CS 425: Computer Networks and Distributed Systems
Solutions for Homework Set 2
Problem 1 Ans:
The probability that an individual computer is down is 4/(24*5+4)~0.03.
Assuming failure independence of machines, the availability is therefor 1-0.033 =
0.999973
Problem 2 Ans:
Process
p
must receive a new group view containing only itself and it must
receive the message it sent. The problem is now in what order these events
should be delivered to
p
. If
p
receives the message first, then that would tell
p
that
q
and
r
received the message but the question implies that they did not
receive it so
p
must receive the group view first.
Problem 3 Ans:
The value timestamp reflects the operations that the replica manager has
applied. Replica managers also need to manage operations that they cannot yet
apply. In particular, they need to assign identifiers to new operations and they
need to keep track of which updates they have received in gossip messages,
whether or not that have applied them yet. The replica timestamp reflects
updates that the replica manager has received, whether or not it has applied
them all yet.
Problem 4 Ans:
The replica manager with timestamp (4,6,8) is the only one that can satisfy the
query or incorporate an update because it is the only one that can satisfy the
condition prev != valueTS.
Problem 5 Ans:
(i) The range of latencies for a read is 80-320 ms. A read with only block if all
three replica mangers are down. Therefore, the blocking probability of a
read is P(all replicas are dead) = 0.013.
(ii)The latency of a write is 320 ms. A write will block if any of the replica
mangers is down. Therefore, the blocking probability of a write is 1-P(all
replica managers are alive) = 1-0.993.
