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A progarm drive Data Transfer results overhead of 6 instruction per byte.System uses 100MHZ clock and need 4 clock on an average for any instruction . what is maximum data transfer rate (Apprx) ?

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one instruction needs 40 ns

overhead is 6 instruction in transfering byte

overhead=240 ns

in 240 ns is byte is transfered

data rate=1/240ns

              =4.16Mbps

data rate is approximatelly equal to 4 Mbps
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I dont know why i am not getting it is simple, but please correct me :(

A program data transfer mean the one which is responsible for sending data to and from memory takes a overhead of 6 instruction per byte.

Since the processor speed is given it mean it takes 100MHZ.

that is  no of clock cycles 100*(10) clock cycle executed takes 1 sec 

10cycle in one sec

then time taken by 1 cycle is =(1/108)=  10 nanosecond 

The data transfer rate mean how much number of words are send 1 sec !

Therefore for a complete instruction to execute it will take (4 clock cycle + overhead of 6 instruction per byte)--------------------------(1)

Hence 4 clock cycle time = 4*(1 clock cycle time )= 4*10= 40 nano second

for 1 instruction it take 4 clck= 40 nano sec 

then for 6 instruction it will take = 240nsec

that is take 240 nsec for 1 byte (1 word )

then in 1 sec it can atrsnfer  4.16 * 106  words which is 4MBPS ==> data transfer rate  

Am i right please help me :(

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