Question

Hosts A and B are each connected to a switch S via 1-Gbps links as below. The propagation delay on each link is 20 μs. S is a store-and-forward device; it begins retransmitting a received packet 35μs after it has finished receiving it. Calculate the total time required to transmit 10,000 bits from A to B in the following two cases:

• When all the 10,000 bits are transmitted in a single packet?
• When the 10,000 bits are transmitted in two 5,000 bits packets sent one right after the other.

Answer 1

Link Bandwidth = 1 Gbps

Propagation delay = 20 micro seconds

Answer 1

Single packet will take time = number of links* (Propagation time + Transmission time) + switch delay

= 20 $\mu s$ + (10000/ 10⁹) * 10⁶ $\mu s$ + 35 $\mu s$

= 2*(20 $\mu s$ + 10 $\mu s$) + 35 $\mu s$

= 95 $\mu s$

Answer 2

Both packets will be transmitted after each other.

First bit of first packet will reach to switch in 20 $\mu s$

First complete packet of 5000 bits will be received at switch S in

20 $\mu s$ + (5000/10⁹) * 10⁶ $\mu s$ = 25 $\mu s$

Switch S will start re-transmission after 35 $\mu s$ and Host B will start receiving packet B (first bit) after 20 $\mu s$

25 $\mu s$ + 35 $\mu s$ + 20 $\mu s$ = 80 $\mu s$

Meanwhile packet 2 has started transmission.

Similarly second packet of 5000 bits will be received at switch S in

= time when first packet is transmitted fully + transmission time for second packet = 25 $\mu s$ + 5 $\mu s$ = 30 $\mu s$

Packet 2 will wait at switch S till first packet is not transmitted fully. S will finish first packet re-transmission at time = 60 $\mu s$ + 5 $\mu s$ = 65 $\mu s$.

S will start sending second packet at 65 $\mu s$. Host B will receive full first packet after 80 $\mu s$ + 5 $\mu s$ = 85 $\mu s$. Host B will receive second packet after 85 $\mu s$ + 5 $\mu s$ = 90 $\mu s$

So two packets take lesser time than single packet.