35. IGMP is a protocol run only between the host and its first-hop multicast router.
IGMP allows a host to specify (to the first-hop multicast router) the multicast group it wants to join. It is then up to the multicast router to work with other multicast routers (i.e., run a multicast routing protocol) to ensure that the data for the host-joined multicast group is routed to the appropriate last-hop router and from there to the host.
36. In a group-shared tree, all senders send their multicast traffic using the same routing tree. With source-based tree, the multicast datagrams from a given source are routed over s specific routing tree constructed for that source; thus each source may have a different source-based tree and a router may have to keep track of several source-based trees for a given multicast group.
architecture, once all circuits are set up, they will never change. Thus, the signaling overhead is negligible over the long run.
Problem 2
a) Maximum number of VCs over a link = 28 = 256.
b) The centralized node could pick any VC number which is free from the set {0,1,…,28-1}. In this manner, it is not possible that there are fewer VCs in progress than 256 without there being any common free VC number.
c) Each of the links can independently allocate VC numbers from the set {0,1,…,28 -1}. Thus, a VC will likely have a different VC number for each link along its path. Each router in the VC’s path must replace the VC number of each arriving packet with the VC number associated with the outbound link.
Problem 3
For a VC forwarding table, the columns are : Incoming Interface, Incoming VC Number, Outgoing Interface, Outgoing VC Number. For a datagram forwarding table, the columns are: Destination Address, Outgoing Interface.
Problem 4
a). Data destined to host H3 is forwarded through interface 3 Destination Address Link Interface
H3 3
b). No, because forwarding rule is only based on destination address.
c).
Incoming interface Incoming VC# Outgoing Interface Outgoing VC#
1 12 3 22
2 63 4 18
Note, those two flows (from H1 and H2) must have different VC#s, true for both incoming and outgoing VC#s.
d).
Router B.
Incoming interface Incoming VC# Outgoing Interface Outgoing VC#
1 22 2 24
Router C.
Incoming interface Incoming VC# Outgoing Interface Outgoing VC#
1 18 2 50
Router D.
Incoming interface Incoming VC# Outgoing Interface Outgoing VC#
1 24 3 70
2 50 3 76
Problem 5
c) No VC number can be assigned to the new VC; thus the new VC cannot be established in the network.
d) Each link has two available VC numbers. There are four links. So the number of combinations is 24 = 16. One example combination is (10,00,00,10).
Problem 6
In a virtual circuit network, there is an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; however the routers have no notion of any connections; hence the terminology connection-oriented service.
Problem 7
To explain why there would be no input queuing, let’s look at a specific design. For simplicity suppose each packet is the same size. We design the switch with time division multiplexing: time is broken into frames with each frame divided into n slots, with one slot needed to switch a packet through the fabric, and with one slot per frame devoted to each input line. Since at most one packet can arrive on each input line in each frame, the switching fabric will clear all packets in each frame.
Problem 8
The minimal number of time slots needed is 3. The scheduling is as follows.
Slot 1: send X in top input queue, send Y in middle input queue.
Slot 2: send X in middle input queue, send Y in bottom input queue Slot 3: send Z in bottom input queue.
Largest number of slots is still 3. Actually, based on the assumption that a non-empty input queue is never idle, we see that the first time slot always consists of sending X in the top input queue and Y in either middle or bottom input queue, and in the second time
slot, we can always send two more datagram, and the last datagram can be sent in third time slot.
NOTE: Actually, if the first datagram in the bottom input queue is X, then the worst case would require 4 time slots.
Problem 9 a)
Prefix Match Link Interface
11100000 00 0 11100000 01000000 1
1110000 2
11100001 1 3
otherwise 3
b) Prefix match for first address is 5th entry: link interface 3 Prefix match for second address is 3nd entry: link interface 2 Prefix match for third address is 4th entry: link interface 3
Problem 10
Destination Address Range Link Interface 00000000
through 0
00111111 01000000
through 1
01011111 01100000
through 2
01111111 10000000
through 2
10111111 11000000
through 3
11111111
number of addresses for interface 0 = 26 =64 number of addresses for interface 1 = 25 =32
number of addresses for interface 2 = 26 +25 =64+32=96 number of addresses for interface 3 = 26 =64
Problem 11
Destination Address Range Link Interface 11000000
through (32 addresses) 0
11011111 10000000
through(64 addresses) 1
10111111 11100000
through (32 addresses) 2
11111111 00000000
through (128 addresses) 3 01111111
Problem 12 223.1.17.0/26 223.1.17.128/25 223.1.17.192/28
Problem 13
Destination Address Link Interface
200.23.16/21 0
200.23.24/24 1
200.23.24/21 2
otherwise 3
Problem 14
Destination Address Link Interface
11100000 00 (224.0/10) 0 11100000 01000000 (224.64/16) 1 1110000 (224/8) 2 11100001 1 (225.128/9) 3
Problem 15
ny IP address in range 128.119.40.128 to 128.119.40.191
our equal size subnets: 128.119.40.64/28, 128.119.40.80/28, 128.119.40.96/28,
Problem 16
rom 214.97.254/23, possible assignments are ) Subnet A: 214.97.255/24 (256 addresses)
29 (128-8 = 120 addresses)
Subnet D: 214.97.254.0/31 (2 addresses)
) To simplify the solution, assume that no datagrams have router interfaces as at
Router 1
otherwise 3
A F
128.119.40.112/28
F a
Subnet B: 214.97.254.0/25 - 214.97.254.0/
Subnet C: 214.97.254.128/25 (128 addresses)
Subnet E: 214.97.254.2/31 (2 addresses) Subnet F: 214.97.254.4/30 (4 addresses) b
ultim e destinations. Also, label D, E, F for the upper-right, bottom, and upper-left interior subnets, respectively.
ongest Prefix Match Outgoing Interface
11010110 01100001 11111111 Subnet A
2
L
11010110 01100001 11111110 0000000 Subnet D 11010110 01100001 11111110 000001 Subnet F Router
ongest Prefix Match Outgoing Interface
11010110 01100001 11111111 0000000 Subnet D
L
11010110 01100001 11111110 0 Subnet B 11010110 01100001 11111110 0000001 Subnet E
Router 3
ongest Prefix Match Outgoing Interface
11010110 01100001 11111111 000001 Subnet F
Problem 17
he maximum size of data field in each fragment = 680 (because there are 20 bytes IP
L
11010110 01100001 11111110 0000001 Subnet E 11010110 01100001 11111110 1 Subnet C
T
header). Thus the number of required fragments =⎡2400−20⎤=4 680 ⎥⎥
⎢⎢
Each fragment will have Identification number 422. Each fragment except the last one ch
Problem 18
P3 file size = 5 million bytes. Assume the data is carried in TCP segments, with each
Number of datagrams required =
will be of size 700 bytes (including IP header). The last datagram will be of size 360 bytes (including IP header). The offsets of the 4 fragments will be 0, 85, 170, 255. Ea of the first 3 fragments will have flag=1; the last fragment will have flag=0.
M
TCP segment also having 20 bytes of header. Then each datagram can carry 1500-40=1460 bytes of the MP3 file
1460 3425 10
5 6
⎥=
⎥
⎢ ⎤
⎢
=⎡ × . All but the last datagram will be 1,500 bytes; the last datagram will be 960+40 = 1000 bytes. Note that here there is not
Problem 19
ses: 192.168.1.1, 192.168.1.2, 192.168.1.3 with the router interface being
NAT Translation Table Side 00 1 3345
fragmentation – the source host does not create datagrams larger than 1500 bytes, and these datagrams are smaller than the MTUs of the links.
a) Home addres 192.168.1.4 b)
WAN Side LAN 24.34.112.235, 40 92.168.1.1,
24.34.112.235, 4001 192.168.1.1, 3346
24.34.112.235, 4002 192.168.1.2, 3445 24.34.112.235, 4003 192.168.1.2, 3446 24.34.112.235, 4004 192.168.1.3, 3545 24.34.112.235, 4005 192.168.1.3, 3546
Problem 20
IP packets are sent outside, so we can use a packet sniffer to record all IP
n he
or more practical algorithms, see the following papers.
“A Technique for Counting NATted Hosts”, by Steven M. Bellovin, appeared in th and end-system characteristics.”
b. owever, if those identification numbers are not sequentially assigned but
won’t
Problem 21
is not possible to devise such a technique. In order to establish a direct TCP connection AN
Problem 22
-x-w-u, y-x-w-v-u,
y-w-v-x-u,
w-v-x-u,
Problem 23 -y-w-z, a. Since all
packets generated by the hosts behind a NAT. As each host generates a sequence of IP packets with sequential numbers and a distinct (very likely, as they are randomly chosen from a large space) initial identification number (ID), we ca group IP packets with consecutive IDs into a cluster. The number of clusters is t number of hosts behind the NAT.
F
IMW’02, Nov. 6-8, 2002, Marseille, France.
“Exploiting the IPID field to infer network pa
Weifeng Chen, Yong Huang, Bruno F. Ribeiro, Kyoungwon Suh, Honggang Zhang, Edmundo de Souza e Silva, Jim Kurose, and Don Towsley.
PAM'05 Workshop, March 31 - April 01, 2005. Boston, MA, USA.
H
randomly assigned, the technique suggested in part (a) won’t work, as there be clusters in sniffed data.
It
between Arnold and Bernard, either Arnold or Bob must initiate a connection to the other. But the NATs covering Arnold and Bob drop SYN packets arriving from the W side. Thus neither Arnold nor Bob can initiate a TCP connection to the other if they are both behind NATs.
y-x-u, y-x-v-u, y
y-w-u, y-w-v-u, y-w-x-u, y-w-x-v-u,
w-u, w-v-u, w-x-u, w-x-v-u,
x to z:
x-y-z, x
x-w-z, x-w-y-z,
x-v-w-z, x-v-w-y-z, z to u:
z-w-x-u, z-w-v-x-u, z-w-x-v-u, z-w-y-x-u, z-w-y-x-v-u,
x-u, z-y-w-y-x-v-u to w:
-w, z-y-x-w, z-y-x-v-w, z-y-x-u-w, z-y-x-u-v-w, z-y-x-v-u-w
Problem 24
tep N’ D(t),p(t) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z)
0 x ∞ ∞ 3,x 6,x 6,x 8,x
Problem 25 .
N’ D(x), p(x) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z)
0 t ∞ 2,t 4,t ∞ 7,t ∞
x
.
N’ D(x), p(x) D(t),p(t) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z) x-u-w-z, x-u-w-y-z,
x-u-v-w-z, x-u-v-w-y-z
z-w-u, z-w-v-u,
z-y-x-u, z-y-x-v-u, z-y-x-w-u, z-y-x-w-y-u, z-y-x-v-w-u, z-y-w-v-u, z-y-w-x-u, z-y-w-v-x-u, z-y-w-x-v-u, z-y-w-y-z
z-w, z-y
S
1 xv 7,v 6,v 3,x 6,x 6,x 8,x 2 xvu 7,v 6,v 3,x 6,x 6,x 8,x 3 xvuw 7,v 6,v 3,x 6,x 6,x 8,x 4 xvuwy 7,v 6,v 3,x 6,x 6,x 8,x 5 xvuwyt 7,v 6,v 3,x 6,x 6,x 8,x 6 xvuwytz 7,v 6,v 3,x 6,x 6,x 8,x
a Step
1 tu ∞ 2,t 4,t 5,u 7,t ∞ 2 tuv 7,v 2,t 4,t 5,u 7,t ∞ 3 tuvw 7,v 2,t 4,t 5,u 7,t ∞ 4 tuvwx 7,v 2,t 4,t 5,u 7,t 15, 5 tuvwxy 7,v 2,t 4,t 5,u 7,t 15,x 6 tuvwxyz 7,v 2,t 4,t 5,u 7,t 15,x
b Step
u ∞ 2,u 3,u 3,u ∞ ∞ ut ∞ 2,u 3,u 3,u 9,t ∞ utv 6,v 2,u 3,u 3,u 9,t ∞ utvw 6,v 2,u 3,u 3,u 9,t ∞ utvwx 6,v 2,u 3,u 3,u 9,t 14,x utvwxy 6,v 2,u 3,u 3,u 9,t 14,x utvwxyz 6,v 2,u 3,u 3,u 9,t 14,x
c.
Step N’ D(x), p(x) D(u),p(u) D(t),pt) D(w),p(w) D(y),p(y) D(z),p(z) v 3,v 3,v 4,v 4,v 8,v ∞ vx 3,v 3,v 4,v 4,v 8,v 11,x vxu 3,v 3,v 4,v 4,v 8,v 11,x vxut 3,v 3,v 4,v 4,v 8,v 11,x vxutw 3,v 3,v 4,v 4,v 8,v 11,x vxutwy 3,v 3,v 4,v 4,v 8,v 11,x vxutwyz 3,v 3,v 4,v 4,v 8,v 11,x
d.
Step N’ D(x), p(x) D(u),p(u) D(v),p(v) D(t),p(t) D(y),p(y) D(z),p(z)
w 6,w 3,w 4,w ∞ ∞ ∞ wu 6,w 3,w 4,w 5,u ∞ ∞ wuv 6,w 3,w 4,w 5,u 12,v ∞ wuvt 6,w 3,w 4,w 5,u 12,v ∞ wuvtx 6,w 3,w 4,w 5,u 12,v 14,x wuvtxy 6,w 3,w 4,w 5,u 12,v 14,x wuvtxyz 6,w 3,w 4,w 5,u 12,v 14,x
e.
Step N’ D(x), p(x) D(u),p(u) D(v),p(v) D(w),p(w) D(t),p(t) D(z),p(z) y 6,y ∞ 8,y ∞ 7,y 12,y yx 6,y ∞ 8,y 12,x 7,y 12,y yxt 6,y 9,t 8,y 12,x 7,y 12,y yxtv 6,y 9,t 8,y 12,x 7,y 12,y yxtvu 6,y 9,t 8,y 12,x 7,y 12,y yxtvuw 6,y 9,t 8,y 12,x 7,y 12,y yxtvuwz 6,y 9,t 8,y 12,x 7,y 12,y
f.
Step N’ D(x), p(x) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(t),p(t) z 8,z ∞ ∞ ∞ 12,z ∞ zx 8,z ∞ 11,x 14,x 12,z ∞ zxv 8,z 14,v 11,x 14,x 12,z 15,v zxvy 8,z 14,v 11,x 14,x 12,z 15,v zxvyu 8,z 14,v 11,x 14,x 12,z 15,v zxvyuw 8,z 14,v 11,x 14,x 12,z 15,v zxvyuwt 8,z 14,v 11,x 14,x 12,z 15,v
Problem 26
Cost to
u v x y z
v ∞ ∞ ∞ ∞ ∞
From x ∞ ∞ ∞ ∞ ∞
z ∞ 6 2 ∞ 0
Cost to
u v x y z v 1 0 3 ∞ 6
From x ∞ 3 0 3 2
z 7 5 2 5 0
Cost to
u v x y z v 1 0 3 3 5
From x 4 3 0 3 2
z 6 5 2 5 0
Cost to
u v x y z
v 1 0 3 3 5
From x 4 3 0 3 2
z 6 5 2 5 0
Problem 27
The wording of this question was a bit ambiguous. We meant this to mean, “the number of iterations from when the algorithm is run for the first time” (that is, assuming the only information the nodes initially have is the cost to their nearest neighbors). We assume that the algorithm runs synchronously (that is, in one step, all nodes compute their distance tables at the same time and then exchange tables).
At each iteration, a node exchanges distance tables with its neighbors. Thus, if you are node A, and your neighbor is B, all of B's neighbors (which will all be one or two hops from you) will know the shortest cost path of one or two hops to you after one iteration (i.e., after B tells them its cost to you).
Let d be the “diameter” of the network - the length of the longest path without loops between any two nodes in the network. Using the reasoning above, after iterations, all nodes will know the shortest path cost of d or fewer hops to all other nodes. Since any path with greater than d hops will have loops (and thus have a greater cost than that path with the loops removed), the algorithm will converge in at most iterations.
−1 d
−1 d
ASIDE: if the DV algorithm is run as a result of a change in link costs, there is no a priori bound on the number of iterations required until convergence unless one also specifies a bound on link costs.
Problem 28
a. Dx(w) = 2, Dx(y) = 4, Dx(u) = 7 b.
First consider what happens if c(x,y) changes. If c(x,y) becomes larger or smaller (as long as c(x,y) >=1) , the least cost path from x to u will still have cost at least 7. Thus a change in c(x,y) (if c(x,y)>=1) will not cause x to inform its neighbors of any changes.
If c(x,y)= δ<1, then the least cost path now passes through y and has cost δ+6.
Now consider if c(x,w) changes. If c(x,w) = ε ≤ 1, then the least-cost path to u continues to pass through w and its cost changes to 5 + ε; x will inform its neighbors of this new cost. If c(x,w) = δ > 6, then the least cost path now passes through y and has cost 11;
again x will inform its neighbors of this new cost.
c. Any change in link cost c(x,y) (and as long as c(x,y) >=1) will not cause x to inform its neighbors of a new minimum-cost path to u .
Problem 29