Competence Based Education – Internet Protocols E XAMPLES AND P ROBLEMS

Competence Based Education – Internet Protocols Example 1 4 hops between 2 terminal nodes; 3200 message bits; transmission rate 9600bps on all links; 24 overhead bits for each packet; 1024 bits fixed packet size;1ms per hop signal propagation delay, 1sec call set up time for circuit switched connection across 4 hops. What is the total time to send the complete message using circuit switching and packet switching? What is the transmission delay for each bit between the terminal nodes. Circuit switching: 3200 bits at 9600 bps= 3200/ sec message duration total propagation delay =0.004 sec total time for message is =1.337 sec

Competence Based Education – Internet Protocols Total delay is sec since bits are not transmitted during call set up. Packet switching: Number of packets=4; ( =1000 bits of message data for I st packet, 4 th packet has 200 bits of message data and 800 dummy bits) Packet duration = 1024/9600=0.107 sec. Entire 1024 bits packet received by each node from proceding node is 1024/ =0.108 Total message time is =4* *0.107=0.753 sec since there are 4 hops and 3 packets in sucession after Ist complete packet is received. transmission delay is 4*(0.001)+3(0.107)=0.325 sec Example 1 - Solution – fixed packet size/Transmission rate

Competence Based Education – Internet Protocols There are 3 multimedia streams involved in a mm application. After running for 30 min., it has been observed that there is a constant delay of 3ms in stream B and relative delay 0.2 ms in stream C. In the packet for stream C is generated at rate of 14 kb/sec. Find rate of change of delays in stream C. No. of streams =3 Delay in stream B, d b = 3ms(constant) Delay in stream C, d c = 0.2 ms (relative) Generation rate from stream C=r c =14 kb/sec so no. of bits generated from C in 30 min =14*30*60=25200 Kbits and constant delay for stream C= 3±0.2 so Bits not received: In case of d c =3+02=14*(3+0.2)=44.8 bits In case of d c =3-0.2=14*(3-0.2)=39.2 bits Rate of change of delay in C=d(d c )/dt=0.2ms/3ms= Example 2.

Competence Based Education – Internet Protocols Example 3 Size of Media Block (Granularity parameter G) G=3; Data transfer rate from disc is r c =2 sec; playback rate is r p = 0.5 sec. How do you place media block Playback Duration= S+G/r p <= G/r c S+G/2 <=G/0.5 S+G<=12 S<=9 Time to skip over a gap and to read next media block is <= Than duration of play back

Competence Based Education – Internet Protocols In a Multitasking system at an instant of time, there are 8 multimedia applications and 12 non-multimedia applications are catered with transfer rate of 10 Mbps. The consumption rate of all multimedia application is at uniform rate bytes/sec and If the non-multimedia data block size is 20 kbps. Find block size of multimedia data which could enable smooth functioning of the system. No. of non-MM application =12 No. of MM application=8 transfer rate =r t =10 Mbps Consumption rate r c =44100 bps non-MM data block= S b =20 Kb So for smooth functioning of system MM data block= [ns b ] /[r t /r c ] - n c = 12*20*2 10 / [10*2 20 /44,100]-8 =1.045 kbytes Example 4 -

Competence Based Education – Internet Protocols EXAMPLE 5 How long does it take to transfer 128 kb from disk to memory assuming the data is found sequentially on one track (Assume disk still must seek and rotate to find start of data? We are give seek time =15 ms, 6000 RPM is = 100 RPS or 10ms/rotation on average distance must wait for half of rotation or 5 ms. Transfer time is time to read the whole track from disk in 10 ms. However transfer bandwidth is constrained by the I/O bus so transfer is ms (128KB MB/s is 31.25ms) Total time for disk to read 125 KB of sequential data is t seek +t rotation + t transfer 15ms +5 ms ms =51.25 ms

Competence Based Education – Internet Protocols In a Playout system the disk read is providing multimedia data at the rate of 20 Mbytes per second. It was found that the contents of buffer is 30 Mbytes at an arbitrary time t 1 and after 4 seconds of that it was 46 Mbytes. Find the consumption rate of this playout system. Data transmission rate=20 Mbps Buffer content at time t 1 =30 Mbytes After 4 sec, at time (t 1 +4) buffer content=46 Mbytes. Let us assume consumption rate =r c Mbps

Competence Based Education – Internet Protocols than within 4 sec, let us see the buffer position Transmission rate-consumption rate +old buffer content=Present Buffer content =20*4-x*4+30=46 20*4-x*4= =4x x=64/4=16 Consumption rate x=16 Mbps

Competence Based Education – Internet Protocols In a multimedia system which has been utilizing a disk system has transfer rate r t =2.42 Mbps and consumption rate =1.2 Mbps, pause for every read-6ms and delay of 17ms and sector sixe of 512 bytes. find the Buffer requirement for smooth functioning of application. transfer rate=2.42Mbps consumption rate=1.2 Mbps total delay=6+17=23 ms Maximum number of buffers required for smooth running of application n=(d max +S s /r t )*r c /S s =57.021

Competence Based Education – Internet Protocols In a multimedia system, if consumption rate is lesser than transfer rate find the buffer requirement. consumption function is 1.5 Mbps and transfer rate is 2.5 Mbps, sector size is 512 bytes and data playout t r is completed at 1.2 ms. Umax=S s +t r (r t -r c )S s.

Competence Based Education – Internet Protocols In a multimedia system, if consumption rate is greator than transfer rate find the buffer requirement. consumption function is 1.5 Mbps and transfer rate is 2.5 Mbps, sector size is 512 bytes and data playout t r is completed at 1.2 ms. Umax=[t r (r c -r t )]+S s / r c