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Communication Networks Problem Set 4 Solutions: IP Addressing, TCP RTT Estimation, Assignments of Organizational Communication

University of Illinois - Urbana-ChampaignOrganizational Communication

Solutions to problem set 4 of the cs/ece 438: communication networks course, covering topics such as ip addressing, tcp round-trip time estimation, ip fragmentation, and the use of networking utilities like ifconfig and arp.

Typology: Assignments

Pre 2010

Uploaded on 03/16/2009

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CS/ECE 438: Communication Networks Fall 2007
Problem Set 4 Solutions
1. IP Addressing
Suppose a router has the following routing table:
64.20.44.0/24 if1
64.20.44.100/30 if2
64.20.32.0/20 if3
64.20.46.0/23 if4
64.0.0.0/10 if5
0.0.0.0/0 if6
(a) For each entry in the table, list the range of IP addresses that will be matched by that entry
(ignoring any overlap between entries) and state the number of addresses in each group
i. 64.20.44.1 – 64.20.44.254 (254)
ii. 64.20.44.101 – 64.20.44.102 (2)
iii. 64.20.32.1 – 64.20.47.254 (4094)
iv. 64.20.46.1 – 64.20.47.254 (510)
v. 64.0.0.1 – 64.64.255.254 (4194302)
vi. All addresses
I excluded the IP addresses for the network address and the broadcast address in these calculations.
(b) Which interface will be used for each of the following addresses? Remember that routers use the
rule with the longest matching prefix.
i. 64.20.44.102 — if2
ii. 64.20.45.102 — if3
iii. 64.20.47.102 — if4
iv. 64.20.48.102 — if5
2. TCP RTT Estimation One difficulty with the original TCP SRTT estimator is the choice of an
initial value. In the absence of any special knowledge of network conditions, the typical approach is to
pick an arbitrary value, such as 2 seconds, and hope this will converge quickly to an accurate value. If
this estimate is too small, TCP will perform unnecessary retransmissions. If it is too large, TCP will
wait a long time before retransmitting if the first segment is lost. Also, the convergence might be slow.
(a) Choose α= 0.75 and SRTT(0) = 2 seconds, and assume all measured RTT values = 1 second
with no packet loss. What is SRTT(15)? Recall,
SRT T (k+ 1) = α∗S RT T (k) + (1 −α)∗RT T (k+ 1).
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Download Communication Networks Problem Set 4 Solutions: IP Addressing, TCP RTT Estimation and more Assignments Organizational Communication in PDF only on Docsity!

CS/ECE 438: Communication Networks Fall 2007

Problem Set 4 Solutions

  1. IP Addressing Suppose a router has the following routing table: 64.20.44.0/24 if 64.20.44.100/30 if 64.20.32.0/20 if 64.20.46.0/23 if 64.0.0.0/10 if 0.0.0.0/0 if

(a) For each entry in the table, list the range of IP addresses that will be matched by that entry (ignoring any overlap between entries) and state the number of addresses in each group i. 64.20.44.1 – 64.20.44.254 (254) ii. 64.20.44.101 – 64.20.44.102 (2) iii. 64.20.32.1 – 64.20.47.254 (4094) iv. 64.20.46.1 – 64.20.47.254 (510) v. 64.0.0.1 – 64.64.255.254 (4194302) vi. All addresses I excluded the IP addresses for the network address and the broadcast address in these calculations. (b) Which interface will be used for each of the following addresses? Remember that routers use the rule with the longest matching prefix. i. 64.20.44.102 — if ii. 64.20.45.102 — if iii. 64.20.47.102 — if iv. 64.20.48.102 — if

  1. TCP RTT Estimation One difficulty with the original TCP SRTT estimator is the choice of an initial value. In the absence of any special knowledge of network conditions, the typical approach is to pick an arbitrary value, such as 2 seconds, and hope this will converge quickly to an accurate value. If this estimate is too small, TCP will perform unnecessary retransmissions. If it is too large, TCP will wait a long time before retransmitting if the first segment is lost. Also, the convergence might be slow.

(a) Choose α = 0.75 and SRTT(0) = 2 seconds, and assume all measured RTT values = 1 second with no packet loss. What is SRTT(15)? Recall,

SRT T (k + 1) = α ∗ SRT T (k) + (1 − α) ∗ RT T (k + 1).

t SRT T (t) 0 2 1 1. 2 1. 3 1. 4 1. 5 1. 6 1. 7 1. 8 1. 9 1. 10 1. 11 1. 12 1. 13 1. 14 1. 15 1. (b) Now let SRTT(0) = 1 second and assume RTT values = 2 seconds and no packet loss. What is SRTT(15)? t SRT T (t) 0 1 1 1. 2 1. 3 1. 4 1. 5 1. 6 1. 7 1. 8 1. 9 1. 10 1. 11 1. 12 1. 13 1. 14 1. 15 1.

  1. IP Fragmentation

Consider two hosts, A and B, each on a separate shared Ethernet with MTU=1500 bytes. In addition to these LAN’s, the route connecting host A to host B through the Internet contains an additional hop over a point-to-point link between a router on A’s Ethernet and a second router on B’s Ethernet. The point-to-point link has MTU=400 bytes. Recall that MTU is the maximum amount of data that can be sent in a frame at the physical layer and thus includes all TCP and IP headers (each of which occupies 20 bytes). Also recall that IP fragmentation breaks data along 8 byte boundaries.

(a) For each packet that arrives at host B, show the part of the header that concerns fragmentation, i.e. the ident, more-fragments, and offset fields, as well as the payload length. The packet will be fragmented into 4 fragments. The fragmentation is done at the IP layer, so there will be 1480 bytes of IP-layer payload to split among 4 packets. Packet 1: Ident: X, More-Fragments: Yes, Offset: 0, Length: 376. The payload will include 20 bytes of TCP header and 356 bytes of TCP payload. The length will be 376 bytes and not 400 due to having to round the offset up to a multiple of 8. Packet 2: Ident: X, More-Fragments: Yes, Offset: 47 (·8 = 376), Length: 376

ece438 [dcllnx23:~]> /sbin/arp

  • sunserv2.ews.uiuc.edu ether 00:03:BA:C6:4F:A1 C eth Address HWtype HWaddress Flags Mask Iface
  • newshepherd.ews.uiuc.ed ether 00:03:BA:32:16:5D C eth
  • uiuc-ewsl-vlan1.gw.uiuc ether 00:0C:DB:25:12:A0 C eth
  • sunserv4.ews.uiuc.edu ether 00:03:BA:33:53:95 C eth
  • lackey.ews.uiuc.edu ether 00:03:BA:31:F7:45 C eth
  • sunserv1.ews.uiuc.edu ether 00:03:BA:C6:ED:AD C eth
  • lnxapps.ews.uiuc.edu ether 00:03:BA:31:0F:51 C eth