IPv4 Header Length Constraints and IHL Field Limits
Short answer
B 64B is impossible because the IPv4 IHL field is limited to 4 bits, capping the header at 60 bytes (15 * 4 bytes).
Question
Key Takeaways
The trap involves confusing decimal multiples with the 4-byte word alignment; 64B is a clean multiple of 4 but requires a 5-bit field to represent its IHL value (16), which does not exist in IPv4.
Identify invalid IPv4 packet header lengths based on the 4-bit IHL field constraints. The correct answer is B (64B), as it exceeds the maximum 60-byte limit defined by RFC 791.
Common mistake: Choosing 60B (C) as incorrect because candidates mistakenly believe the maximum header size is strictly 20 or 32 bytes, ignoring that the full 15-value IHL allows for up to 60 bytes of options.
Expert Analysis
The IPv4 header length is governed by the Internet Header Length (IHL) field, which is a 4-bit value. This field specifies the header size in units of 32-bit (4-byte) words. The minimum valid IHL value is 5, representing the mandatory 20 bytes of the standard header. The maximum valid IHL value is 15, representing 60 bytes when including all possible options.
To determine validity, divide the byte count by 4. If the result is between 5 and 15 inclusive, it is a valid header length.
- Option D (20B): 20 / 4 = 5. Valid.
- Option A (32B): 32 / 4 = 8. Valid.
- Option C (60B): 60 / 4 = 15. Valid.
- Option B (64B): 64 / 4 = 16. Invalid, as 16 cannot be represented in 4 bits.
Comments highlight this binary limitation, noting that while 60B is the maximum, 64B exceeds the capacity of the 4-bit IHL field.
Exam Strategy
When analyzing protocol headers, always check the bit-width of control fields. For IPv4, memorize that the IHL is 4 bits, meaning the maximum multiplier is 15. Any question asking for invalid sizes should immediately flag values resulting in an IHL > 15 or < 5.
Frequently Asked Questions
Why is 60B the maximum IPv4 header size?
Is 20B always the minimum header size?
Community Votes
67% of anonymous learners picked answer B. Votes are pick records left by other test-takers — they are not the verified answer.
Discussion 3 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 3
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Fmoura17 👍 1 Answered BThe correct answer is B. 64B. Checking each option: A. 32B ✅ (Valid) 32 ÷ 4 = 8 → IHL = 8, which is within the valid range (5–15) ✅ B. 64B ❌ (Invalid) 64 ÷ 4 = 16 → IHL = 16, which exceeds the maximum value of 15 that a 4-bit field can represent. Therefore 64 bytes is impossible ❌ C. 60B ✅ (Valid) 60 ÷ 4 = 15 → IHL = 15, which is the maximum valid value ✅ D. 20B ✅ (Valid) 20 ÷ 4 = 5 → IHL = 5, which is the minimum valid value (fixed header, no options) ✅
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Heidorn 👍 1 Answered CThe IPv4 packet header is only up to 60Bytes, 32 is also a valid value in its step.
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amiie00 👍 1 Answered B64 cannot
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