1. Introduction, OSI Reference, Network Services/Applications
1. OSI seven layer model
1. Transport layer establishes communication between  individual process to individual process communication on different machines.  Yes No
2. SSL is a session layer protocol which runs on top of TCP. Yes No
3. Data Link layer checks if the frame is garbled or not. Yes No
2. PDU encapsulation/decapsulation

How does the TCP layer protocol entity know the payload of a TCP segment should be delivered to an application layer  protocol entity such as web server?
1. From the type field in the Ethernet  header Yes No
2. From the MAC address in the IP header Yes No
3. From the Protocol field in the TCP header Yes No
4. From the Protocol field in the IP header Yes No
5. From the Protocol field in the application layer header Yes No
3. Network  address and their format
1. IP address of IP protocol version 6 contains 4 bytes. Yes No
2. The terminal server of an ISP dynamically assigns one address from a pool of IP addresses to a host when the user dials in. It helps to reduce the usage of IP addresses.  Yes No.
3. Can a ftp server be setup without using a new IP address?  Yes No
4. The IP packet content sent by a machine will not be changed by the intermediate router.  Yes No
5. If the IP packet content sent by a machine is not changed by the intermediate routers, what are changed in the frames that allow those frames to be sent from one router to the other?

The IP packet is changed by the itermediate routers (TTL is decremented) but I'll try to answer the question anyway... The router looks at the destination IP address and tries to determine if it belongs inside or outside the router boundary (or which branch to send it to). The IP router then sends the packet to the correct NIC. The NIC (or NIC software/firmware/hardware) wraps the IP packet into an ethernet frame and changes the destination MAC address to where the IP packet is destined. This destination MAC address is probably obtained from an ARP session. This new MAC address corresponds to the next router the IP packet is destined for.

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The source MAC address and destination MAC address of the frame will be chagned.

4. Domain Name System (DNS)
1. To speed up the performance, the name server caches the results return from another DNS server. Yes No
2. After  changing a machine to a new IP address, other name servers in Internet may still have old entries in their cache and will answer name query with the old IP address.  Yes No
3. Why happens after the DNS request sent to the primary (first) name server never gets replied and the machine times out?
   If specified, the machine will attempt a secondary name server. Otherwise the machine reports to the user that the DNS server has timed out.

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   The DNS client of the machine will send the request to the secondary DNS server. If still not response, the request will be sent to the tenary DNS.

5. Basic web access steps
1. An Ethernet card  accept an Ethernet frame with its destination MAC (Medium Access Control) address equal to its burnin Ethernet address.  Yes No
2. An Ethernet card  accepts an Ethernet frame with broadcast address.  Yes No
3. What protocol is used by a machine to find out the MAC address of another machine in the same subnet? 

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   ARP

4. A web server can run on port 443. Yes No
5. A web client typically runs on a port number larger than 1023. Yes No
6. IP Address and Routing.
   Given the following routing table on archie.uccs.edu.
   Kernel IP routing table
   Destination  Gateway      Genmask         Flags   MSS Window irtt Iface
   128.198.60.0 0.0.0.0      255.255.255.192 U       40  0      0    eth0
   127.0.0.0    0.0.0.0      255.0.0.0       U       40  0      0    lo
   0.0.0.0      128.198.60.1 0.0.0.0         UG      40  0      0    eth0
   
   1. From the first entry of the routing table, we have destination IP address 128.198.60.0 and genmask (or called netmask) 255.255.255.192 that identifies the subnet which archie is in.
      a. What is the equivalent Classless IP addess notattion for this subnet? 128.198.60.0/

      grade (3 pts):

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      26
      Hint: Specify the size of net address.
      b. How many IP addresses can be assigned to machines in this subnet (assume one is used by the router, 128.198.60.1)?
      

      grade (3 pts):

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      61

   2. Why in a traceroute session, the round trip time to a router at 13 hops can be shorter than that to a router at 17 hops?
      If the 13 hop trip has more net congestion than the 17 hop trip, it is very possible for the 17 hop trip to take less time than the 13 hop trip.

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      Because the traffic volume is heavier, when traceroute sends ICMP echo request messages to the router 13 hops away, than that when traceroute sends ICMP echo request messages to the router 17 hops away plus the propagation delay, transmission delay, and message processing delay on links between the 13th hop router and the 17th hop router.
      

2. Protocol Engineering-specification, verification
1. Alternating Bit Protocol and Protocol Simulation
1. Can the Alternating Bit Protocol deliver messages in sequence even when the data gets lost or corrupted frequently?
   Yes. The protocol prevents delivering or receiving new data until the previous data just delivered or received has been acknowledged. This prevents messages from getting lost. As far as corrupted, the alternating bit protocol does not perform any error checking within the payload. This protocol does not check if data is corrupted.

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   yes. So long as some of the packets get through.

2. Reachability Analysis. For the following two machines,


1. What is the buffer size of C1?

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   1

2. Is there a deadlock? Yes No
3. Is there an unspecified reception? Yes No
4. Is there non-executable state? Yes No
5. Is there non-exeutable transiton? Yes No
6. How many global states generated?

   grade (3 pts):

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   7

3. Protocol Programming/Implementation
1. Unix Domain Socket vs. Internet Domain Socket
1. Internet Domain Socket cannot send messages to a different machine. Yes No
2. Unix Domain Socket can only send message to a different machine. Yes No
2. Stream vs. Datagram
1. The Stream socket does not deliver message boundary.  It is possible to read two packets sent by the sender  in single read() or recv() function call if we specifies a buffer size bigger than that of two or three messages sent. If the message size is fixed, then we can set the buffer size to the message size and read one message one at a time.. Yes No
2. Stream connection guarantees packets to be delivered in sequence without loss..  Yes No
3. The two subsequent packets reads from a datagram socket can be from different senders. Yes No
3. Forking or not?

Given that creating a child process using fork() takes 10 milliseconds,
1. From system throughput point of view, is it a good choice for services, that require retrieving data from another Database server, to use fork() and child process to serve the requests?  Yes No
2. Servers do simple table lookup for return data (to be done in 10 microseconds) should not create child processes using fork() to serve the requests. Yes No
4. Select()
1. Select() allows us to implement timeout and retransmission of messages in Unix system. Yes No
2. After return from the select() call, the value of the 2nd parameter (readfdset) may be changed.  Yes No
4. Transmission Fundamental
1. If we use a 32-signal set with QAM modulation method over  a TV channel with 8 MHz bandwidth and S/N ratio of 40dB, what is the maximum bit rate we can achieve?

R = 2*W*M = 2*(8 MHz)*(log2(32)) = 2*8*5 = 80 SNR = 10^(SNR(dB) / 10) = 10^(40/10) = 10000 C = W*log2(1+SNR) = 80 * log2(10001) = 80*13.288 = 1063 Mbps

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Without noise, using 8MHz bandwidth (16M signals/seconds) and 5(=log2(32)) bits per signal, we can transmit 16M*5=80 Mbps. 40dB of SNR= 10000 SNR With noise, the maximum bit rate is decided by the Shannon Theorem, C=Wlog2(1+SNR)=8M*log2(1+10000)=106.304 Mbps. 106.304Mbps < 80Mbps. The maximum bit rate is 80Mbps.

2. Given the following signal constellation diagram (16 signal set) and their bit pattern assignments.
   

   1. If the modem is sending signals at the rate of 4800 baud (4800 signals per second), what is the bit rate this modem can achieve?

   4800*log2(16) = 19,200 Kbps

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   With 16 signal set, we can send log2(16)=4 bits per signal. C=4*4800=19200 bps.

   2. If the sending modem receives a byte with 11110000 bit pattern from the sending computer, what are analog signals that will be generated by the sending modem?
   2A, 7/4 pi, A, 1/8 pi

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   1111 bit pattern will be sent with analog signal where amplitude=2A and phase=7/4 pi. 0000 bit pattern will be sent with analog signal where ampltitude=A and phase=1/8 pi.
   

   3. If the receiving modem receives two analog signal: one with amplitude of 2A and phase angle of 3/4pi followed by a signal of amplitude of A and phase angle of 5/4p, what bit patterns should it sends to the receiving computer?

   01111001

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   Look up the signal constellation diagram, signal with 2A and 3/4pi corresponds to bit pattern 0111. Singal with A and 5/4pi, corresponds to bit pattern 1001. The bit patterns to be sent to the receiving computer is 01111001.

If you feel some of the questions are ambiguous, state the problem # and your assumptions on the answers.
1.c.4 Do you consider the IP header to be part of the "content" of the IP packet? I sure do. The router decrements the TTL in the IP packet so it changes the content but not the payload. 3.d.1 Select() does this on the Unix platform as well as any other platform that compiles C.

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