Consider the network of Fig. 5-12(a). Distance vector routing is used, and the f
ID: 3706267 • Letter: C
Question
Consider the network of Fig. 5-12(a). Distance vector routing is used, and the following
vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D: (16, 12,
6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The cost of the links from C to B, D, and E,
are 6, 3, and 5, respectively. What is C’s new routing table? Give both the outgoing
line to use and the cost.
PLEASE type answer. PLEASE do not hand write. Thank you very much.
State Packets B 2 C 4 Seq Seq. Seq. Seq. Seq. Seq. Age Age AgeAge Age Age 7 Figure 5-12. (a) A network. (b) The link state packets for this network.Explanation / Answer
To find the final routing table of C, we will calculate intermediate routing table of C via B,D and E respectively. Then we will find which line is best to follow in order to reach other nodes. Given cost of links from C to B,D,E are 6,3,5 respectively so CB=6, CD=3, CE=5.
Routing table of C via B : (11,6,14,18,12,8)
CA=CB+BA= 6+5=11, CB=CB+BB= 6+0= 6, CC=CB+BC= 6+8= 14, CD=CB+BD= 6+12= 18,
CE=CB+BE= 6+6=12, CF=CB+BF= 6+2=8
Routing table of C via D : (19,15,9,3,12,13)
CA=CD+DA= 3+16=19, CB=CD+DB= 3+12= 15, CC=CD+DC= 3+6= 9, CD=CD+DD= 3+0= 3,
CE=CD+DE= 3+9=12, CF=CD+DF= 3+10=13
Routing table of C via E : (12,11,8,14,5,9)
CA=CE+EA= 5+7=12, CB=CE+EB= 5+6= 11, CC=CE+EC= 5+3= 8, CD=CE+ED= 5+9= 14,
CE=CE+EE= 5+0=5, CF=CE+EF= 5+4=9
Now observe the minimum value of CA,CB,CC,CD,CE,CF in the calculated routing table, except CC, because it will be 0.
Final routing table of C= (11,6,0,3,5,8) and the links to use are (B, B, -, D, E, B)