Node A is using distance vector routing. The current distance table and routing table at Node A are shown below. Current distance table at Node A (rows: to; columns: via B C D E) to B: via B = 2 to C: via C = 1 to D: – to E: via E = 8 Current routing table at Node A dest | next hop | cost B | B | 2 C | C | 1 E | E | 8 Node A receives the following vector from neighbour Node B [A, 2 C, 2 D, 2 E, 5] Show the updated distance table at A after it processes this vector.
Figure required: Figure 1 (network): nodes A, B, C, D, E with link costs A–B 2, A–C 1, B–C 2, B–D 2, D–E 3, A–E 8. The distance and routing tables for Node A are transcribed into the question text.
Try an answer before revealing the guidance below.
Key Concepts
- Bellman-Ford update
- Distance table column “via B”
- c(A, B) = 2
- Neighbour vector
Answer Approach
- Only the “via B” column changes when a vector arrives from B.
- Each entry is c(A, B) plus B’s advertised distance to that destination.
- Leave the other columns as they were.
Full Answer
Answer status: Draft answer (unofficial). Revision notes, not an official marking scheme.
Via B column = c(A,B) + D_B(y) with c(A,B) = 2: to B = 2, to C = 2 + 2 = 4, to D = 2 + 2 = 4, to E = 2 + 5 = 7. Via C: to C = 1 (unchanged). Via E: to E = 8 (unchanged). New minimums: B 2 (via B), C 1 (via C), D 4 (via B), E 7 (via B), so the routing table becomes B→B 2, C→C 1, D→B 4, E→B 7.
D^A(y, B) = c(A, B) + D_B(y)Shortcuts: K concepts · A approach · F answer · R reviewed · B bookmark · ← / → previous / next