Khalid Sarwar Engineer NGN HCTE RWP # 03135003545 s1

Khalid Sarwar Engineer NGN HCTE RWP # 03135003545

Lab 1.1: Calculating Bandwidth

Objectives: After completing this lab, you will be able to calculate bandwidth requirements for a live event that is streamed by using Microsoft® Windows Media™ Services.

Scenario : You are the network administrator for an Internet Service Provider (ISP). Your company has been asked to provide Windows Media Services for a live event produced by one of your clients.

You have been asked to calculate the total data transfer required for the event. You have also been asked to verify that your Internet connection can support the maximum estimated concurrent listeners. You have a 45-megabits- per-second (Mbps) connection to the Internet (T3).

Event details : Time of the event: 6:00 P.M.-8:00 P.M. , Encoding rate: 22 kilobits per second (Kbps) , Average length of play: 45 minutes , Total listeners: 1,500 , Concurrent listeners: 1,000

Exercise 1: Calculating Total Data Transfer

In this exercise, you will calculate the total data transfer for a live event.

To calculate total data transfer

1. Calculate the data transfer per minute.

Answer : 22 Kbps multiplied by 60 seconds per minute=1,320 kilobits per minute.

2. Multiply the kilobits per minute by the average length of play.

Answer : 1,320 kilobits per minute multiplied by 45 minutes=59,400 kilobits.

3. Convert the total kilobits to kilobytes.

Answer : 59,400 kilobits divided by 8 bits per byte =7,425 kilobytes.

4. Convert the total kilobytes to megabytes.

Answer :7,425 kilobytes divided by 1,024=7.25 MB.

Exercise 2

Calculating the Required Internet Connection

In this exercise, you will calculate the total data transfer for a live event.

To calculate the required Internet connection

1. Multiply the maximum concurrent listeners by the encoding rate.

Answer : 1,000 users multiplied by 22 Kbps=22,000 Kbps.

2. Convert the maximum kilobits per second to megabits per second.

Answer : 22,000 Kbps divided by 1,024=21.5 Mbps. SE NG HCTE S/Town RWP 051-4425075 Khalid Sarwar Engineer NGN HCTE RWP # 03135003545

3. Can your Internet connection support the maximum number of concurrent users?

Answer : Yes, a T3 connection can support bursting of 21.5 Mbps

Bandwidth Planning

Bandwidth allocation and usage are important aspects of any streaming technology.

Bandwidth Management Strategies :Windows Media Services provides tools to manage resource allocation proactively. The critical scaling factor is the number of concurrent users expected, both on an ongoing basis and at peak times. A well-positioned site might regularly receive a few hundred concurrent user connections, and could peak with many thousands of users during a special event.

One strategy is to plan a server configuration that is designed to handle only a normal sustained load and then to implement contingent services to meet peak-time demands. In addition, you can lower required capacity by offering lower bit rate streams or by using a scalable codec to encode the ASF stream.

Bandwidth and Client Limitations : Windows Media Services limits the maximum number of concurrent client connections to the smaller of either the maximum clients configured or the effective number of client connections based on the maximum bandwidth configured. Both of these settings are configured on the General tab of the CONFIGURE SERVER - Server Properties screen in Windows Media™ Administrator.

These limits affect the entire Windows Media Services system; specifically, all streams delivered by the server running Windows Media Services are affected, including both live and on-demand content.

You can also limit client connections or bandwidth for a specific publishing point. To do this, modify Maximum client limit or Maximum bandwidth limit on the CONFIGURE SERVER - Edit Broadcast Publishing Point screen or the CONFIGURE SERVER - Edit Unicast On-Demand Publishing Point screen for a specific publishing point.

The following tables describe the server and client properties that you can set.

Server/client properties Description Maximum clients The maximum number of concurrent clients that can be connected to the server at any given time. Minimum is 0; maximum is unlimited. Maximum bandwidth The maximum amount of bandwidth that the server streams at any given time. This number is measured in kilobits per second (Kbps). The upper limit should not exceed 85 percent of the capacity of the network adapter. Minimum is 1; maximum is unlimited. Maximum file bitrate The maximum amount of bandwidth at which any one file can be streamed. This number is measured in Kbps. Minimum is 1; maximum is unlimited.

Publishing point properties Description Maximum client limit The maximum number of concurrent clients that can be connected to the publishing point at any given time. Minimum is 0; maximum is unlimited. SE NG HCTE S/Town RWP 051-4425075 Khalid Sarwar Engineer NGN HCTE RWP # 03135003545

Maximum bandwidth limit The maximum amount of bandwidth that the publishing point streams at any given time. Minimum is 1; maximum is unlimited.

Note Clients connected to Windows Media Services that are not receiving streams are still allocated bandwidth. If Microsoft Windows Media™ Player has allocated bandwidth but does not send a command during the client inactivity time-out period (60 seconds), that bandwidth is de-allocated. .9 INTERNET BANDWIDTH CALCULATION 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 Pakistan Ø 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” = Modem speed/capacity (Kbps) x UAR = 32Kbps x 10% = 3.2 Kbps It 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 other frames 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 in this case: · Satellite delay which is of 250 m sec. · Mean holding time of equipment If 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 or capacity. 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 means that we have not to wait to transmit our frame and T = b/c. If the link is busy then r=1, this cause T = ¥.

SE NG HCTE S/Town RWP 051-4425075 Khalid Sarwar Engineer NGN HCTE RWP # 03135003545

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 Mbps At present total bandwidth requirement for Internet users of Pakistan is 30 Mbps.

SE NG HCTE S/Town RWP 051-4425075