a)Given a 100M line program with the following cycle breakdowns: 50% R-type 25% load-store 25% branch. How long would it take to execute on a single cycle processor running at 1MHz
b)How long would it take to execute on a pipleine processor running at 10MHz, where: 50% of R type instructions incur a 1 cycle stall 50% of load-store incur a 3 cycle stall all branches incur a 1 cycle stall.
a) Given a 100M line program with the following cycle breakdowns: 50% R-type 25% load-store 25% branch. How long would it take to execute on a single cycle processor running at 1MHz
Given:
Total lines in program = 100 x 106
Assume CPI ie clocks per instruction = 1
Total cycles required for program to execute = 100 x 106
For the Given single cycle processor:
Frequency = 1MHz = 106 Hz
So,
Time required to execute whole program = total cycles required / frequency = 100 x 106 / 106 = 100 seconds
b) How long would it take to execute on a pipeline processor running at 10MHz, where: 50% of R type instructions incur a 1 cycle stall 50% of load-store incur a 3 cycle stall all branches incur a 1 cycle stall.
New frequency = 10 MHz = 10 x 106
From previous part,
Total cycles required for program to execute = 100 x 106
50% are R-type, 25% are load-store, 25% are branch.
Rc = R- type instruction count = 50% = 50 x 106
LSc = load-store instruction count = 25% = 25 x 106
Bc = branch instruction count = 25% = 25 x 106
Assume CPI for simple instructions = 1
50% of R-type instructions incur a 1 cycle stall
CPIR = Average CPI for R-type instructions = .5( 1) + .5( 1+1 ) = 1.5
50% of load-store instructions incur 3 cycle stall
CPILS = Average CPI for load-store instructions = .5(1) + .5( 1+3 ) = 2.5
all branch instructions incur a 1 cycle stall
CPIB = Average CPI for branch instructions = 1+1 = 2
Tc = Total cycles required to execute whole program = Rc x CPIR + LSc x CPILS + Bc x CPIB
=> Tc = 1.5 x Rc + 2.5 x LSc + 2 x Bc = 1.5 x 50 x 106 + 2.5 x 25 x 106 + 2 x 25 x 106
=> Tc = 75 x 106 + 62.5 x 106 + 50 x 106 = 187.5 x 106
So,
Time required to execute whole program = total cycles required / frequency = 187.5 x 106 / 10 x 106 = 18.75 seconds
//do comment if any problem arises
a)Given a 100M line program with the following cycle breakdowns: 50% R-type 25% load-store 25% branch....
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