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<h3>Frame relay link dimensioning</h3>

<p>The Gb interface capacity must equal the estimated GPRS
traffic during the busy hour, plus the overhead generated on the Gb interface.
The required number of PCM timeslots can be defined by dividing the estimated
GPRS peak traffic by 64 kbit/s.</p>


<p>The capacity of the Frame Relay (FR) connection can be
configured in steps of 64 kbit/s, from one 64 kbit/s PCM timeslot to 31 timeslots.
Consequently, each logical FR connection has a capacity of n*64 kbit/s, where
n can be 1-31 (from 64 kbit/s to 1.984 Mbps).</p>


<p>Furthermore, we have to consider the uplink and downlink
traffic and calculate the constraints involved, as follows:</p>


<ul>
<li><p><em>Constraints from user traffic for Uplink</em> (G<sub>b1</sub>)</p>


<p>G<sub>b1</sub> = Ceiling (TR<sub>BSCSGSN</sub>  / TR<sub>Gb</sub>)</p>
</li>
<li><p><em>Constraints from user traffic for Downlink</em> (G<sub>b2</sub>)</p>


<p>G<sub>b2</sub> = Ceiling (TR<sub>SGSNBSC</sub>  / TR<sub>Gb</sub>)</p>
</li>
</ul>


<p>The number of interfaces required can now be calculated with the following
formula:</p>


<p>NGb = MAX (G<sub>b1</sub>, G<sub>b2</sub>)</p>


<p>In this calculation we dimension by using the largest possible
amount of traffic on the uplink and downlink. It is possible to do so since
the Gb links are bi-directional.</p>


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<h6>Example</h6>



<p>In this example, the following values are used:</p>


<ul>
<li><p>Subscribers: 1000</p>
</li>
<li><p>TR<sub>DownLink</sub> per user: 3 Kbps</p>
</li>
<li><p>TR<sub>UpLink</sub> per user: 0.2 Kbps</p>
</li>
<li><p>Packet Size: 512 octets</p>
</li>
<li><p>OH % signalling: 5% </p>
</li>
</ul>



<p>With a packet size of 512 octets, the overhead values are:</p>


<ul>
<li><p>OH%<sub>min</sub>   = 5,86%</p>
</li>
<li><p>OH%<sub>max</sub>  = 21,09%. </p>
</li>
</ul>



<p>An intermediate value is chosen for the calculation: OH<sub>data</sub>%
=13,5%. </p>



<p>We then have (for user traffic alone):</p>



<p>TR<sub>BSCSGSN</sub> = TR<sub>UL</sub> * NUM<sub>USR</sub>
= 200 kbps</p>



<p>TR<sub>SGSNBSC</sub> = TR<sub>DL</sub> * NUM<sub>USR</sub>
= 3 Mbps</p>



<p>Next, take into account the protocol overheads and signalling
traffic, where Oh<sub>dataUL</sub>= Oh<sub>dataDL</sub>= OH<sub>data</sub>
= 13,5%:</p>



<p>TR<sub>UL</sub> = TR<sub>BSCSGSN</sub> * (1+OH<sub>dataUL</sub>)
* (1+ OH<sub>signalling</sub>) = 238 Kbps</p>



<p>TR<sub>DL</sub> = TR<sub>SGSNBSC</sub> * (1+OH<sub>dataDL</sub>)
* (1+ OH<sub>signalling</sub>) = 3,6 Mbps</p>



<p>G<sub>b1</sub> = Ceiling (TR<sub>UL</sub> / TR<sub>Gb</sub>)
= Ceiling (238 kbps/ 64kbps) = 4</p>



<p>G<sub>b2</sub> = Ceiling (TR<sub>DL</sub> / TR<sub>Gb</sub>)
= Ceiling (3,6Mbps/ 64Kbps) = 57</p>



<p>Finally, calculate the total number of Gb links (FR timeslots)
required:</p>



<p>N<sub>Gb</sub>= Max (G<sub>b1</sub>, G<sub>b2</sub>) =
57</p>
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