Pré-visualização35 páginas
```0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
probability
Ef
fic
ie
nc
y
N=15
N=20
N=30

Slotted ALOHA
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
probability
Ef
fic
ie
nc
y
N=15
N=20
N=30

Problem 13

The length of a polling round is
)/( polldRQN + .

The number of bits transmitted in a polling round is NQ . The maximum throughput
therefore is
Q
Rd
R
dRQN
NQ
pollpoll +
=+
1
)/(

Problem 14

a), b) See figure below.

192.168.3.003
99-99-99-99-99-99
192.168.3.002
88-88-88-88-88-88
192.168.3.001
77-77-77-77-77-77
192.168.2.003
55-55-55-55-55
192.168.2.001
44-44-44-44-44-44
192.168.2.002
33-33-33-33-33-33
A
B
Router 1 LAN
C
D
F
192.168.1.001
00-00-00-00-00-00
192.168.1.002
22-22-22-22-22-22
192.168.1.003
11-11-11-11-11-11
E
192.168.2.004
66-66-66-66-66
LAN Router 2 LAN
c)
1. Forwarding table in E determines that the datagram should be routed to interface
192.168.3.002.
2. The adapter in E creates and Ethernet packet with Ethernet destination address 88-
88-88-88-88-88.
3. Router 2 receives the packet and extracts the datagram. The forwarding table in
this router indicates that the datagram is to be routed to 198.162.2.002.
4. Router 2 then sends the Ethernet packet with the destination address of 33-33-33-
33-33-33 and source address of 55-55-55-55-55-55 via its interface with IP
5. The process continues until the packet has reached Host B.

d)
ARP in E must now determine the MAC address of 198.162.3.002. Host E sends out an
ARP query packet within a broadcast Ethernet frame. Router 2 receives the query packet
and sends to Host E an ARP response packet. This ARP response packet is carried by an
Ethernet frame with Ethernet destination address 77-77-77-77-77-77.

Problem 15

a). No. E can check the subnet prefix of Host F\u2019s IP address, and then learn that F is on
the same LAN. Thus, E will not send the packet to the default router R1.
Ethernet frame from E to F:
Source IP = E\u2019s IP address
Destination IP = F\u2019s IP address
Source MAC = E\u2019s MAC address
Destination MAC = F\u2019s MAC address

b). No, because they are not on the same LAN. E can find this out by checking B\u2019s IP
Ethernet frame from E to R1:
Source IP = E\u2019s IP address
Destination IP = B\u2019s IP address
Source MAC = E\u2019s MAC address
Destination MAC = The MAC address of R1\u2019s interface connecting to Subnet 3.

c).
Switch S1 will broadcast the Ethernet frame via both its interfaces as the received ARP
1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update
its forwarding table to include an entry for Host A.

Yes, router R1 also receives this ARP request message, but R1 won\u2019t forward the
message to Subnet 3.

obtained from A\u2019s query message.

Once switch S1 receives B\u2019s response message, it will add an entry for host B in its
forwarding table, and then drop the received frame as destination host A is on the same
interface as host B (i.e., A and B are on the same LAN segment).

Problem 16
Lets call the switch between subnets 2 and 3 S2. That is, router R1 between subnets 2 and
3 is now replaced with switch S2.

a). No. E can check the subnet prefix of Host F\u2019s IP address, and then learn that F is on
the same LAN segment. Thus, E will not send the packet to S2.
Ethernet frame from E to F:
Source IP = E\u2019s IP address
Destination IP = F\u2019s IP address
Source MAC = E\u2019s MAC address
Destination MAC = F\u2019s MAC address

b). Yes, because E would like to find B\u2019s MAC address. In this case, E will send an ARP
This query packet will be re-broadcast by switch 1, and eventually received by Host B.
Ethernet frame from E to S2:
Source IP = E\u2019s IP address
Destination IP = B\u2019s IP address
Source MAC = E\u2019s MAC address

c).
Switch S1 will broadcast the Ethernet frame via both its interfaces as the received ARP
1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update
its forwarding table to include an entry for Host A.

Yes, router S2 also receives this ARP request message, and S2 will broadcast this query
packet to all its interfaces.

obtained from A\u2019s query message.

Once switch S1 receives B\u2019s response message, it will add an entry for host B in its
forwarding table, and then drop the received frame as destination host A is on the same
interface as host B (i.e., A and B are on the same LAN segment).
Problem 17
Wait for 51,200 bit times. For 10 Mbps, this wait is

12.5
1010
102.51
6
3
=×
×
bps
bits msec

For 100 Mbps, the wait is 512 \u3bc sec.
Problem 18
At 0=t A transmits. At , 576=t A would finish transmitting. In the worst case, B
begins transmitting at time t=324, which is the time right before the first bit of A\u2019s frame
arrives at B. At time t=324+325=649 B 's first bit arrives at A . Because 649> 576, A
finishes transmitting before it detects that B has transmitted. So A incorrectly thinks that
its frame was successfully transmitted without a collision.
Problem 19
Following the same reasoning as in last problem, we need to make sure that one end of
the Ethernet is able to detect the collision before it completes its transmission of a frame.
Thus, a minimum frame size is required.

Let BW denote the bandwidth of the Ethernet. Consider the worst case for Ethernet\u2019s
collision detection:
1. At t=0 (bit times): A sends a frame
2. At t=dprop-1 (bit times): B sends a frame right before it can sense A\u2019s first bit.
3. At t=2dprop-2 (bit times): If A finishes its transmission of its last bit just before B\u2019s
frame arrives at A, then, A won\u2019t be able to detect a collision before finishing its
transmission of a frame. Thus, in order for A to detect collision before finishing
transmission, the minimum required frame size should be >= 2dprop-1 (bit times).

Assume that the signal propagation speed in 10BASE-T Ethernet is 1.8*108m/sec.
As dprop=d/(1.8*108)*BW(here, we need to convert the propagation delay in seconds to
bit times for the specific Ethernet link), we find the minimum required frame size is
2*d/(1.8*108)*BW-1 (bits). Or approximately we choose 2*d/(1.8*108)*BW (bits).
If d=2km, then minimum required frame size is: 222 bits.
Problem 20
Based on the solution of last problem, you know that a higher link speed requires a larger
minimum required packet size. If you cannot change packet size, then you can add
switches or routers to segment your LAN, in order to make sure each LAN segment\u2019s
size is sufficiently small for small frame size.
Problem 21
Time, t Event
0 A and B begin transmission
245 A and B detect collision
293 A and B finish transmitting jam signal
293+245 = 538
538+96=634
B 's last bit arrives at A ; A detects an idle channel
A starts transmitting

293+512 = 805

634+245=879

B returns to Step2
B must sense```