Computer Networks · TEE · papers
Q4(a) · Distance vector at z
2024-02-06 · AI-only Sem V variant
Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.
Q4(a). Initially each node knows only neighbor costs. For undirected edges u–v=1, u–y=2, v–x=3, y–x=3, v–z=6, x–z=2, apply distance vector and show distance-table entries at z. [10 marks]
Answer
Given
At z, neighbor costs c(z,v)=6 and c(z,x)=2; all original links are stated in the question.
To find
Distance table at z over synchronous exchanges.
Formula and why
LaTeX: D_z(d)=\min\{6+D_v(d),\ 2+D_x(d)\}Add the local link cost to each neighbor’s advertised distance. Round0 below uses the neighbors’ initialized direct-link vectors; next rounds use their previous completed exchange.
Solution
Step 1. Evaluate both candidate routes for every destination
| Exchange input round | Destination | Via v | Via x | Minimum |
|---|---|---|---|---|
| 0 | u | 7 | ∞ | 7 |
| 0 | v | 6 | 5 | 5 |
| 0 | x | 9 | 2 | 2 |
| 0 | y | ∞ | 5 | 5 |
| 1 | u | 7 | 6 | 6 |
| 1 | v | 6 | 5 | 5 |
| 1 | x | 9 | 2 | 2 |
| 1 | y | 9 | 5 | 5 |
| 2 | u | 7 | 6 | 6 |
| 2 | v | 6 | 5 | 5 |
| 2 | x | 9 | 2 | 2 |
| 2 | y | 9 | 5 | 5 |
| 3 | u | 7 | 6 | 6 |
| 3 | v | 6 | 5 | 5 |
| 3 | x | 9 | 2 | 2 |
| 3 | y | 9 | 5 | 5 |
Step 2. Read the converged routes
From z: u=6 viax→v→u; v=5 viax→v; x=2 direct; y=5 viax→y; z=0. The route to u throughx→y→u costs7 and is therefore not the minimum.