HP0-J73 · Question #127
A system administrator wants to add five external juke boxes with four disks each to the wide SCSI controller of a server. The internal boot disks and CD-ROM are installed on a second controller. How
The correct answer is C. 20. Each jukebox presents its four disks as individual LUNs, so 5 jukeboxes multiplied by 4 disks each yields 20 LUNs visible on the wide SCSI controller.
Question
A system administrator wants to add five external juke boxes with four disks each to the wide SCSI controller of a server. The internal boot disks and CD-ROM are installed on a second controller. How many additional LUNs will be visible on the server after connecting the juke boxes?
Exhibit
Options
- A4
- B5
- C20
- D25
How the community answered
(39 responses)- A8% (3)
- B15% (6)
- C72% (28)
- D5% (2)
Why each option
Each jukebox presents its four disks as individual LUNs, so 5 jukeboxes multiplied by 4 disks each yields 20 LUNs visible on the wide SCSI controller.
4 would represent only the disks in a single jukebox, ignoring the remaining four enclosures connected to the controller.
5 would represent only the number of jukebox enclosures, not the individual disk LUNs each enclosure exposes to the host.
In SCSI architecture, a jukebox (multi-disk enclosure) connects as a single SCSI target ID on the bus and exposes each internal disk as a distinct Logical Unit Number (LUN). With 5 jukeboxes each containing 4 disks, the server sees 5 x 4 = 20 LUNs. The internal boot disks and CD-ROM on the second controller are irrelevant to this count because they reside on a separate SCSI bus.
25 incorrectly adds the 5 jukebox units themselves to the 20 disk LUNs; in SCSI, the enclosure is the SCSI target and its disks are the LUNs - the enclosure does not appear as a separate addressable LUN alongside its disks.
Concept tested: SCSI LUN addressing with multi-disk enclosures
Source: https://learn.microsoft.com/en-us/windows-hardware/drivers/storage/introduction-to-storage-class-drivers
Topics
Community Discussion
No community discussion yet for this question.
