1.) A group of N stations share a 56-kbps pure ALOHA channel. station outputs a 1000-bit, every 100 sec. maximum value of N?
     usable bandwidth is 0.184 x 56 kbps = 10.3 kbps. N = 10300/10 = 1030 stations.
2.) Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is less?
     transmission can start instantly. At low load, no collisions are expected, the transmission will be successful.With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay.
3.)What is the length of a contention slot in CSMA/CD for a) a 2-km twin-lead cable? b)a 40-km multimode fiber optic cable
  a.) Signal propagation speed in twin lead is 2.46 × 10^8 m/sec. Signal propagation time for 2 km is 8.13 µ sec. So, the length of contention slot is 16.26 µ sec.      b.) a 40-km multimode fiber optic cable.
4.)How long does a station, s, have to wait in the worst case before it can start transmitting its frame over a LAN that uses the basic bit-map protocol?
   Assuming total N station and frame-size is of d bits. The worst case is all stations want to send, s is the highest-numbered station out of the N stationsand s becomes ready to send just after the bitmap has passed. Station s must wait (N – 1) × d bit times for the transmission before new cycle begins. N bit times for a contention bitmap, and (N – 1) × d bits for transmission of other frames before its frame starts. The total is N + 2 (N – 1) d bit times.
5.)In the binary countdown protocol, explain how a lower-numbered station may be starved from sending a packet.
    If a higher-numbered station and a lower-numbered station have packets to send at the same time, the higher-numbered station will always win the bid. Thus, a lower-numbered station will be starved from sending its packets if there is a continuous stream of higher-numbered stations ready to send their packets.
7.) Six stations, A through F, communicate using the MACA (Multiple Access with Collision Avoidance) protocol. Is it possible for two transmissions to take place simultaneously?
   Yes.Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F.
8.) A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200 m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead…..What is the effective data rate, excluding overhead, assuming that there are no collisions?
   Transmitter seizes cable => 10 μ sec.Delivery or, transmit data needs 25. 6 μ sec. Propagation delay for last bit to get to the end will need 5.0 μ sec. Receiver seizes cable => 10 μ sec. Acknowledgement sent 3.2 μ sec. Propagation delay for last bit to get to the end will need 5.0 μ sec. the sum of these is 58. 8 μ sec. In this period, 224 data bits (since, 256 -32 = 224) are sent, therefore the effective data-rate is 224bits / 58.8 μ sec, or, about 3.8bits/μ sec, or, 3.8Mbps.
9.) In the given Figure, four stations, A, B, C, and D, are shown. Which of the last two stations do you think is closest to A and why? Figure: Virtual channel sensing using CSMA/CA
Station C is the closest to A since it heard the RTS and responded to it by asserting its NAV signal. D did not respond, so it must be outside A’s radio range.
10.) Give two reasons why networks might use an error-correcting code instead of error detection and retransmission.
    First reason is the need for real-time quality of service. If an error is discovered, there is no time for a retransmission. The show must go on. Forward error correction can be used here. Second reason is that on very low-quality lines the error rate can be so high that practically all frames would have to be retransmitted.
11.) (a) Briefly describe the difference between store-and-forward and cut-through switches. (b) Store-and-forward switches have an advantage over cut-through switches with respect to damaged frames. Explain what it is.
(a) A store-and-forward switch stores each incoming frame in its entirety, then examines it and forwards it. A cut-through switch starts to forward incoming frames before they have arrived completely. As soon as the destination address is in, the forwarding can begin.   (b) Store-and-forward switches store entire frames before forwarding them. After a frame comes in, the checksum can be verified. If the frame is damaged, it is discarded immediately. With cutthrough, damaged frames cannot be discarded by the switch because by the time the error is detected, the frame is already gone. Trying to deal with the problem is like locking the barn door after.
12.) Consider the extended LAN connected using bridges B1 and B2 in the given Figure. Suppose the hash tables in the two bridges are empty. List all ports on which a packet will be forwarded for the following sequence of data transmissions: (a) A sends a packet to C. (b) E sends a packet to F. (c) F sends a packet to E. (d) G sends a packet to E. (e) D sends a packet to A. (f) B sends a packet to F.
(a) B1 will forward this packet on ports 2, 3, and 4. B2 will forward it on 1, 2 and 3.
(b) B2 will forward this packet on ports 1, 3, and 4. B1 will forward it on 1, 2 and 3.
(c) B2 will not forward this packet on any of its ports, and B1 will not see it.
(d) B2 will forward this packet on port 2. B1 will not see it.
(e) B2 will forward this packet on port 4 and B1 will forward it on port 1.
(f) B1 will forward this packet on ports 1, 3 and 4. B2 will forward it on port 2.