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CS/ECE 457 Fall 2008
Homework 2 Solution
Problem 1. (1 point):
Let’s say the FCC allocated a total of 40Mhz for a new service and divided this into two 20Mhz bands, one for the upstream direction and the second for the downstream direction. This service is bidirectional, with each communication ses- sion requiring one 20khz channel in each direction. It is an FDMA/FDD system. Assume that 20 channels are set aside in each direction for control messages (which means these channels are unavailable to carry user data). Assume that a service provider was able to purchase the full 40Mhz band for a coverage area of 96 cells to provide this service. The service provider chooses a reuse factor of 4. What is the maximum number of simultaneous communication sessions that can be supported by this ser- vice provider?
Answer : channels. Of these 20 channels are used for control. This leaves 980 channels for com-
munication sessions. In the system with a reuse factor of 4, these 980 channels can be repeatedly used times.
Thus the maximum number of simultaneous bidirectional communication sessions is.
Problem 2. (1 point):
Consider a video stream, which is created using a resolution of (640 x 480) pixels per frame, 8 bits per pixel, and a frame rate of 20 frames/sec. What is the smallest-rate SONET signal required to carry this video stream?
Answer: The compressed video stream rate is b/s which is 49.152Mbps. This signal fits into a SONET
OC1 signal because the User data rate of an OC1 signal is 49.536Mbps.
Problem 3. [2 points]
Consider the packet-based multiplexer shown in Fig. 1. The output link transmission rate is 10Mbps. Assume packets des- tined to this output link arrive as shown in Table 1.
Fill in the departure times of packets under the two scheduling schemes:
a. First-Come First-Served (FCFS)
b. Priority queueing (1: high priority; 2: low priority)
Packet 1 and 3 each have an emission delay is sec, while packet 3 has an emission delay of
Table 1: Packet arrivals and departures
Packet Priority Length in bytes
Time of arrival (packet has fully arrived at the listed time instant; from this instant, it awaits transmission on the output link)
Time of departure (when emission of the whole packet is complete) under FCFS
Time of departure under Priority queueing
1 1 9000 t = 0.2 sec 0.2072 0. 2 2 64000 t = 0.202 sec 0.2584 0. 3 1 9000 t = 0.205 sec 0.2656 0.
( 20 Mhz ) ⁄ ( 20 khz )= 1000 96 ⁄ 4 = 24 980 × 24 = 23520
640 × 480 × 8 × 20
Fig 1 Packet-based multiplexer
Link 1
Link L
Link 2 Output link: 10Mbps
( 8 × 9000 ) ⁄ ( 10 Mbps )=0.
sec. Under FCFS, packet 2 transmission can only begin after packet 1 is fully emitted (i.e.,
at 0.2072) even though it arrives at 0.202. Packet 3 transmission starts at 0.2584 since packet 2 transmission completes then.
Under Priority queueing, because both packets 2 and 3 are in queue when packet 1 completes transmission, packet 3 is trans-
mitted first as it is of higher priority.
Problem 4. (1 point):
Assume that the three switches shown in Fig. 2 are connectionless packet switches. All the links are bidirectional SONET OC3 links. Ports within each switch are labeled ‘a’, ‘b’, etc. Routing tables at the three switches are shown below. We assume there are no summarized addresses; instead the complete address of the host interfaces are stored in the destination address column. Also, we omit the next-hop node in the routing tables and instead show only the output port (interface) corresponding to each destination address. For each packet listed in Table 3, determine the route, i.e., identify the switches and ports through
which each packet will traverse.
Problem 5. [2 points]
Assume that the switches shown in Fig. 2 are SONET circuit switches. All switches operate at the OC1 crossconnect rate. Switch SW1 consists of multiplexers, demultiplexers, and a 3-stage non-blocking space switch. The space switch consists of five input-stage switches, five output-stage switches and a minimum number of intermediate-stage switches. Answer the fol-
Table 2: Routing tables
Routing table at switch SW
Routing table at switch SW
Routing table at switch SW
Destination address
Output port
Destination address
Output port
Destination address
Output port H1 a H1 a H1 b H2 b H2 a H2 b H3 c H3 a H3 b H4 e H4 a H4 d H5 e H5 a H5 c H6 d H6 c H6 b
Table 3: Routes taken by packets
Packets Routes (list all the switches & ports on the path taken by each packet) From H3 to H5 SW1 (c, e); SW3 (b-c) From H6 to H4 SW2 (c,a); SW1 (d,e); SW3 (b-d)
( 8 × 64000 ) ⁄ ( 10 Mbps )=0.
Host H
Host H
Host H
Host H
a b c
d e
b
a
c
d
a (^) b
c
SW
SW
SW
Host H
Host H
OC
OC
OC
OC
OC
OC
OC
OC
OC
Fig. 2 Network of switches and hosts
Answer:
For routes with a choice, either answer is acceptable.
Problem 7. (3 points)
Assume that the bridge IDs are the numbers shown in the figure, i.e., the bridge ID of bridge 1 is 1, and so on.
Table 4: Steps of Dijkstra’s shortest path algorithm
Iteration M
0 {c} 3, {c-b} 1, {c-e} 6, {c-f}
1 {c,e} 3, {c-b} 2, {c-e-d} 1, {c-e} 3, {c-e-f}
2 {c,e,d} 4, {c-e-d-a} 3, {c-b} or (c-e-d-b} 2, {c-e-d} 1, {c-e} 3, {c-e-f} or {c-e-d-f} 3 {c,e,d,b,f} 4, {c-e-d-a} or 4, {c-b-a}
3, {c-b} or (c-e-d-b} 2, {c-e-d} 1, {c-e} 3, {c-e-f}or {c-e-d-f} 4 {c,e,d,b,f,a} 4, {c-e-d-a} or 4, {c-b-a}
3, {c-b} or (c-e-d-b} 2, {c-e-d} 1, {c-e} 3, {c-e-f}or {c-e-d-f}
Table 5: Routing table at switch c
Destination address Next-hop
a e or b b b or e d e e e f e
a
b
c
d
e f 1
Fig. 3 An example network
Da Db Dd De Df
∞ ∞
∞
Bridge 3
LAN B
LAN A
LAN D
Bridge 1 Bridge 4
Bridge 2
1
1 2
2
(^112)
2
LAN C
LAN E
H 1
H 2
3
a. What are the initial BPDUs sent by the bridges at t=0? Use the class notation for BPDUs (i.e., root ID, root path cost, bridge ID). Fill in the table below (one row is filled in as an example):
b. Upon receiving BPDUs (those sent at t=0) from its neighbors, in what state (forwarding or blocked) does bridge 3 place its port 2? Answer: Forwarding state. Bridge 3 receives [4,0,4] on its port 2, but since its BPDU is lower, it assumes it is the designated bridge for LAN A (which is connected to its port 2), and hence it places this port in a forwarding state.
c. What BPDUs do the bridges send out at the next time instant, say t=1, after computing a new root and root path cost?
d. After the spanning tree converges, in what state (forwarding or blocked) does bridge 3 place its port 2?
Answer: Blocked state. The spanning tree is as shown below (not required for the answer). Since Bridge 4 is one- hop away from the root, while Bridge 3 is two hops away, Bridge 4 is the designated bridge for LAN A, and hence bridge 3’s port2, which connects to LAN A needs to be blocked.
Table 6: BPDU generated at t=
Bridge BPDU Bridge 1 [1,0,1] Bridge 2 [2,0,2] Bridge 3 [3,0,3] Bridge 4 [4,0,4]
Table 7: BPDUs sent at t=
Bridge BPDU sent at t= Bridge 1 [1,0,1] Bridge 2 [1,1,2] Bridge 3 [2,1,3] Bridge 4 [1,1,4]
Bridge 1
Bridge 4
Bridge 3
Bridge 2
LAN C
LAN B
LAN E LAN A
LAN D
Problem 9. (5 points):
Consider the network shown in Fig. 5. All the subnets in this network are Ethernet LANs with subnet mask /24, except for the subnet 193.41.50.128 whose subnet mask is 255.255.255.192, as shown in Fig. 5. The subnet IDs for the Ethernet hubbed LANs are shown next to each LAN, and the host IDs are shown next to the host and router interfaces. For example, the IP address of host H1’s NIC is 158.67.9.17.
The routing tables in the IP routers and at some of the hosts are shown in the tables below. The /32 rows, corresponding to each router’s and host’s own interfaces, are not shown in the tables below, but assume that these rows are present in the routing tables. Note that the default gateway row is not mandatory. For example, there is no such row in the routing table at IP router, R2.
Table 10: Routing table at IP router R
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
189.90.43.0 255.255.255.0 G=0 189.90.43.2 189.90.43.
129.15.16.0 255.255.255.0 G=0 129.15.16.1 129.15.16.
193.41.50.128 255.255.255.192 G=0 193.41.50.129 193.41.50.
198.80.94.0 255.255.255.0 G=1 189.90.43.1 189.90.43.
198.80.94.90 255.255.255.255 G=1 189.90.43.3 189.90.43.
158.67.9.0 255.255.255.0 G=0 158.67.9.1 158.67.9.
0.0.0.0. /0 G=1 189.90.43.3 189.90.43.
Table 11: Routing table at IP router R
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
193.41.50.128 255.255.255.192 G=0 193.41.50.131 193.41.50.
Router R
Router R
Router R
Host H
Host H
Host. H
Host H
198.80.94.
Fig. 5 Network of IP routers, Ethernet hubbed LANs and hosts
195.67.89.0 255.255.255.0 G=0 195.67.89.1 195.67.89.
143.56.78.0 255.255.255.0 G=0 143.56.78.1 143.56.78.
198.80.94.90 255.255.255.255 G=0 143.56.78.1 143.56.78.
189.90.43.0 255.255.255.0 G=0 189.90.43.3 189.90.43.
Table 12: Routing table at IP router R
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
198.80.94.0 255.255.255.0 G=0 198.80.94.1 198.80.94.
167.18.19.0 255.255.255.0 G=0 167.18.19.1 167.18.19.
198.80.94.90 255.255.255.255 G=1 189.90.43.3 189.90.43.
189.90.43.0 255.255.255.0 G=0 189.90.43.1 189.90.43.
0.0.0.0 /0 G=1 189.90.43.3 189.90.43.
Table 13: Routing table at Host H
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
158.67.9.0 255.255.255.0 G=0 158.67.9.17 158.67.9.
0.0.0.0 /0 G=1 158.67.9.1 158.67.9.
Table 14: Routing table at Host H
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
198.80.94.0 255.255.255.0 G=0 198.80.94.80 198.80.94.
198.80.94.90 255.255.255.255 G=1 198.80.94.1 198.80.94.
0.0.0.0 /0 G=1 198.80.94.1 198.80.94.
Table 11: Routing table at IP router R
Destination IP address Subnet mask Flags
“Gateway” - Next hop IP address OR IP address of directly connected interface
Interface
