In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.
17 address lines and 8 data lines. 2^17=128k
2^32 is amount of blocks that address bus could locate. and each blocks is 64bit because data bus has 64 lines. then maximum number of bits stored in memory is (2^32)*64 bit. By: Mohammad Saghafi Email: mohammads1364@yahoo.com
Family Lines System ended in 1983.
They do not. A set of lines can also be considered as a system of linear equations. But the fact that there is such a system does not mean that the lines intersect.
If the lines intersect, then the intersection point is the solution of the system. If the lines coincide, then there are infinite number of the solutions for the system. If the lines are parallel, there is no solution for the system.
17 address lines and 8 data lines. 2^17=128k
A 2K X 8 memory requires 11 address lines and 8 data lines
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.
It takes 23 address lines to address 8 mb of memory.
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..
for 16 MB memory has 24 address lines
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11
Firstly we need to convert Mb's into bits i.e 1Mb=1024x1024 = 210x210 =220 That means there are 220 memory locations and we will need 20 address lines.
ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.