Download Homework, sheet 10

Homework, sheet 10
July 4, 2013
Exercise 1. Consider a single-sender with the CDMA code (1, 1, −1, 1, 1, −1, −1, 1).
What would be the sender’s output for the input bits (1, 0)?
Answer: Using Zi,m = di · cm , where c = (1, 1, −1, 1, −1, −1, 1) and d =
(1, −1), as an output we get (1, 1, −1, 1, 1, −1, −1, 1) and (−1, −1, 1, −1, −1, 1, 1, −1)
corresponding to bits 1 and 0 respectively.
Exercise 2. Codes c1 , . . . , cn are orthogonal if ci ·cj = 0 for i 6= j and if ci ·ci =
const for all i. If two CDMA senders have codes (1, 1, 1, −1, 1, −1, −1, −1) and
(1, −1, 1, 1, 1, 1, 1, 1), would the corresponding receivers be able to decode the
data correctly? Justify.
What is the number of orthogonal codes of length N ?
Answer: No, because the CDMA codes are not orthogonal as c1 · c2 = 1 · 1 +
1 · (−1) + 1 · 1 + (−1) · 1 + 1 · 1 + (−1) · 1 + (−1) · 1 + (−1) · 1 = −2.
The number of orthogonal codes of length N is N .
Exercise 3. Suppose an 802.11b station is configured to always reserve the
channel with the RTS/CTS sequence. Suppose this station suddenly wants to
transmit 1,000 bytes of data, and all other stations are idle at this time. Assume
a transmission rate of 11 Mbps. As a function of SIFS and DIFS, and ignoring
propagation delay and assuming no bit errors, calculate the time required to
transmit the frame and receive the acknowledgement.
Answer: A frame without data is 32 bytes long. Assuming a transmission
rate of 11 Mbps, the time to transmit a control frame (such as RTS frame, a
CTS frame, or an ACK frame) is (256 bits)/(11 Mbps) = 23 µsec. The time
required to transmit the data frame is (8256 bits)/(11 Mbps) = 751 µsec.
DIFS + RTS + SIFS + CTS + SIFS + FRAME + SIFS + ACK = DIFS + 3
SIFS + (3 · 23 + 751) µsec = DIFS + 3 SIFS + 820 µsec.
Exercise 4. One proposed solution that allowed mobile users to maintain their
IP addresses as they move among foreign networks was to have a foreign network
advertise a highly specific route to the mobile user and use the existing routing
infrastructure to propagate this information throughout the network. We identified scalability as one concern. Suppose that when a mobile user moves from
one network to another, the new foreign network advertises a specific route to
the mobile user, and the old foreign network withdraws its route. Consider how
routing information propagates in a distance-vector algorithm (particularly for
the case of interdomain routing among networks that span the globe).
a. Will other routers be able to route datagrams immediately to the new foreign
network as soon as the foreign network begins advertising its route?
b. Is it possible for different routers to believe that different foreign networks
contain the mobile user?
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c. Discuss the timescale over which other routers in the network will eventually
learn the path to the mobile users.
Answer:
a. No. All the routers might not be able to route the datagram immediately.
This is because the Distance Vector algorithm (as well as the inter-AS routing protocols like BGP) is decentralized and takes some time to terminate.
So, during the time when the algorithm is still running as a result of advertisements from the new foreign network, some of the routers may not be able
to route datagrams destined to the mobile node.
b. Yes. This might happen when one of the nodes has just left a foreign network
and joined a new foreign network. In this situation, the routing entries from
the old foreign network might not have been completely withdrawn when the
entries from the new network are being propagated.
c. The time it takes for a router to learn a path to the mobile node depends
on the number of hops between the router and the edge router of the foreign
network for the node.
Exercise 5. Two communication nodes 1 and 2 send files to a third node 3.
Files from 1 and 2 require on average R1 and R2 time units for transmission,
respectively. Node 3 processes a file of node i (i = 1, 2) in an average of Pi time
units and then requests another file from either node 1 or node 2 (the rule of
choice is left unspecified). If λi is the throughput of node i in files sent per unit
time, what is the region of all feasible throughput pairs (λ1 ,λ2 ) for this system?
Answer:
We represent the system as shown in the figure. The number of files in the
entire system is exactly one at all times. The average number in node i is λi Ri
and the average number in node 3 λ1 P1 + λ2 P2 . Therefore the throughput pairs
(λ1 , λ2 ) must satisfy (in addition to nonnegativity) the constraint λ1 (R1 + P1 ) +
λ2 (R2 + P2 ) = 1.
If the system were slightly different and queueing were allowed at node 3, while
nodes 1 and 2 could transmit at will, a different analysis would apply. The
transmission bottleneck for the files of node 1 implies that λ1 ≤ R11 .
Similarly for node 2 we get that λ2 ≤ R12 .
Node 3 can work on only one file at a time. If we look at the file receiving service
at node 3 as a system and let N be the average number receiving service at node
3, we conclude from Little’s theorem that λ1 P1 + λ2 P2 = N and N ≤ 1. This
implies that λ1 P1 + λ2 P2 ≤ 1.
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