1. Amdahl's Law

2. 𝑆𝑝𝑒𝑒𝑑𝑢𝑝= 𝑂𝑙𝑑 𝑇𝑖𝑚𝑒 𝑁𝑒𝑤 𝑇𝑖𝑚𝑒
Speed up
𝑆𝑝𝑒𝑒𝑑𝑢𝑝= 𝑂𝑙𝑑 𝑇𝑖𝑚𝑒 𝑁𝑒𝑤 𝑇𝑖𝑚𝑒
A program used to take 20 minutes to run. Now it takes 15. What is the speedup?
1.33 times or 33%
𝑆𝑝𝑒𝑒𝑑𝑢𝑝= =1.33

3. Gene Amdahl

4. Amdahl's Law
Describes overall speedup of a system when we speed up one part of a system
𝑆= 1 1−𝑓 + 𝑓 𝑘

5. Amdahl's Law
Describes overall speedup of a system when we speed up one part of a system
f : fraction of time part is limiting factor
𝑆= 1 1−𝑓 + 𝑓 𝑘

6. Amdahl's Law
Describes overall speedup of a system when we speed up one part of a system
f : fraction of time part is limiting factor
k : speedup of that part
𝑆= 1 1−𝑓 + 𝑓 𝑘

7. Amdahl's Law
Describes overall speedup of a system when we speed up one part of a system
f : fraction of time part is limiting factor
k : speedup of that part
1 – f : fraction of time doing other stuff
𝑆= 1 1−𝑓 + 𝑓 𝑘

8. Amdahl's Law
Describes overall speedup of a system when we speed up one part of a system
f : fraction of time part is limiting factor
k : speedup of that part
1 – f : fraction of time doing other stuff
S : speed up
𝑆= 1 1−𝑓 + 𝑓 𝑘

9. Example
On a large system, suppose we can upgrade its disk drives for $7,000 to make them 150% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service.
What would the expected speedup be?

10. Example 𝑆= 1 1−𝑓 + 𝑓 𝑘 f : fraction of time part is limiting factor
…upgrade its disk drives for $7,000 to make them 150% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service
f : fraction of time part is limiting factor
k : speedup of that part
1 – f : fraction of time doing other stuff
S : speed up
𝑆= 1 1−𝑓 + 𝑓 𝑘

11. Example 𝑆= 1 1−.30 + .30 2.50 150% increase on 100% = 250%
…upgrade its disk drives for $7,000 to make them 150% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service
f : fraction of time part is limiting factor
k : speedup of that part
1 – f : fraction of time doing other stuff
S : speed up
𝑆= 1 1−
150% increase on 100% = 250%

12. Example 𝑆= 1 .70+.12 𝑆= 1 .82 𝑆=1. 21951 Speed up is ~ 22%
…upgrade its disk drives for $7,000 to make them 150% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service
Speed up is ~ 22% 𝑆=
𝑆= 1 .82 𝑆=

13. Example
On a large system, suppose we can upgrade its disk drives for $7,000 to make them 150% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service.
Or we could spend $10,000 to upgrade the CPU to be 50% faster…

14. Example
…upgrade its cpu to run 50% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service
f : fraction of time part is limiting factor
k : speedup of that part
1 – f : fraction of time doing other stuff
S : speed up
𝑆= 1 1−

15. Example 𝑆= 1 .30+.46 6 𝑆= 1 .76 6 𝑆=1.304 Speed up is ~ 30%
…upgrade its cpu to run 50% faster. Processes spend 70% of their time running in the CPU and 30% of their time waiting for disk service
Speed up is ~ 30% 𝑆= 𝑆=
𝑆=1.304

16. Choices Disks : 22% speedup for $7,000 ~ $318 per %
CPU : 30% speedup for $10,000 ~ $333 per %

17. Non IO Protein String Matching Code Which is the better tradeoff?
4 days execution time on current machine
20% of time doing integer instructions
35% percent of time doing I/O
Which is the better tradeoff?
Compiler optimization that reduces number of integer instructions by 25%
Hardware optimization that reduces the latency of each IO operations from 6us to 5us.

18. Non IO Reduce instructions by 25% = 33% speedup (1/.75)
Reduce IO from 6us to 5us = 20% speedup (6/5)
𝑆= 1 1−
𝑆= 1 1− 𝑆= 𝑆=

19. Limitations of Amdahl's Law
Amdahl's Law doesn't take into account interactions
Speeding up one part of process may change the % of time it is limiting factor