CS HW #3
a) If there are no routers to connect the 20 computers and there needs to be a clique
graph, then the Gauss formula must be used. This means that there must be N(N
1)/2 wires, or (N1)+(N2)+(N3)…+3+2+1 wires. In this case, there must be
19+18+17…+3+2+1 or 190 wires.
The total number of connections is 26: one connection between each router and
computer, and 6 amongst the routers, connecting them all. The best case of hops
from one computer to another is if both computers are attached to the same router,
or two hops. The worst case is if they are connected to two separate routers, or 3
Since 1 byte equates to 8 bits, then 4 bytes equals 32 bits. Therefore, there are 32 bits in
an IP address. If the first 22 bits out of the 32 are the same in the system of computers,
then there are 10 bits left that can vary. Each bit has two choices, 0 or 1, so there are 2
or 1,024 different possible computers.
a) Each tweet consists of 140 bytes. If each person tweets about 10 times a day,
assuming the worstcase scenario, the LoC would need 1,400 bytes per person
each day (10 x 140). This would mean 511,000 bytes per year (365 x 1,400).
b) One person will need 30,660,000 bytes of storage for 60 years (511,000 x 60).
This means that it would take 9,198,000,000,000,000 bytes of storage, or 9.198 x 10 bytes to compensate for 3,000,000 U.S. Citizens for 60 years each
(30,660,000 x 300,000,000).
c) Working with 9.198 x 10 , the total number of GB of storage for the tweets is
9,198,000GB [9.198 x 10 /10 (byte▯gigabyte conversion)]. Since 10GB cost a
dollar, then it would cost $919,800 to store the tweets (9,198,000GB/$10 per GB).
d) The rate of price decrease is too great to cause a problem for LoC, even with the
rate of user increase at 1.5% per year. Therefore, the LoC should not have a
financial problem storing the tweets.
e) The global population increase rate falls somewhere between 1 and 2%, which is