Solution to HW1
Problem 1
Need to find shortest path from a single source s to a
single destination d.
Have a condition in the Dijkstra algo loop which
terminates the process whenever v is inducted into the
set S.
Problem 2
Whenever some node v is inducted into S such that d(v) = ,
then we know the graph is disconnected.
When a vertex v is inducted into S, d(v) is the shortest
distance from s to v. If this shortest distance is infinity,
then s is disconnected to v.
Problem 3
All answers were correct!
Problem 4
Given a graph, construct a new graph with same nodes
and edges, weight of an edge = negative of weight of
the corresponding edge in the given graph.
Minimum spanning tree in this new graph is
maximum weight spanning tree in the old one.
Problem 5
MST s are identical if a0, otherwise they may be different
(two MSTs have same number of edges).
Shortest paths are identical if a0 and b=0.
Yes, with reciprocal edge weights, minimum weight
spanning tree becomes maximum weight spanning tree.
Now the light weight edge crossing a cut in one is the
heavy weight edge crossing the cut.
By choosing the heavy weight one we get maximum
weight spanning tree.
By choosing the light weight one we get minimum
weight spanning tree.
The same sequence of choices in Kruskals leads to
minimum spanning tree in one and max. spanning
tree in another.
So, minimum spanning tree in one is identical to the
maximum spanning tree in another.
Problem 6
We know all the vertices in the Steiner tree given the
intermediate ones.
Need to find a minimum spanning tree in the induces
subgraph.
Problem 7
No.
All counter examples were valid!
Problem 8
The system can be represented with the queue lengths of
different sessions in different buffers.
The state process is a markov chain.
Let the current state of the markov chain be (…..Bi1(t),
Bi2(t),….. )
There is a positive probability of 0 arrival in this state
If there is no arrival, then it moves to a state where total
number of packets in the system is lesser, or equal but packets
have moved closer to the destination.
Given a particular state, there is a finite time such that if
there is no arrival in any slot in that time interval, then the
system moves to all zero state.
Given any finite number of slots, there is a positive
probability that the number of arrivals is 0 in each of
these slots.
There is a positive probability path from each state
to the all zero state.
Every state communicates to the all zero state.
All zero state communicates to some states.
All states which have positive probability path from the
zero state, communicate with each other, and does not
communicate to any state which does not have a positive
probability path from the all zero state.
All states which have positive probability path from the zero
state constitute a closed set (along with the all zero state).
All other states communicate to the zero state, and the zero
state does not communicate to them. So all other states are
open.
Problem 9
Sort all the reward values.
R = highest reward value.
(1)Remove all edges with reward value less than R
Find whether source is connected to the destination.
If so, stop.
If not, reduce R to the next highest reward value and go
to step (1)
The final value of R is the highest reward and the final
path is the highest reward path.
Complexity is O(E2)
Can do a binary search on the reward values, complexity
will be O(ElogE)
Problem 10
Unirate
Multicast
The feasible set looks like
ri + rj + ….. rp

Cl (constraint for link l)
……………….
………………………..
For every link constraint, create a link in the new unicast
network, with all sessions appearing in the constraint
traversing the link.
The capacity of this link is the same as the capacity of the
link in the constraint.
Consider any session i.
Join all the links traversed by session i by links
of very high capacity.
Every session is unicast in this network.
The number of sessions is the same in both networks.
The number of links are N(L + 1) in the new network.
The feasibility constraints are identical in both links.
Hence, the feasible sets are identical.
Multirate Multicast Network
The constraints are max(ri , rj
….  Cl
,
….. rp) + max(…..) +
Replace each constraint by linear constraints.
Consider an example:
max(r1 , r2 ) + r3 
Cl
The equivalent linear constraints are:
r1 + r 3 
Cl ,
r2 + r3 
Cl
Consider an example:
max(r1 , r2 ) + max(r3 , r4 ) 
Cl
The equivalent constraints are:
r1 + r3 
Cl ,
r2 + r3 
Cl
r1 + r 4 
Cl ,
r2 + r4 
Cl
If there are L links in the network, there can be at most
mN L linear constraints, where m is the maximum
number of receivers in the network, and N is the
number of sessions.
These constraints are of the form ri + rj + ….. rp

And involve r1 , r2 , ….. rM
Proceeding in the same manner as for unirate
multicast, we can now get a unicast network of M
sessions, mN L(M + 1) links.
A multirate, multicast can be reduced to unicast network,
but the unicast network has exponentially larger size!
Cl