4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits
(frame no * page size) + offset value = physical add where frame value is the value present in the corresponding page number offset value is the last n bits of the logical address page no is the first m-n bits of logical address 2^m is the logical address 2^n is the page size
As was given for a 4 Page, 1024 words & 64 frames (shown below) 4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits The answer for this problem is 13. 8 pages -> 2^3 bits 1024 bytes -> 2^10 bits 32 frames -> 2^5 bits Therefore: Logical memory = 3+10=13 bits (Page + Word) Physical memory = 10 + 5 =15 bits (Word + Frame)
32 bits in a IPv4 address
a.Logical address will have3 bits to specify the page number (for 8 pages) .10 bits to specify the offset into each page (210 =1024 words) = 13 bits.b.For (25) 11 32 frames of 1024 words each (Page size = Frame size)We have 5 + 10 = 15 bits.
You need 20 bits of address bus to address 1 Mb of memory.
How many bits are there in a data link layer ethernet address?
32 bits (for IPv4), or 128 bits (for IPv6).32 bits (for IPv4), or 128 bits (for IPv6).32 bits (for IPv4), or 128 bits (for IPv6).32 bits (for IPv4), or 128 bits (for IPv6).
a TcP IPv4 ip address has 32 bits.
It takes 10 bits.
There are 16 bits in a port address, which gives a port range of 0 - 65535
IPv4 => 32 bits => 4 bytes