<?xml version="1.0" encoding="UTF-8"?><rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:atom="http://www.w3.org/2005/Atom" version="2.0"><channel><title><![CDATA[Topics tagged with cs314]]></title><description><![CDATA[A list of topics that have been tagged with cs314]]></description><link>https://community.secnto.com//tags/cs314</link><generator>RSS for Node</generator><lastBuildDate>Mon, 08 Jun 2026 23:52:55 GMT</lastBuildDate><atom:link href="https://community.secnto.com//tags/cs314.rss" rel="self" type="application/rss+xml"/><pubDate>Invalid Date</pubDate><ttl>60</ttl><item><title><![CDATA[CS314 Handouts pdf]]></title><description><![CDATA[<p dir="auto"><a href="#">CS314 Handout PDF</a></p>
]]></description><link>https://community.secnto.com//topic/1804/cs314-handouts-pdf</link><guid isPermaLink="true">https://community.secnto.com//topic/1804/cs314-handouts-pdf</guid><dc:creator><![CDATA[zaasmi]]></dc:creator><pubDate>Invalid Date</pubDate></item><item><title><![CDATA[CS314 Assignment 1 Solution and Discussion]]></title><description><![CDATA[@moaaz said in CS314 Assignment 1 Solution and Discussion:

The complete channel bandwidth.

Bandwidth is a fixed quantity, so it cannot be changed. Hence, the channel capacity is directly proportional to the power of the signal, as SNR = (Power of signal) / (power of noise). so for example a signal-to-noise ratio of 1000 is commonly expressed as: 10 * log10(1000) = 30 dB.
Capacity = bandwidth * log2(1 + SNR)
Output1 : C = 30000 * log2(1 + SNR) = 30000 * 11.62 = 348600 bps
]]></description><link>https://community.secnto.com//topic/1763/cs314-assignment-1-solution-and-discussion</link><guid isPermaLink="true">https://community.secnto.com//topic/1763/cs314-assignment-1-solution-and-discussion</guid><dc:creator><![CDATA[zaasmi]]></dc:creator><pubDate>Invalid Date</pubDate></item><item><title><![CDATA[CS314 Assignment 3 Solution and Discussion]]></title><description><![CDATA[solution #2
]]></description><link>https://community.secnto.com//topic/1098/cs314-assignment-3-solution-and-discussion</link><guid isPermaLink="true">https://community.secnto.com//topic/1098/cs314-assignment-3-solution-and-discussion</guid><dc:creator><![CDATA[bc170403108 MUHAMMAD WALEED]]></dc:creator><pubDate>Invalid Date</pubDate></item><item><title><![CDATA[CS314 Assignment 2 Solution and Discussion]]></title><description><![CDATA[Answer 1 (a)
Uplink Frequency Band=890 − 915MHz
Downlink Frequency Band = 935 − 960 MHz
Frequency Width for any band =Upper Frequency Limit − Lower Frequency Limit =915−890MHz
=25MHz Width of a frequency channel=200KHz
Maximum Available Channels
= Band Width / Channel Width
[image: Ni9sgA9.png]
However, actually 124 channels are used if you consider the guard bands used between adjacent channels to avoid any inteference
Answer – 1 (b)
Channel Number (C ) = 50
Uplink carrier frequency for channel 50 = F uplink (50)
Fuplink (50)  =Baseuplink frequency + ChannelWidthz * Channel Number
=890MHz+(200KHzx50) =890MHz+(0.2x50MHz )
=890+10MHz
= 900 MHz
Fdownlink (50)=Downlinkcarrier frequency forchannel50+DuplexSpacing =900MHz+45MHz
= 945 MHz
Alternate Method for calculating Downlink carrier frequency for channel 50
Downlink carrier frequency for channel 50 = Fdownlink (50)
Answer 2 (a)
Fdownlink (50)
=Basedownlink frequency+ChannelWidthChannel Number =935+(200KHz50)
= 935MHz+(0.250MHz )
= 935+10MHz
= 945 MHz
Ali is able to receive signals from Zain Network due to Roaming Service. Mobilink does not have coverage in Makkah so in order to provide coverage to its subscribers, Mobilink has an agreement with Zain Network. As per this agreement, Zain Telecom provides all the available services to the roaming customer of Mobilink.
Answer 2(b)
Following network components will be used in this case
• Mobilink’s HLR
• Zain’s MSC and VLR
• Mobilink’s Authentication Center (AuC)
Answer 2©
When Ali makes a call from his phone, the request will first go to Zain Telecom’s BS (Base station) which will direct the request to MSC the Switching Center. MSC has a dedicated VLR (Visitor Location Register) installed which is dedicated for the visitor users. Since, Ali is a visitor, MSC will request VLR to fetch the Ali’s subscription details. If Ali’s subscription is already available, the VLR will authenticate Ali and allow the MSC to proceed with the call. Otherwise, VLR will look up the subscriber information from HLR. HLR before sending the subscriber details, asks AuC (Authentication Center) to authenticate the subscriber. If subscriber is valid and not blocked, then HLR provides the subscriber details to VLR.****
]]></description><link>https://community.secnto.com//topic/728/cs314-assignment-2-solution-and-discussion</link><guid isPermaLink="true">https://community.secnto.com//topic/728/cs314-assignment-2-solution-and-discussion</guid><dc:creator><![CDATA[zareen]]></dc:creator><pubDate>Invalid Date</pubDate></item><item><title><![CDATA[CS314 Assignment 1 Solution and Discussion]]></title><description><![CDATA[@zareen said in CS314 Assignment 1 Solution and Discussion:

What should be the minimum distance kept between two co-channel cells in order to avoid any interference?

Solution
[image: HCY59gC.png]
]]></description><link>https://community.secnto.com//topic/557/cs314-assignment-1-solution-and-discussion</link><guid isPermaLink="true">https://community.secnto.com//topic/557/cs314-assignment-1-solution-and-discussion</guid><dc:creator><![CDATA[zareen]]></dc:creator><pubDate>Invalid Date</pubDate></item></channel></rss>