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Dimas [21]
3 years ago
10

Consider a 512 Mb DRAM with 8 banks, 256k subarrays, and 8 bit word. How are the address bits allocated? How many subarrays are

there and how are they arranged?
Computers and Technology
1 answer:
Softa [21]3 years ago
4 0

Answer:

26 address bits are needed to address 2∧26 locations.

There are 24 = 16 subarrays.

Explanation:

Considering a 512Mb DRAM, it means it can store 512*2∧20 bits = 2∧9X2∧20 = 2∧29. Since it is a 8 bit word, at a time, 8 bits are accessed. Therefore, the capacity can be re-written as 2∧26X 2∧3 = 2∧26X8

It simply means 26 address bits are needed to address 2∧26 locations, since there are 8 banks, 3 address bits are needed to select a bank ( 8 = 2∧3 )

Each subarray is 256K which means 19 bits are needed to select one address from this 256K subarray ( since 256K = 2∧19 )

22 addresses are used up out of a total of 26 addresses (19 for locating a location in 256K subarray and 3 for banks). The remaining addresses are 4 ( 26 - 22 ). These 4 addresses are used to locate a subarray. thus there are 2∧4 = 16 subarrays

There are a total of 8 banks  and within a bank, there are 32 subarrays.

Within each subarray there are 256K locations each of 8 bit size.

Note: " ∧" means raise to power

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Given the following adjacency matrix, what is the approximate rank vector after one iteration of the power iteration method (use
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Note: The matrix referred to in the question is: M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]

Answer:

a) [5/18, 5/18, 4/9]'

Explanation:

The adjacency matrix is  M = \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]

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To get the rank vector for the first Iteration:

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X_1 =  \left[\begin{array}{ccc}1/2&1/3&0\\1/2&1/3&0\\0&1/3&1\end{array}\right]\left[\begin{array}{ccc}1\\1\\1\end{array}\right] \\\\X_1 = \left[\begin{array}{ccc}5/6\\5/6\\4/3\end{array}\right]\\

Multiplying the above matrix by 1/3

X_1 = \left[\begin{array}{ccc}5/18\\5/18\\4/9\end{array}\right]

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