Consider the network of Fig 1a The first four columns of par

Consider the network of Fig. 1(a). The first four columns of part (b) show the delay vectors received from the neighbors of router J. Suppose that J has measured or estimated its delay to its neighbors. A, I, H. and K. as 10, 6, 15, and 5 msec, respectively. Update the new routing label for J.

Solution

To reach A,
via A-10msec
via I-6+24=30
via H-15+20=31
via K-5+21=26
It takes 10msec to reach A via A

To reach B,
via A-10+12=20
via I-6+36=42
via H-15+31=46
via K-5+28=33
It takes 20msec to reach B via A

To reach C,
via A-10+25=35
via I-6+18=24
via H-15+19=34
via K-5+36=41
It takes 24msec to reach C via I

To reach D,
via A-10+40=50
via I-6+27=33
via H-15+8=23
via K-5+24=27
It takes 23msec to reach D via H

To reach E,
via A-10+14=24
via I-6+7=13
via H-15+30=45
via K-5+22=27
It takes 13msec to reach B via I

To reach F,
via A-10+23=33
via I-6+20=26
via H-15+19=34
via K-5+40=45
It takes 26msec to reach F via I

To reach G,
via A-10+18=28
via I-6+31=37
via H-15+6=21
via K-5+31=36
It takes 21msec to reach G via H

To reach H,
via A-10+17=27
via I-6+20=26
via H-15+0=15
via K-5+19=24
It takes 15sec to reach H via H

To reach I,
via A-10+21=31
via I-6+0=6
via H-15+14=29
via K-5+22=27
It takes 6msec to reach I via I

To reach J,
It takes 0msec to reach J

To reach K,
via A-10+24=34
via I-6+22=28
via H-15+22=37
via K-5+0=5
It takes 5msec to reach K via K

To reach L,
via A-10+29=39
via I-6+33=39
via H-15+9=24
via K-5+9=14
It takes 14msec to reach L via K

10	A
20	A
24	I
23	H
13	I
26	I
21	H
15	H
6	I
0	-
5	K
14	K