Computer Organization & Architecture

Datapath and control

Use this for “A program runs 2 × 10⁹ instructions at CPI 1.5 on a 3 GHz CPU. How long does it take?”

Diagram

Instr.memoryRegisterreadALUDatamemoryRegisterwriteclock cyclesone cycle: T = 0.3333 nsi1i2i3i4i5each instruction takes CPI = 1.5 cycles on average
t = IC × CPI × T
t = 2,000,000,000 × 1.5 × 0.3333 ns
t = 1 s
This is an example. Type your own values to start a new problem.
t (s)
=
IC
×
CPI
÷
f (GHz)
T
Clock period
calculated
MIPS
MIPS rating
calculated

Formulas

T
=
1
f
0.3333
=
1
3
t
=
IC
×
CPI
f
×
10
9
1
=
2,000,000,000
×
1.5
3 ×
10
9
MIPS
=
IC
t
×
10
6
2,000
=
2,000,000,000
1 ×
10
6

Assumptions

•
CPI is the average number of clock cycles an instruction takes over the whole program.
•
The clock period is T = 1 ÷ f, so a 1 GHz clock ticks every 1 ns.
•
The CPU runs only this program; time spent waiting for input and output is left out.
•
MIPS counts millions of instructions a second, whatever each instruction does.

Step-by-step

WE KNOW
IC = 2,000,000,000, CPI = 1.5, f = 3 GHz
FIND
clock period (T), CPU time (t), MIPS rating (MIPS)
Step 1 · Find clock period (T)
Use T = 1 ÷ f
The
period
is
one
over
the
clock
rate;
1
1
GHz
is
1
ns.
T
=
1
3
T
=
0.3333
ns
Step 2 · Find CPU time (t)
Use t = IC × CPI ÷ (f × 10⁹)
Instructions
×
cycles
each
gives
the
cycles;
the
clock
runs
f
×
10
9
of
them
a
second.
t
=
2,000,000,000
×
1.5
3 ×
10
9
t
=
3 ×
10
9
3 ×
10
9
t
=
1
s
Step 3 · Find MIPS rating (MIPS)
Use MIPS = IC ÷ (t × 10⁶)
Millions of instructions for each second of CPU time.
MIPS
=
2,000,000,000
10
6
MIPS = 2,000
Check
Put the numbers back in. Both sides should match.
0.3333
=
1
3
✓
1
=
2,000,000,000
×
1.5
3 ×
10
9
✓
2,000
=
2,000,000,000
10
6
✓
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