In order to implement a DNS amplification attack, the attacker must trigger the
ID: 3771165 • Letter: I
Question
In order to implement a DNS amplification attack, the attacker must trigger the creation of a sufficiently large volume of DNS response packets from the intermediary to exceed the capacity of the link to the target organization. Consider an attack where the DNS response packets are 500 bytes in size (ignoring framing overhead). How many of these packets per second must the attacker trigger to flood a target organization using a 0.5-Mbps link? A 2 Mbps link? Or a 10 Mbps link? If the DNS request packet to the intermediary is 60 bytes in size, how much bandwidth does the attacker consumer to send the necessary rate of DNS request packets for each of the three cases?
Explanation / Answer
1byte =8bits
1mbps=10,00000bits
1mbps=1000kbps
1kbps=1000bits
one DNS Response packet size = 500 bytes
=500X8=4000 bits
for 0.5mbps how many packets of DNS
0.5mbps=500000bits=125X4000=5,00,000bits
no of packets=bandwidth/packetsize
no of packet of DNS for 0.5 mbps=5,00,000/4000=125 packets received(downlink)
0.5-Mbps : 125 packets of DNS response
no of packet of DNS for 2 mbps
converting 2-Mbps in to bits=2X10,00000=20,00000bits
one DNS Response packet size = 500 bytes
=500X8=4000 bits
no of packets=bandwidth/packetsize
no of packet of DNS for 2 mbps=20,00,000/4000=500 packets received(downlink)
no of packets for 10mbps
converting 10mbps into bits=10X10,00000=1,00,00000
one DNS Response packet size = 500 bytes
=500X8=4000 bits
no of packets=bandwidth/packetsize
no of packet of DNS for 10 mbps=1,00,00,000/4000=2500 packets received(downlink)
each DNS request packet size is = 60 bytes
= 60X8=480 bits
0.5mbps=0.5*1000kbps=500kbps
bandwidth=no of packets*packet size
s =125X480= 60000bits =60kbps(uplink)
for 0.5 mbps
for 2-Mbps: (500X480)=240000bits=240kbps (uplink)
for 10-Mbps: 2500*480=1200000bits=1200kbps(uplink)
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