Which of the following statements regarding DSCP bits are TRUE? (Choose three)

Question

Options

• AThe three most significant (first) bits define 8 forwarding classes.
• BThe three least significant (last) bits of the DSCP specify the drop probability.
• CThe full 8 bits of the TOS field are used for DSCP.
• DTo convert DSCP to IP precedence, the three most significant (first) bits are matched.
• EDSCP provides for eight drop probabilities.

• A
  90% (53)
• C
  7% (4)
• E
  3% (2)

Explanation

DSCP uses only 6 bits of the 8-bit TOS field - the 3 most significant bits define 8 forwarding classes (2³ = 8), making A correct, and the 3 least significant bits specify the drop probability, making B correct. Because DSCP's first 3 bits are identical in position to the IP Precedence bits in the old TOS model, converting between them is as simple as reading those same 3 bits, making D correct.

C is wrong because DSCP uses only 6 of the 8 TOS bits - the remaining 2 bits are reserved for ECN (Explicit Congestion Notification), not DSCP. E is wrong because although there are 3 drop-probability bits (theoretically allowing 8 values), only 3 drop probability levels are actually defined and used: low, medium, and high.

Memory tip: Think of DSCP as "3+3 out of 6" - first 3 bits = Class (who you are), last 3 bits = Drop (how expendable you are). Since the class bits align exactly with IP Precedence bits, conversion is trivial - just read the top 3. And remember: 6 bits ≠ 8 bits, so C is always a trap.

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