---
title: "Many Cores and the Limit on Speedup"
description: "For a fixed workload with unchanged serial time, Amdahl's ideal model gives speedup 1 / (s + (1 - s) / N), where s is the serial fraction and N is the core count. For positive s, 1 / s is the limiting"
canonical: https://lightmysky.com/learn/computing/many-cores-and-the-limit-on-speedup-mt_dy3-K9dV19
source: https://lightmysky.com/learn/computing/many-cores-and-the-limit-on-speedup-mt_dy3-K9dV19.md
retrieved: 2026-11-12
---

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# Many Cores and the Limit on Speedup

For a fixed workload with unchanged serial time, Amdahl's ideal model gives speedup 1 / (s + (1 - s) / N), where s is the serial fraction and N is the core count. For positive s, 1 / s is the limiting bound. Coordination and resource contention can reduce gains.

Subject: Computing · Area: Computer Systems · Ages 19 to 20
Page: https://lightmysky.com/learn/computing/many-cores-and-the-limit-on-speedup-mt_dy3-K9dV19

## Ready when they can

- Compute ideal finite-core speedup and the limiting bound for a fixed workload that is one tenth serial
- Distinguish work that splits across cores from work that splits across data lanes
- Explain why adding cores can make a program slower

## Lesson: Why ten cores do not always give tenfold speedup

Pipelining overlaps stages inside a core; multiple cores can share a program's work. In Amdahl's model, the workload stays fixed and the serial part takes the same time. If its positive fraction is s, speedup approaches the bound 1 / s as core count grows. A serial tenth gives a tenfold bound, not tenfold speedup at every count.

*(drawing: The serial part keeps its duration. Only the parallel part is divided among cores. Dividing both parts changes Amdahl's assumptions.)*

**Example.** Normalize the one-core time to 1. For N cores, ideal time is s + (1 - s) / N; speedup is 1 divided by that time. This assumes equal sharing of the parallel part and no added overhead. With s = 0.25 and N = 3, time is 0.25 + 0.75 / 3 = 0.5, so speedup is 2.

Multiple cores can execute different instruction streams. SIMD data lanes apply the same operation to several data elements together. Both can be used by one program: several cores might each perform SIMD additions. Lanes do not turn the serial part of a fixed workload into independent tasks.

Actual time includes overhead that the ideal formula leaves out. Dividing work, coordinating results and waiting for shared resources all take time. A lock allows controlled access to a shared resource; workers may have to wait for it. If these costs exceed the time saved by parallel work, more cores can make the program slower.

For a serial tenth on 100 cores, ideal time is 0.1 + 0.9 / 100 = 0.109. Speedup is 1 / 0.109, about 9.17. This is below ten because the parallel part still takes some time. Increasing N reduces that part, but leaves the serial 0.1 unchanged.

**Recap.** For fixed work, unchanged serial time limits ideal speedup. Compute finite-core time before inverting it. Cores and SIMD lanes differ; overhead can erase the gain.

*Original lesson, questions and flashcards written for this dataset.*

## Practice

8 questions on this page, each with its working shown.

## Needs first

- [Pipelining and the Cost of a Hazard](https://lightmysky.com/learn/computing/pipelining-and-the-cost-of-a-hazard-mt_GI_fPY9d6s)

## Opens up

- [What an Operating System Is: Privilege and System Calls](https://lightmysky.com/learn/computing/what-an-operating-system-is-privilege-and-system-calls-mt_0kMavEKxpX)
- [Deep Learning: Convolution, Sequence and Scale](https://lightmysky.com/learn/computing/deep-learning-convolution-sequence-and-scale-mt_karzGN5JnI)
