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Question

____ number of lines are required to select __________ memory locations.

The correct answer is

10, 1024

Memory Selection and Address Lines

In computer architecture and digital electronics, memory is organized into individual storage units, each with a unique address. To access a specific memory location, a set of address lines is used. The number of memory locations that can be uniquely addressed depends directly on the number of these address lines.

Understanding Address Lines

Each address line can carry either a logical '0' or a logical '1'. If there are 'n' address lines, each line can represent two states. Therefore, 'n' lines can represent \(2^n\) unique combinations. Each of these unique combinations corresponds to a distinct memory address, allowing the system to select a particular memory location.

  • Address Lines: These are physical lines (wires) that carry the binary address for a memory location.
  • Memory Locations: These are individual storage units within the memory, each capable of storing a certain amount of data (e.g., 1 byte, 2 bytes, etc.). Every location has a unique address.

Calculating Memory Locations from Address Lines

The fundamental relationship between the number of address lines and the total number of addressable memory locations is given by the formula:

\( \text{Number of Memory Locations} = 2^{\text{Number of Address Lines}} \)

Let's use this formula to determine the values for the given statement.

Step-by-Step Calculation:

  1. Identify the knowns: The question requires finding the number of lines and the corresponding memory locations. The correct option suggests '10' lines and '1024' locations. Let's verify this.
  2. Assume 10 Address Lines: If we have 10 address lines, we can calculate the number of memory locations as follows: \[ \text{Number of Memory Locations} = 2^{10} \]
  3. Perform the Calculation:
    • \(2^1 = 2\)
    • \(2^2 = 4\)
    • \(2^3 = 8\)
    • \(2^4 = 16\)
    • \(2^5 = 32\)
    • \(2^6 = 64\)
    • \(2^7 = 128\)
    • \(2^8 = 256\)
    • \(2^9 = 512\)
    • \(2^{10} = 1024\)
    Thus, 10 address lines can uniquely select 1024 memory locations.

Completing the Statement

Based on the calculation, if 10 address lines are used, then 1024 distinct memory locations can be selected. Therefore, the blanks in the statement are filled as follows:

"10 number of lines are required to select 1024 memory locations."

This aligns perfectly with the understanding of how address lines are used to access memory in digital systems.

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Important Questions from Memory Hierarchy

  1. Suppose that the register A and register K have the bit configuration. Only the three leftmost bits of A are compared with memory words because K has 1's in these positions. Because of its organization, this type of memory is uniquely suited to parallel searches by data association. This type of memory is known as

  2. Consider a system with 2 level cache. Access times of Level 1, Level 2 cache and main memory are 0.5 ns, 5 ns and 100 ns respectively. The hit rates of Level 1 and Level 2 caches are 0.7 and 0.8 respectively. What is the average access time of the system ignoring the search time within cache?

  3. Related to computers, what is 'Wetware'?

  4. A computer has a memory hierarchy consisting of two-level cache (L1 and L2) and a main memory. If the processor needs to access data from memory, it first looks into L1 cache. If the data is not found in L1 cache, it goes to L2 cache. If it fails to get the data from L2 cache, it goes to main memory, where the data is definitely available. Hit rates and access times of various memory units are shown in the figure. The average memory access time in nanoseconds (ns) is ________. (rounded off to two decimal places)

  5. Given a computing system with two levels of cache (L1 and L2) and a main memory. The first level (L1) cache access time is 1 nanosecond (ns) and the “hit rate" for L1 cache is 90% while the processor is accessing the data from L1 cache. Whereas, for the second level (L2) cache, the “hit rate" is 80% and the "miss penalty" for transferring data from L2 cache to L1 cache is 10 ns. The “miss penalty" for the data to be transferred from main memory to L2 cache is 100 ns.

    Then the average memory access time in this system in nanoseconds is ____________ (rounded off to one decimal place)
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