Solution#5.5

Four hops means five routers are involved. The virtual circuit involved requires tying up 5x8 = 40 byte of memory for 1000 sec. The datagram implementation requires transmitting 12x4x200 = 9600 bytes of header over and above that of virtual circuit implementation needs. Thus the question comes down to the cost of 40,000 bytes-sec of memory versus 9600 byte-hops of circuit capacity. If memory is depreciated over 2x52x40x3600 = 1.5 x 10⁷ sec, a byte-sec costs 6.7x10^-8 cents, and 40,000 of them just cost over 2 millicents. If byte-hop costs 10^-6 cents then 9600 of them cost 9.6 millicents. Virtual circuits are cheaper for this set of parameters.

Solution#5.8

Going via B gives (11, 6, 14, 18, 12, 8).
Going via D gives (19, 15, 9, 3, 9, 10).
Going via E gives (12, 11, 8, 14, 5, 9).

Taking the minimum for each destination except C gives (11, 6, 0, 3, 5, 8). The outgoing lines are (B, B, -,D,E,B).

Solution#5.13

(a)The reverse path forwarding algorithm takes five rounds to finish. The packet recipients on these rounds are AC, DFIJ, DEGHIJKN, GHKN, and LMO, respectively. A total of 21 packets are generated.
(b)The sink tree needs four rounds and 14 packets.

Solution#5.14

There are multiple spanning trees possible.

Solution#5.17

The router has to do approximately the same amount of work queueing a packet, no matter how big it is. There is little doubt that processing 10 packets of 100 bytes each is much more work than processing 1 packet of 1000 bytes.

Solution#5.20

Here, Capacity of bucket C=8 Mbits, Arrival rate p=1Mbps, Transmission speed M=6 Mbps and we need to find burst time S.
It is found by substituting given values in the following formula.

S=C/(M-p)
S=1.6 sec.

Solution#5.29

ARP does not provide service to the network layer. It is a part of the network layer and helps provide a service to the transport layer. The issue of IP addressing does not occur in the data link layer. Data link layer protocols are like protocols 1 through 6 in Chap. 3, HDLC, PPP, etc. They move bits from one end of a line to the other.

Solution#5.36

With 16 bytes there are 2¹²⁸ or 3.4x10³⁸ addresses. If we allocate then at a rate of 10¹⁸ per second, they will last for 10¹³ years. This number is 1000 time the age of the universe. Of course, the address space is not flat, so they are not allocated linearly, but this calculation shows that even an allocation scheme that has an efficiency of 1/1000 (0.1 percent), one will never run out.

Solution#9

(a) Following table showsthe entries in the router tables of a,b,c,d and e for g.

(b)The following table shows the router table entries when source is also stored.