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Assume a 24-bit MAR that is organized as follows: Row select lines Column select lines 12 bits 12 bits What is the maximum size of the memory unit on this machine? What are the dimensions of the memory, assuming a square twodimensional organization?

Short Answer

Expert verified
The maximum memory size is 16,777,216 bits; dimensions are 4096 x 4096.

Step by step solution

01

Understand the Organization

The Memory Address Register (MAR) has 24 bits, split into two groups: 12 bits for row select lines and 12 bits for column select lines. This implies that each group can select from a range of addresses independently.
02

Calculate Number of Rows and Columns

For a 12-bit selection, the binary addresses range from 0 to \(2^{12}-1\). Therefore, both the number of rows and columns in the memory can be calculated as \(2^{12}\).
03

Calculate Maximum Memory Size

The total number of memory locations in the memory is the product of the number of rows and columns, which is \(2^{12} \times 2^{12} = 2^{24}\). Hence, the memory can store 16,777,216 bits.
04

Consider Memory Unit Size

Since the memory unit here doesn't specify word size, we assume it is storing bits. Thus, the maximum memory size is 16,777,216 bits, or 2 megabytes if converted into bytes (assuming 8 bits per byte).
05

Determine Memory Dimensions

The memory is organized as a square, which means the number of rows equals the number of columns. Since there are \(2^{12}\) rows and \(2^{12}\) columns, the dimensions are 4096 x 4096.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Memory Organization
The concept of memory organization in computing refers to how memory is structured within a system. Understanding the organization is crucial because it affects how data can be accessed and manipulated. In the context of the problem addressed, the Memory Address Register (MAR) comprises 24 bits, which are divided into two equal sections of 12 bits each. These sections are known as the row select lines and column select lines. This division implies that one set of 12 bits is responsible for selecting rows, while the other set manages column selection.

Each bit in these sections can independently function, allowing for a range of addresses to be selected from. Since we have binary address spaces, a 12-bit section can choose from addresses ranging from 0 to 4095. Organizing memory in this manner allows for efficient data management and retrieval, critical for system performance.
Memory Size Calculation
Calculating the memory size involves understanding the structure provided by the memory organization. Here, given 12 rows and 12 columns, both set by the 12-bit sections on the MAR, we determine the total possible combinations or addresses.

Each 12-bit section can individually select from 4096 possibilities because they can address from 0 to \(2^{12} - 1\), or 4095. Thus, to find the total number of addressable memory locations, you multiply the total rows by the total columns:
\[ 4096 \times 4096 = 16,777,216 \]

This result signifies that the memory can potentially address over 16 million unique positions. Since each position is assumed to hold one bit, the total memory size is 16,777,216 bits, equating to approximately 2 megabytes when converted (using 8 bits per byte). Conversion to bytes helps relate bits to the standard byte-size storage units we often discuss when buying or evaluating the memory capacity of devices.
Two-dimensional Memory Array
In this system, a two-dimensional memory array is set up to structure the data storage as a grid-like model where rows and columns intersect.

Each intersection point on this grid corresponds to a unique memory location. Given the grid is square-shaped, and the MAR bit allocation provides for \(2^{12}\) different options for both rows and columns, the array becomes perfectly square with dimensions of 4096 x 4096.
This layout is beneficial because it offers an organized, logical approach to accessing memory. By understanding the concept of rows and columns, processors can quickly identify and access data within the large pool of bits. The efficiency of data retrieval and storage is thereby significantly improved, making computing processes faster and more reliable.

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Most popular questions from this chapter

A student was asked to translate the following algorithmic operation into machine language, where \(x\) and \(y\) were stored in locations 500 and 501 , respectively: Set \(x\) to the value of \(y+19\) Here is what was produced: LOAD 501 ADD 19 STORE 500 Is this translation correct? If not, describe the error and explain how to correct it.

Explain what use read-only memory (ROM) serves in the design of a computer system. What type of information is kept in ROM, and how does that information originally get into the memory?

Assume that our MAR contains 20 bits, enabling us to access up to \(2^{20}\) memory cells, which is \(1 \mathrm{MB}\), but our computer has \(4 \mathrm{MB}\) of memory. Explain how it might be possible to address all \(4 \mathrm{MB}\) memory cells using a MAR that contains only 20 bits.

Assume that we are receiving a message across a network using a modem with a rate of 56,000 bits/second. Furthermore, assume that we are working on a workstation with an instruction rate of 500 MIPS. How many instructions can the processor execute between the receipt of each individual bit of the message?

Assume that the variables \(a\) and \(b\) are stored in memory locations 300 and 301 , respectively. Also assume that the three integer values \(+1\), \(-1\), and 0 are stored in memory locations 400 , 401 , and 402 , respectively. Finally, assume that the code sequence you are writing begins in memory location 50 . Using any of the machine language instructions shown in Section 5.2.4, translate the following algorithmic operations into their machine language equivalents. a. Set \(a\) to the value of \(a+b-1\) b. if \(a>0\) Set \(b\) to the value \(+1\)

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