Q$3$:

$\backslash$table$[[$Subnet Number,Subnet Mask,Next Hop$],[128\mathrm{.}96\mathrm{.}170\mathrm{.}0,255\mathrm{.}255\mathrm{.}254\mathrm{.}0,$Interface $0],[128\mathrm{.}96\mathrm{.}168\mathrm{.}0,$Interface $1,],[128\mathrm{.}96\mathrm{.}166\mathrm{.}0,$R$2,],[128\mathrm{.}96\mathrm{.}164\mathrm{.}0,255\mathrm{.}255\mathrm{.}254\mathrm{.}0,$R$3]]$

Suppose a router has built up the routing table shown in the Table shown above. The router can deliver packets directly over interfaces $0$ and $1,$ or it can forward packets to routers R$2$ and R$3.$ Assume the router does the longest prefix match. Describe what the router does with a packet addressed to each of the following destinations:

$255\mathrm{.}255\mathrm{.}254\mathrm{.}0=11111111\mathrm{.}11111111\mathrm{.}11111110\mathrm{.}00000000$

$255\mathrm{.}255\mathrm{.}252\mathrm{.}0=11111111\mathrm{.}11111111\mathrm{.}11111100\mathrm{.}00000000$

$1-128\mathrm{.}96\mathrm{.}171\mathrm{.}92,(10000000\mathrm{.}01100000\mathrm{.}10101011\mathrm{.}01011100)$

Applying the subnet mask $255\mathrm{.}255\mathrm{.}254\mathrm{.}0,$ we get $128\mathrm{.}96\mathrm{.}170\mathrm{.}0.$ Use interface $0$ as the next hop.

$2-128\mathrm{.}96\mathrm{.}167\mathrm{.}151(10000000\mathrm{.}01100000\mathrm{.}10100111\mathrm{.}10010111)$

Applying subnet mask $255\mathrm{.}255\mathrm{.}254\mathrm{.}0,$ we get $128\mathrm{.}96\mathrm{.}166\mathrm{.}0($next hop is Router $2).$ Applying subnet mask $255\mathrm{.}255\mathrm{.}252\mathrm{.}0,$ we get $128\mathrm{.}96\mathrm{.}164\mathrm{.}0($next hop is Router $3).$ However, $255\mathrm{.}255\mathrm{.}254\mathrm{.}0$ is a longer prefix. Use Router $2$ as the next hop.

$3-128\mathrm{.}96\mathrm{.}169\mathrm{.}192(10000000\mathrm{.}01100000\mathrm{.}10101001\mathrm{.}11000000)$

Applying subnet mask $255\mathrm{.}255\mathrm{.}254\mathrm{.}0,$ we get $128\mathrm{.}96\mathrm{.}168\mathrm{.}0.$ Use interface $1$ as the next hop.

$4-128\mathrm{.}96\mathrm{.}165\mathrm{.}121(10000000\mathrm{.}01100000\mathrm{.}10100101\mathrm{.}01111001)$

Applying subnet mask $255\mathrm{.}255\mathrm{.}252\mathrm{.}0,$ we get $128\mathrm{.}96\mathrm{.}164\mathrm{.}0.$ Use Router $3$ as the next hop.

$3$