# Segmentation in OS | Practice Problems

## Segmentation in OS-

Before you go through this article, make sure that you have gone through the previous article on Segmentation in OS.

We have discussed-

• Like Paging, Segmentation is another non-contiguous memory allocation technique.
• Segmentation divides the process into meaningful segments.
• Segment table stores the information about each segment of the process.

In this article, we will discuss a practice problem based on segmentation.

## Problem-

Consider the following segment table-

 Segment No. Base Length 0 1219 700 1 2300 14 2 90 100 3 1327 580 4 1952 96

Which of the following logical address will produce trap addressing error?

1. 0, 430
2. 1, 11
3. 2, 100
4. 3, 425
5. 4, 95

Calculate the physical address if no trap is produced.

## Solution-

In a segmentation scheme, the generated logical address consists of two parts-

1. Segment Number
2. Segment Offset

We know-

• Segment Offset must always lie in the range [0, limit-1].
• If segment offset becomes greater than or equal to the limit of segment, then trap addressing error is produced.

### Option-A: 0, 430-

Here,

• Segment Number = 0
• Segment Offset = 430

We have,

• In the segment table, limit of segment-0 is 700.
• Thus, segment offset must always lie in the range = [0, 700-1] = [0, 699]

Now,

• Since generated segment offset lies in the above range, so request generated is valid.
• Therefore, no trap will be produced.
• Physical Address = 1219 + 430 = 1649

### Option-B: 1, 11-

Here,

• Segment Number = 1
• Segment Offset = 11

We have,

• In the segment table, limit of segment-1 is 14.
• Thus, segment offset must always lie in the range = [0, 14-1] = [0, 13]

Now,

• Since generated segment offset lies in the above range, so request generated is valid.
• Therefore, no trap will be produced.
• Physical Address = 2300 + 11 = 2311

### Option-C: 2, 100-

Here,

• Segment Number = 2
• Segment Offset = 100

We have,

• In the segment table, limit of segment-2 is 100.
• Thus, segment offset must always lie in the range = [0, 100-1] = [0, 99]

Now,

• Since generated segment offset does not lie in the above range, so request generated is invalid.
• Therefore, trap will be produced.

### Option-D: 3, 425-

Here,

• Segment Number = 3
• Segment Offset = 425

We have,

• In the segment table, limit of segment-3 is 580.
• Thus, segment offset must always lie in the range = [0, 580-1] = [0, 579]

Now,

• Since generated segment offset lies in the above range, so request generated is valid.
• Therefore, no trap will be produced.
• Physical Address = 1327 + 425 = 1752

### Option-E: 4, 95-

Here,

• Segment Number = 4
• Segment Offset = 95

We have,

• In the segment table, limit of segment-4 is 96.
• Thus, segment offset must always lie in the range = [0, 96-1] = [0, 95]

Now,

• Since generated segment offset lies in the above range, so request generated is valid.
• Therefore, no trap will be produced.
• Physical Address = 1952 + 95 = 2047

Thus, Option-(C) is correct.

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Summary
Article Name
Segmentation in OS | Practice Problems
Description
Practice Problems based on Segmentation in OS. Segmentation in OS is a non-contiguous memory allocation technique. Unlike Paging, Segmentation divides the process into segments.
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Publisher Name
Gate Vidyalay
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