KCNA CPU Request Practice Question
A pod has resource requests set to 'cpu: 500m' and 'memory: 256Mi'. The node has 2 CPU cores and 4Gi memory. How many pods with the same resource requests can be scheduled on that node, assuming no other pods?
⚠ Common exam trap
Candidates often mistakenly calculate the maximum number of pods based on memory (16) instead of CPU (4), since memory appears less restrictive. However, CPU is the tighter constraint in this scenario.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
4
Each pod requests 0.5 CPU cores (500m) and 256 MiB of memory. The node has 2 CPU cores, so the CPU limit allows 2 / 0.5 = 4 pods. The node has 4 GiB of memory (4096 MiB), so the memory limit allows 4096 / 256 = 16 pods. The tighter constraint is CPU, which permits exactly 4 pods. Option B is correct.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
2
Why it's wrong here
Two pods consume 1000m CPU and 512Mi memory, leaving capacity for eight pods total, so this undercounts by a factor of four. It is tempting because 2 CPU cores suggests two pods, but the scheduler compares requests (500m each), not whole cores.
- ✓
4
Why this is correct
Scheduling is governed by requests, not limits. Memory allows 4Gi ÷ 256Mi = 16 pods; CPU allows 2000m ÷ 500m = 4 pods. The CPU request is the binding constraint, so exactly 4 pods fit on the node.
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8
Why it's wrong here
CPU permits four pods (4 x 500m = 2000m), but memory limits the count to sixteen (16 x 256Mi = 4Gi), so the binding constraint is CPU at four. The figure eight ignores that requests are summed per resource and the tighter of the two caps applies.
- ✗
16
Why it's wrong here
Sixteen pods would need 8 CPU cores, exceeding the node's 2 cores; memory alone would allow sixteen, but CPU is the binding constraint. It is tempting because 4Gi divided by 256Mi equals sixteen, ignoring the CPU request entirely.
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Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
This KCNA practice question is part of Courseiva's free CNCF certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the KCNA exam.