Setting SCSI IDs with jumpers: SCSI IDs start at 0. What jumper pattern selects the fourth SCSI ID on a 3-bit ID block (weights 4-2-1)?
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A011
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B110
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C010
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D101
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E001
Answer
Correct Answer: 011
Explanation
Introduction / Context:SCSI devices use binary ID selection. Understanding how to set jumpers avoids bus conflicts and ensures proper device ordering on the chain.
Given Data / Assumptions:
- SCSI IDs are zero-based: 0, 1, 2, 3, ...
- Three jumpers represent binary weights 4, 2, and 1 (from MSB to LSB).
- We want the fourth ID counting from 0.
Concept / Approach:
Counting from 0: the fourth ID is 3 in decimal (0,1,2,3). Convert 3 to binary using weights 4-2-1: 3 = 2 + 1 = 011. Therefore, enable the 2 and 1 jumpers, leave the 4 jumper off.
Step-by-Step Solution:
Enumerate IDs: 0, 1, 2, 3 → fourth is 3.Convert decimal 3 to binary with weights 4-2-1.3 = 2 + 1 → set jumpers for 2 and 1 → pattern 011.Verify on device label/silk-screen and terminate bus ends as needed.Verification / Alternative check:
Most SCSI devices print a truth table on the chassis. Selecting ID 3 aligns with pattern 011 on 3-bit selectors.
Why Other Options Are Wrong:
- 110: equals 6 (4 + 2), not 3.
- 010: equals 2 only.
- 101: equals 5 (4 + 1).
- 001: equals 1.
Common Pitfalls:
Misreading bit order (left-to-right vs. right-to-left), forgetting the host adapter also consumes one ID, and neglecting termination rules.
Final Answer:
011