Project ForgeSuppose the weights of the two previously-unused links (R1-R2 and R4-R6) are changed to 2 each, and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?
Explanation
With R1-R2=2 and R4-R6=2, recomputing all shortest paths: R1-R2's direct edge (2) is now cheaper than via R3 (5) — R1-R2 becomes used. R4-R6's direct edge (2) is now cheaper than via R5 (5) — R4-R6 becomes used. But this also makes R5-R6's shortest route go via R4 (1+2=3) instead of direct (4) — so the direct R5-R6 edge becomes unused instead. Net result: exactly 1 link (R5-R6) remains unused.
Source: GATE CS 2010 (2010)
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