# Amdahl's Law > Amdahl's law tells you how much faster a whole program gets when you speed up only part of its work, and why the part you leave alone caps the gain. Page: https://formulabro.yunusemrevurgun.com/computer-science/amdahls-law/ Category: Computer Science Published: 2026-10-22 Updated: 2026-10-22 ## Formula $$ S = \frac{1}{(1 - p) + \frac{p}{s}} $$ Read it out loud: "S equals one divided by the quantity one minus p, plus p divided by s." ## What each symbol means - $S$: The overall speedup: how many times faster the whole job runs. A value of 1 means no change and a value of 2 means twice as fast. - $p$: The fraction of the work that can be sped up, from 0 to 1. For example $p = 0.9$ means 90% of the job. - $s$: The speedup of that part alone. With 8 cores on a perfectly split job, $s = 8$. - $1 - p$: The fraction of the work that stays the same, the part that cannot be sped up. - $\frac{p}{s}$: The time the sped-up part now takes, as a fraction of the original total time. ## What the formula is doing Imagine a program whose total running time is 1 unit. A fraction $p$ of that time is work that can be split across several processors. The rest, $1 - p$, must run one step after another and cannot be split. Without any speedup the program takes $p + (1 - p) = 1$ unit. With the speedup, the splittable part takes $\frac{p}{s}$ of the original time, and the part that cannot be split still takes $1 - p$. The new total time is $(1 - p) + \frac{p}{s}$. Speedup is the old time divided by the new time. The old time is 1, so the speedup is $S = \frac{1}{(1 - p) + \frac{p}{s}}$. That is the formula. The shape explains a hard limit. As $s$ grows, $\frac{p}{s}$ shrinks toward 0, but the $1 - p$ part never shrinks. So the speedup can never be bigger than $\frac{1}{1 - p}$, however many processors you add. ## An everyday picture Think of cooking a holiday dinner. Chopping the vegetables can be shared among four helpers, which speeds that part up a lot. The oven still needs its one hour, and no helper can make it bake faster. The dinner cannot be ready sooner than the oven allows, however many people chop. ## Worked example: A program that is 90% parallel on 8 cores A program spends 90% of its time on work that can be split across cores, so $p = 0.9$. That part runs $s = 8$ times faster on 8 cores. The other 10% is serial and stays the same. 1. Serial part: $1 - p = 1 - 0.9 = 0.1$ of the original time. 2. Parallel part on 8 cores: $\frac{p}{s} = \frac{0.9}{8} = 0.1125$ of the original time. 3. New total time: $0.1 + 0.1125 = 0.2125$, compared with 1 before. 4. Speedup on 8 cores: $S = \frac{1}{0.2125} \approx 4.71$. 5. Limit with unlimited cores: the parallel part takes almost no time, so the total time is close to $0.1$ and $S = \frac{1}{0.1} = 10$. **Answer:** On 8 cores the program runs about **4.71 times faster**, not 8 times faster. Even with unlimited cores it can never get more than **10 times** faster, because the 10% serial part still takes its full time. ## Common mistakes - Assuming 8 cores give 8 times the speed. Only the parallel part gets faster, so the total speedup is always smaller than the number of cores. - Entering $p$ as a percentage, such as 90, instead of a fraction, such as 0.9. The formula needs $p$ between 0 and 1. - Forgetting the serial part. The $1 - p$ term is what limits the speedup, and it is easy to leave out of the calculation. - Reading the law as saying parallel work is useless. It says the gain is limited by the serial fraction. More cores still help, but each extra core helps a little less. ## Where you will meet it - Deciding how many CPU cores or threads a program should use, and when adding more stops helping. - Planning parallel jobs on supercomputers and GPUs, to judge whether a program is worth rewriting for parallel hardware. - Performance tuning, where engineers measure a program and speed up the part that takes the most time first. - Comparing multicore chip designs, where it shows how much a serial part limits the benefit of extra cores. ## Questions people ask ### Does Amdahl's law mean parallel computing is pointless? No. It means the serial part sets a ceiling. When $p$ is small, the gain is small, and when $p$ is close to 1, many cores pay off well. The law tells you where the ceiling is, so you know which part to work on. ### What is a serial part? Work that must be done in order, where each step needs the result of the step before it, such as a calculation that builds on the previous answer. ### Is the speedup ever equal to the number of cores? Only when the whole job can be split perfectly, which is rare. For any job with a serial part, the speedup is smaller than the number of cores. ## Related formulas - [Little's Law](https://formulabro.yunusemrevurgun.com/computer-science/littles-law/) - [Binary Search Steps](https://formulabro.yunusemrevurgun.com/computer-science/binary-search-steps/) ## Sources - [Amdahl's law, Wikipedia](https://en.wikipedia.org/wiki/Amdahl%27s_law) - [Parallel computing, Wikipedia](https://en.wikipedia.org/wiki/Parallel_computing) From Formula Bro (https://formulabro.yunusemrevurgun.com/), a side project developed by yunusemrevurgun.com.