A packet of size L bits is transmitted from a source to a destination through…

Concept

In store-and-forward packet switching, every node along the path must receive the entire packet before it begins transmitting it onward. The time to push one packet of L bits onto a single link of rate R is the transmission delay L/R seconds. So the total transmission delay equals the number of links along the path multiplied by L/R, independent of propagation, processing, and queuing delays.

Applying it here

1. Identify the path: the packet travels source → R1 → R2 → … → RN → destination, where R1…RN are the N intermediate routers.

2. Count the links on that path: source-to-R1, each router-to-next-router, and finally RN-to-destination. That chain of nodes is joined by N+1 links.

3. Apply store-and-forward at each link: the full L-bit packet is transmitted once per link, each taking L/R seconds.

4. Add the per-link delays, which happen sequentially rather than overlapping: total = (N+1) × (L/R) = (N+1)L/R.

Cross-check

Units: L bits divided by R bits-per-second gives seconds, multiplied by a dimensionless link count, so the result is in seconds. Boundary: with no intermediate routers (N = 0, a single direct link) the formula reduces to L/R, exactly one link's transmission delay, as expected.