On page 362 of the textbook, figure 12.9 shows LLC PDU in a generic MAC frame. What is the highest and lowest percentage of the length of the MAC frame could be LLC PDU data?
The lowest length for LLC PDU data varies from 16 to 20 octets. Encrypted and framed LLC PDU data has the highest (maximum) length of 1560 octets. Octets is the first name for byte.
In the Ethernet MAC frame, the data segment has a minimum length of 46 byte. Why do we have to have a minimum size of data in this frame?
A data segment on the Ethernet MAC frame link is usually stated as Ethernet packets. These Ethernet packets carry Ethernet frame as its packet format (payload). Ethernet frame internet structure is detailed in IEEE 802.3. This internet structure consists of: Ethernet header, Minimum Data Portion and Minimum Ethernet Frame Size. Ethernet header has a value of 18 bytes, minimum data portion – 46 bytes and frame size is equal to 64 bytes. Minimum length was set in order to compensate Pad field. CSMA station should get a false statement of successful frame transfer if this frame had a pile-up with other frames. If collision appeared it takes twice the time of delay to get information if transmission was successful. If short frame was send by a station, it can release the Ethernet without getting data that there was a crash.
Everyone probably has a wireless router at home and/or at work. Do you think a wireless router is a pure layer 3 device? Why?
Wireless router has a purpose of providing a connection among different networks and relocate (transfer) packets which were primarily set for other networks. Wireless router can be said as layer 3 device because its packets forwarding options is established on layer 3 device IP destination information. It is called layer 3 device because only at layer 3 router see the contents of data which packet carries and then makes a decision where it should be forwarded.
If a carrier signal uses a combination of modulation methods, how many bits of data can one signal carry with BPSK, QPSK, and 8 possible different amplitude?
If a carrier signal uses a combination of modulation methods, following bits will be carried by one signal carry: BPSK – 1 bps, QPSK – 2 bps, 8-PSK – 3 bps.
A packet with the size of 2K bytes goes through a WAN. If the packet goes through 10 store-and-forward switching/routing devices, all of which have the transmitting rate of 2Mbit/s, what is the total delay the packet experiences in this WAN? Please ignore the processing delay and propagation delay.
Total delay of the packet experiences is:
2Mbit/s we transfer to bit/s = 2x106.We know that packet goes through 10 devices
2000 Bytes ×8bits2×106 bps×10= 8 ms
In a highly populated area such as big cities, a cell of a cellular network may have way more user than a cell in other areas. What methods are available for crowded areas?
One of available methods for crowded methods is to divide the total occupied area by users into particular parts. The procedure go in this way: We count number of people in different parts or put them in squares, then we average them and the last step would be multiplying and adding the area (squares) they occupy. Every time we add more small parts to our average, then every time our number would get more accurate.
In a CDMA network, two users with codes 110110 (user A) and 100101 (user B). What would be the bit sequence for each user in the encoded message “1010”?
User A: 1101101010 = 512+256+64+32+8+2= 874
User B: 1001011010 = 512+64+16+8+2= 602
In a network system the administrator wants to monitor the traffic going through each switch point, and control the routing accordingly to balance traffic going through each link. Please create one distributed solution, one centralized solution, and briefly describe under what conditions one of the two solutions are preferable.
We can easily monitor traffic through switch points. To do that we have to obtain Ethernet switches which have monitor mode. This mode bestow a port to which you connect your device. We have to select two monitoring ports and set rights to attach a certain port to our monitor. The example of distributed solution would be building two host connected to switch. If we need to capture data from specific host we use this method. The advantage of this method that it is easy use, but we face serious disadvantage other traffic is not possible to capture. The centralized solution would be building two hosts connected to a switch. Then the switch is connected to a computer. Thus the computer can monitor traffic to both switch points.
References:
Radio-electronics.com,. (2015). Ethernet 802.3 Frame Format / Structure :: Radio-Electronics.Com. Retrieved 28 June 2015, from http://www.radio-electronics.com/info/telecommunications_networks/ethernet/ethernet-data-frames-structure-format.php
Stallings, W. (2005). Wireless Communications and Networks (2nd ed.). Retrieved from http://www.slideshare.net/vavichi/wireless-networks-pdf
Stevens, A., & Pro, I. (2015). Layer 2 and Layer 3 switches. IT PRO. Retrieved 28 June 2015, from http://www.itpro.co.uk/88699/layer-2-and-layer-3-switches
