Question

a)Given a 100M line program with the following cycle breakdowns: 50% R-type 25% load-store 25% branch....

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.

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Answer #1

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

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