### Table 8c Finiteness in declaratives and questions: Daniel

### Table 8.1. Evolution of Concepts to FEDORA (from Daniel amp; Lagoze (1997b)).

2001

Cited by 1

### Table A.1: Results for the Daniel and Martin problem, using full Newton iterations

### Table A.2: Results for the Daniel and Martin problem, repeated 10 times, using full Newton iterations

### Table A.3: Results for the Daniel and Martin problem, using damped Newton itera- tions

### Table A.4: Results for the Daniel and Martin problem, repeated 10 times, using damped Newton iterations

### Table 1 shows average link utilization for the measurement interval when the number of connections is 7. We can see that RDVRP more evenly distributes the data messages over the whole network than SEGAL does by investigating the link utilization over the whole network. And also, the total and the average of the link utilizations over the whole network indicate that RDVRP uses network resources more efficiently than SEGAL does.

"... In PAGE 6: ... Table1 : Link Utilization: (Measurement interval: 50,000 msec, Number of Connections: 7) 8 FUTURE WORK We have many possible directions for future work. The routing protocol proposed here is based on flat network architectures.... ..."

### Table 2.1 Quantiles for the average from the bootstrap, the sad- dlepoint and the saddlepoint mixture approximation for Daniels and Young apos;s data.

1997

Cited by 1

### Table 2: Variance of estimators of E( m) k Truncated perio. Daniell Tukey-Hanning Var( m)=k

"... In PAGE 15: ... Then the initial positive sequence estimator is the m which gives the rst negative ?m. Thus (19) can be written as b S(0) = 1 2 2m+1 X j=?2m?1 b j (24) The results are given in Table2 (along with the results for the other types of windows). To investigate the dependence of the variance of the estimator versus the length of the sample, we computed S(0)=k for all windows using a 500 long... In PAGE 20: ...221. Table2 gives us qS(0)=k = 5:531 10?4. Thus the 95% CI for m is [0:221 ? 0:889 5:531 10?4; 0:221 + 1:141 5:531 10?4] = [0:2205; 0:2216].... ..."

### Table 4: Variance of estimators of E( f) k Truncated perio. Daniell Tukey-Hanning Var( f)=k

"... In PAGE 20: ...1000 samples, 95 % CI). For a 1000-long sample, the expectation of f is -0.301. Table4 gives us qS(0)=k = 0:00689. Thus the 95% CI for f is [?0:301 ? 0:782 0:00689; ?0:301 + 1:383 0:00689] = [?0:306; ?0:291].... ..."