Refer to the partial command output shown. Which two statements are correct regarding the router hardware? (Choose two.)
A.
Total RAM size is 32 KB.
B.
Total RAM size is 16384 KB (16 MB).
C.
Total RAM size is 65536 KB (64 MB).
D.
Flash size is 32 KB.
E.
Flash size is 16384 KB (16 MB).
F.
Flash size is 65536 KB (64 MB).
I think A and E are the correct answer. Please check.
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Just read the output
RAM = 64 MB
FLASH = 16 MB
NVRAM = 32 KB
==> answer C, E
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C and E
RAM: 53248k + 12288k = 65536 K
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The line “Cisco 2621 (MPC860) processor (revision 0×600) with 53248K/12288K bytes of memory” tells how much RAM in your router. The first parameter (53248) specifies how much Dynamic RAM (DRAM) in your router while the second parameter (12288K) indicates how much DRAM is being used for Packet memory (used by incoming and outgoing packets) in your router. Therefore you have to add both numbers to find the amount of DRAM available on your router -> C is correct.
Note: Cisco 4000, 4500, 4700, and 7500 routers have separate DRAM and Packet memory, so you only need to look at the first number to find out the DRAM in that router.
The flash size is straightforward from the line “16384K bytes of processor board system flash (Read/Write)” -> E is correct.
(Reference: http://www.cisco.com/en/US/products/sw/iosswrel/ps1834/products_tech_note09186a00800fb9d9.shtml)
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