Answers of Exercise 12Answers of Exercise 12

1. Using nslookup utility to get IP address of www.k.hosei.ac.jp. Find out how many name servers in domain k, hosei, ac and jp, respectively. Answer: IP address of ww.k.hosei.ac.jp is 133.25.253.152. Numbers of name servers (tested at 4:00PM in June 21, 2001) Domain k: 2 Domain hosei: 2 Domain ac: 6 Domain jp: 6

2. A host can dynamically get an IP address by means of exchanging information with a DHCP server using TCP/IP protocols. However, the host has no IP address before getting the IP address. How does the host communicate with DHCO server when having no IP address? Furthermore, the host can only hold the issued IP address with finite lease time such as one hour. Why? What method in DHCP is used to renew the release to hold the IP address more than one hour? Answer: When a host first requests IP address from the DHCP server, it uses this computer IP address (0.0.0.0) and hardware broadcast address (all 1s). The lease time is used to improve IP address efficiency of usage. A host will automatically apply to renew lease time of the current IP address after each 50% lease time so that it can keep the IP address so long as it is always attached to the network.

Answers of Exercise 12Answers of Exercise 12

3. Data of parameters and states called objects of a host/router is stored in MIB (management information database). Each object in MIB has a unique identifier represented in hierarchical ASN.1 name scheme. Explain the meaning of ipForwarding object of a router, and give its identifier. Answer: The ipForwarding object keeps a value of number of datagrams forwarded by a router. Its MIB identifier is 1.3.6.1.2.1.4.1 4. SNMP uses ASN.1 to represent data of an object. Give the SNMP representations of string data “SNMP” and IP address “133.25.252.22” in hexadecimal format, respectively. Answer: ASCII codes of SNMP: 53 4E 4D 50 The data representation: 04 04 53 4E 4D 50 Hexadecimal format of 133.25.252.22: 83 19 FC 16 The SNMP data representation: 40 04 83 19 FC 16

5. Explain why the digital signature method shown in the lecture note can guarantee both authentication and confidence. Answer: When the message is encrypted using A’s private key, the encrypted message can only be decrypted using A’s public key. Thus when we success to decrypt a message using A’s public key, we can make sure that the message is sent from A. If the message is further encrypted using B’s public key, no one except B can decrypted the message. Thus, the message can be kept confidential.