Question 1

a)
The consultants proposed scheme indeed produces a viable link between the two locations. The explanation is given below.
The distance between the locations are 40 meters.
The transmission power is Pt = 80mW
The sensitivity of the receiver is Pr = -80dBm If we convert dBm into mW we have:
Pr = -80dBm = 10^(-80/10) = 10^-8 mW
The operating frequency of IEEE 802.11g protocol is 2.4 GHz and in Hz is 2.4*10^9. So frequency f = 2.4*10^9 Hz
The transmission range of a wireless node with the about characteristics is in a free space model without any obstacles in the line of sight is:
The transmission range d is given from the following formula d = (Pt/Pr)*c2 4πf c is the speed of light and it equals to 3*10^8 m/s
If we replace the values of the above formula we get that transmission range is d = 889.70 meters
That means that the transmitters of the proposed solution can ideally transmit a packet up to a distance of 889.70 meters and the packet will be received by the receiver successfully. In the case scenario, given that the distance between the transmitter and the receiver is only 40 meters, we can assume that the proposed solution provides a very strong link and data can travel between the two locations.

b)
IEEE 802.11g protocol claims a data rate of 54 Mb/s. We assume that there is a Bit Error Rate (BER) of 1*10^-3 at the receiver which means that for every 1000 bits of data received 1 bit has an error.
Within the period of 1 second 802.11 protocol can receive 54Mbits or 54.000.000 bits. When the BER is applied to that link, the number of error bits is 54.000.000*10^3 = 54.000 error bits. So from the 54.000.000 bits that are sent only (54.000.000 - 54.000) = 53.946.000 bits are successfully received giving us a rate of 53,946 Mb/s.
Time to transmit without the BER applied:
20GB of data = 160Gbits of data = 160.000 Mbits The rate is 54Mbit/sec which means the transmission time is t = Amount of data/ data rate = 160.000 / 54 = 2963 seconds ≅ 49 minutes.
After the BER applies to the transmission the new transmission time t will equal to: t = 160.000 Mbits / 53,946 Mbits/sec = 2966 seconds. The difference is minimal because the BER is small.
c)
The capacity of the new scheme could further be improved if we used directional antennas instead the non-directional antennas since there are no object between the two transmitters. Directional antennas have the ability create beams of strengthened signals and direct them to other receivers providing improved performance and also reducing the received interference. Also if we invested on some better quality receivers then the BER could be decreased and provide a more reliable link.
d)
Other networking technologies that could be deployed and increase the capacity of the original design are:
Optical wireless use laser lights to transmit digital signals between two transceivers. They provide a high capacity link of about 1Gb/sec full duplex as long as there are no obstacles between the transceivers (clear line of sight). They are industry standards and they don't require any expensive equipment in order to be deployed.
Other technology that could be deployed is the IEEE 802.11n standard. This protocol is a standard provides multiple streams over a channel and has an advertised throughput of 600 Mb/s compared to the IEEE 802.11g and 802.11a protocols. Apart from that the protocol includes error correction mechanisms that enable it to provide a reliable and high-throughput links.
e)
Two reasons that the real throughput is lower than nominal capacity of the link are:
In case of errors due to interference or other sources of noise, depending on the protocol and networking technology used, there will be retransmissions taking place and these reduce the available for data transfer bandwidth. Especially wireless transmissions are prone to errors and should be carefully designed and implemented.
Another reason is the protocol overhead. Any packet sent over a link includes the useful user data of a file and extra information that are used to maintain a reliable link, route the data and perform other controls during a transmission. In case of wireless networks channel coordination used up some of the available bandwidth.
Question 2

a)
Fragmentation and reassembly processes ensure interconnection of networks with different maximum transmission links. Fragmentation of an Internet datagram is required when a large amount of data needs to be transmitted from a network that can fit it to its MTU (Maximum transmission unit) to a network with a smaller MTU.
When a datagram needs to be fragmented, the receiving device must be provided with the appropriate information so that it will be able to identify the fragments and reassemble them into the original datagram. This information lies within the IP datagram header and is created by the device that carries out the fragmentation.
Some of the most important parts of the information are the total length field which indicates the length of each fragment, the identification field which is 16 bits wide and is uniquely assigned to each of the fragments, the more fragments field which has a flag set to 1 for all of the fragments apart from the last one which has a flag set to 0 so that the end device is able to know that it received the last fragment and the fragment offset field which specifies where is the place of each fragment in the overall message.
For our current network:

Datagrams generated from the file server are able to fit into the MTU limit of the wireless network while Datagrams generated from the wireless network need to be fragmented in order to fit into the MTU of the file server.

The first fragment is created by taking the first 1500 bytes of the 2272 byte IP datagram. This includes the original header, which becomes the IP header of the first fragment. Therefore 1480 bytes of data are in the first fragment. This leaves 772 bytes to encapsulate (2252 minus 1480). The remaining 772 bytes are placed into the second fragment, with a 20 byte header. Therefore the message sent by the wireless client will be 2292 (20b larger due to the header added in fragment 2).
After the fragments are received by the file server, it must collect them and reassemble them into the original message by using the information provided inside the header. Although fragmentation allows protocols to optimize performance for high bandwidth connections and higher level protocols to be unconcerned with the characteristics of the transmission channel, and to send data in conveniently sized pieces it has also many drawbacks.

According to Kent and Mogul “the arguments against fragmentation fall into three categories:

· Fragmentation causes inefficient use of resources: Poor choice of fragment sizes can greatly increase the cost of delivering a datagram. Additional bandwidth is used for the additional header information, intermediate gateways must expend computational resources to make additional routing decisions, and the receiving host must reassemble the fragments.
· Loss of fragments leads to degraded performance: Reassembly of IP fragments is not very robust. Loss of a single fragment requires the higher level protocol to retransmit all of the data in the original datagram, even if most of the fragments were received correctly.
· Efficient reassembly is hard: Given the likelihood of lost fragments and the information present in the IP header, there are many situations in which the reassembly process, though straightforward, yields lower than desired performance”.
b)
Path MTU Discovery is a technique for determining the path MTU between two IP hosts. It works by setting the DF (Don't Fragment) option in the IP headers of outgoing packets. Any device along the path whose MTU is smaller than the packet will drop such packets and send back an ICMP "Destination Unreachable" message containing its MTU. This information allows the source host to reduce its assumed path MTU appropriately. The process repeats until the MTU becomes small enough to traverse the entire path without fragmentation. In our case if the wireless client sends a packet of 2272B with the don’t fragment bit set to 1, the access point will discard the packet and reply with an ICMP “Destination unreachable” forcing the wireless client to reduce its path MTU.
c)
When messages are received from the wireless link the MTU size of the packets is 2272 bytes and the wireless bridge will have to fragment them to packets of MTU 1500 bytes. Due to fragmentation more packets are created and these have to be sent over the wired link. If the BER of the wired part was the same as the wireless, then there would be an increase to the number of the dropped packets. Since the BER of the wired network is much smaller that means although the transmitted packets have increased there will not essentially be a corresponding increase to the number of dropped fragments. The outcome of all these would be that the average throughput is not affected.
When MTU path discovery is attempted all along the line the wireless there will be a common MTU size for both the wired and the wireless part. Since the MTU of the wired network is 1500 bytes then the wireless will not be able to use its MTU of 2272 and will have to keep sending packets of MTU 1500. This means more packets to be sent, its packet will have its own header and the average throughput is dropping.

Question 3
a)
Cycle 1
The server receives the 140 bytes but is able to send only 40 bytes to the application, leaving 100 bytes in the buffer. When the acknowledgement is sent back, the server can reduce its send window by 100 bytes leaving 260 bytes. When the client receives this segment from the server it will see the acknowledgment of the 140 bytes sent and slides its window 140 bytes to the right. However, as it slides this window, it reduces its size to only 260 bytes. The new window ensures that the server receives a maximum of 260 bytes from the client which will fit in the 260 bytes remaining in the server’s buffer.
Cycle 2
The client sends a segment of 180 bytes but the server cannot remove any of them. It could buffer the 180 bytes and in the acknowledgment sends for those bytes, reduces the window size by the same amount down to 80 bytes. When the client receives the acknowledgment for 180 bytes it would see the window size had reduced by 180 bytes. The client then slides its window by the same amount as the window size was reduced.
Cycle 3
After the send window is reduced to 80 the client sends a third request of 80 bytes but the server is still busy. The server then reduces its window to 0 telling the client to stop transmission of data entirely.

b)
Increasing the size of the windows enhances the bandwidth and makes the connection more efficient. Considering the basic functionality of the window size, that is holding the data by a device from its peer at any time, increasing the size will then make the device to hold more data in the buffer at a time. This will enhance efficiency of communication.

Increasing the size of the windows enhances the bandwidth and makes the connection more efficient. Considering the basic functionality of the window size, that is holding the data by a device from its peer at any time, increasing the size will then make the device to hold more data in the buffer at a time. This will enhance efficiency of communication.

Let's say we have to transfer 150MB of data. Our TCP window size is 65KB and the round trip delay is 2ms.
For the 150MB of data and the 65KB window size we will need:
150.000.000 bytes / 65.000bytes = 2307,7 ACKS to be sent.
The total waiting time for the ACKS to arrive are, if the roundtrip is 2ms, meaning one way delay is 0.1ms:
2307.7 * 0.001 seconds = 2,307 seconds
The throughput of the connection for a 2 ms delay and 65KB windows size is:
(65.000 KB * 8) / 0.002 seconds = 52000 / 0.002 = 260.000.000 bits/second = 260 Mb/second
The total transmission time for the file will be:
(150.000.000 * 8) / 260.000.000 = 1.200.000.000 / 260.000.000 = 4.61 seconds (including the waiting time for the ACKs.)
We can conclude from the above that although our available bandwidth is 1000 Mb/second we utilise only 260 Mbit/sec, a utilisation factor of:
260.000.000 / 1.000.000.000 = 0.26 %
The 65KB (2^16) window size is a default value for most operating systems but the RFC 1323 describes that the window can increase up to 2^30. If we further increase that then we can increase the throughput of the line and consequently utilising better the line's available capacity since now it is only up to 26%.
c)
Selective Acknowledgement (or SACK) is this technique implemented as a TCP option that can help reduce unnecessary retransmissions on the part of the sender. If the TCP connection has negotiated the use of SACK (through the use of the TCP header option fields), the receiver can offer feedback to the sender in the form of the selective acknowledgement option. The receiver reports to the sender, which blocks of data, have arrived. This list of blocks in the SACK option tells the sender which contiguous byte stream blocks it has received. At maximum, four SACK blocks can be sent in one TCP segment because of the maximum size of the options field in a TCP head is 40 bytes and each block report consists of 8 bytes plus the option header field of 4 bytes (for a total of 36 bytes). SACK information is advisory information only. The sender cannot rely upon the receiver to maintain the out-of-order data. The performance gain is to be had when the receiver does queue and re-order data that has been reported with the SACK option so that the sender limits its retransmissions. (J. Kristof)

Question 4

a)
In trying to locate a certain server on the internet, a computer communicates to a series of DNS servers around the internet. The process involves the domain resolver determining the most appropriate domain name through queries that start with the top level label of the domain.
The process entails some steps as follows; - * On sending the message after the composition, the email client connects to the domains SMTP server * The senders email communicates SMTP with email address of the recipient. It hands over the message in the email and the address. * The SMTP processes the email address of the recipient in the same domain. When the names correspond, the message is directly routed to the POP3 server or the IMAP server. In this case, it is routed to the SMTP. In the case of the local client who is sending an email over the network, there will be no communication with another domain server. This communication will be from the internal mail server 10.1.0.10 from 10.1.0.10/8. * To get the server of the recipient, SMTP will communicate to the DNS. It will take the email domain name of the recipient and translate it to an IP address. * On getting the email address, the SMTP server can connect to the recipient SMTP server. For the local client, this communication is maintained within the same server. For sending an email over the internet, the information is passed over to the border router. * The email is the sent along with a series of SMTP servers until it arrives. On arrival, the server scans it and if it happens to understand the source, it forwards the message to the IMAP server.

b)

A DNS lookup occurs when a device supporting an IP asks the DNS server for the IP address that is associated with a certain domain name. When looking for postmaster@myertor.com, the series of lookups that will take place include; the primary DNS server will be seeking to establish the IP address for postmaster@myertor.com. The first request will be to a DNS root server handling .com. the reply generated will be the IP address for .com. It will then ask it to give out the IP for myetor.com. On finding the IP for myetor.com, it will then inquire from this IP, the IP for postmaster@ myetor.com. It will then be able to proceed with the right channel after this.

c) zone " myertor.com" { type master; file " myertor.com.db";
}
; myertor.com

@ IN SOA postmaster@myertor.com’. root. myertor.com. ( 2006020201 ; Serial 604800 ; Refresh 86400 ; Retry 2419200 ; Expire 604800); Negative Cache TTL
;
@ IN NS ns1 IN MX 10 mail IN A 89.16.175.2 ns1 IN A 89.16.175.2 mail IN A 89.16.175.3 www IN A 10.0.0.0/8

d)

External access to the mail servers can be enabled by enabling through the border router. Rights that allow access can be activated. A good example is through the firewall. This has the advantage of effective communication but is risky to the company in that it’s easy to hack into the system

References

1) Kent and Mogul, Fragmentation considered harmful, 1987 2) J. Kristof, TCP congestion Control, available at: http://condor.depaul.edu/jkristof/technotes/congestion.pdf