In order to find the capacity of bandwidth on the international network, we have to find:Ø Total number of Internet subscribers - As there are 1,25,000 Internet users at present in PakistanØ Elementary Internet Traffic of one subscriber i.e.,20m Earlang (Standard)Ø User Activity Rate (UAR= Duration of effective transmission/Holding time)Ø Mean rate of effective user “on line” (The rate of 1Earlang) As we know that data received is greater than the data send in case of Internet. Also it has been calculatedafter number of trials that the effective duration of Internet usage is 1.2 minutes out of 12 minutes. For example while the request for the download of picture on Internet, it takes only 1.2 minutes to send a request and receive the information and the remaining of 12 minutes are only wasted in seeing the picture. During seeing/observing the picture there will be no transmission takes place. So UAR becomes:User Activity Rate = UAR= 1.2/12 = 10%Let the Modem speed is 32 Kbps (which is taken on average) for an ISP. It is also supposed that there is only one Internet access provider having 125000 subscribers. This done in order to calculate the international bandwidth required for Pakistan.The mean rate of an effective user “on line” = Modemspeed/capacity (Kbps) x UAR = 32Kbps x 10% = 3.2 KbpsIt means that during one Earlang of traffic (full utilization) the computer uses 3.2 Kbps.Let if the access provider receives many calls at a time. Let 20E traffic is received from the area having 30 circuits. Then the rate used (for 1 PCM for one Earlang) = Internet traffic x rate of one earlang /Efficiency.Let the efficiency is 70%, and the internet traffic is equal to the product of elementary traffic one Internet subscriber and total number of subscribers, which comes out 2500 earlang (20x10-3 x 125000)So the rate (used) = 2500 x 3.2Kbps / 0.7 =11.5Mbps It means that we have to carry 11.5 Mbps traffic from the 30 circuits. We have to determine the quality of service before the access provider circuits, so we have to find the failure rate.Capacity required to carry 11.5 Mbps:Consider there is only one processor and software is running on the computer and one user to use the link at the same time. We know that the frame length of IP frame is 320 bits. Then the transmission time ‘T’ of frame to transmit at the rate of 32 Kbps is 10-m sec. (320bits/32 Kbps). As we know that we have to share the capacity because there are large number of otherframes are coming from other places also. It means that we have to wait for 10 m sec. x no. Of frames which are ahead of us.As T = Length of frame (b)/capacity(c)There are two quality factors, which are involved inthis case:· Satellite delay which is of 250 m sec.· Mean holding time of equipmentIf these factors are taken into account, then the value of ‘T’ becomes:T = b/c (1-r), where r is similar to UAR, which is equal to the ratio of used rate/Physical rate orcapacity. For example the capacity of 64 Kbps and used rate of 32 Kbps, Then r=0.5.When we are going to use free link then r=0. It meansthat we have not to wait to transmit our frame and T = b/c. If the link is busy then r=1, this cause T = ¥. It is clear from the graph that it is dangerous to use the r (Charge) higher than 0.5 because quality decreases abruptly after 0.5. For an ordinary Internet user we use the value of r equal to 40%. Then for 11.5 Mbps the capacity required is determined as:r = Used rate / Capacity =11.5 Mpbs/ 0.4 = 28.75 Mbps@ 30 MbpsAt present total bandwidth requirement for Internetusers of Pakistan is 30 Mbps.