
4
3
5
×
2
4
3
5
2.6
1
5.204
3.6
3.6
3.6
3.6
»
-
1
1
2.617
2
2
2
2
1.617
1
1
1
1
z
1
=
1.617z + 1
z + 1
z + 1
z + 1
z + 1
1
1
1
1
1
1
1
2.617
2
2
2
2
1
1
2.617
2
2
2
2
Figure 9.15: u
31
becomes a variable z
We can express the sum of the squares of the errors as
2
+
2
+
1−(z + 1)
2
+
4−(z + 1)
2
2−(1.617z + 1)
3−(z + 1)
Note that there is no contribution from the element in the second column of
the third row, since this element is blank in M . The expression simplifies to
(1−1.617z)
2
+ (2−z)
2
+ (−z)
2
+ (3−z)
2
The usual process of setting the derivative to 0 gives us
−2×
1.617(1−1.617z) + (2−z) + (−z) + (3−z)
= 0
whose solution is z = 6.617/5.615 = 1.178. The next estimate of the decompo-
sition U V is shown in Fig. 9.16.
2
9.4.4
Optimizing an Arbitrary Element
Having seen some examples of picking the optimum value for a single element in
the matrix U or V , let us now develop the general formula. As before, assume
