1. Justin Andrews<br />ACSG 520<br />Assignment #4<br />2. Indirect, as the network masks would be identical otherwise.<br />4. <br />Network AddressNet-hop AddressInterface192.16.7.0-----M0111.0.0.0-----M1Default:111.30.31.18M0170.14.0.0111.25.19.20M0145.80.0.0111.25.19.20M0<br />6. 145.80.14.26 becomes 10010001 01010000 00001110 00011010, which is shifted 28 bits to the right, becoming 00000000 00000000 00000000 00001001, which is 9, which is class B. The resulting network address is 145.80.0.0, and finding it, proceeds to forward the packet on toward its destination.<br />8. 145.14.192.71 becomes 145.14.192.0 after applying the subnet mask. The packet is then delivered through the use of ARP, with the next-hop address being 145.14.192.71, and the outgoing interface being m3.<br />10. 201.4.16.70 is first given the mask /26, resulting in 201.4.16.64, which doesn’t match the given network address. The same process is done with the mask /25, with the result being 201.4.16.0, again resulting in no matches. The third time, the mask /24 is used, with the result being 201.4.16.0, which also results in no matches. The final mask is applies, /22, with the resulting address being 201.4.16.0, and as there’s a match here, the next-hop address and the interface number, m1, are passed on to ARP.<br />12. <br />Network AddressNext-hop AddressInterface107.0.0.0-----M0<br />14. Star topology is used here.<br />16. A router can receive a packet with the destination address 140.24.7.42 if it’s part of the subnet 140.24.7.0/26, as it can be forwarded to the appropriate destination afterward.<br />18. <br />MaskNetwork addressNext-hop addressInterface/20120.14.64.0-----M0/23120.14.64.0-----M5/23120.14.78.0-----M6<br />20.<br />MaskNetwork addressNext-hop addressInterface/20120.14.112.0-----M4/24120.14.112.0 (to 16)-----M4<br />