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tamaranim1 [39]
1 year ago
8

Which operation is not efficiently supported by heaps?

Computers and Technology
1 answer:
vodka [1.7K]1 year ago
6 0

In the case above, the operation is not efficiently supported by heaps is Find.

<h3>What are heap operations?</h3>

The operation that are known to often use heaps are:

  • Heapify
  • Find-max (or Find-min)
  • Insertion, etc.

Therefore, In the case above, the operation is not efficiently supported by heaps is Find.

See full question below

Which operation is not efficiently supported by heaps?

a. DeleteMin

b. Find

c. FindMin

d. Insert

e. all of the above are sufficiently supported

Learn more about operation from

brainly.com/question/24214198

#SPJ11

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Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and
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Answer:

for the 5 segments, the utilization is 3.8%

Explanation:

Given the data in the question;

segment size = 1090 bytes

Receiver window size = 5,718 bytes

Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps

propagation delay = 22.1 ms

so,

Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms

we determine the total segments;

Total segments = Receiver window size / sender segment or segment size

we substitute

Total segments = 5718 bytes / 1090 bytes

Total segments = 5.24587 ≈ 5

Next is the throughput

Throughput = Segment / Round trip

Throughput = 1090 bytes / 44.2 ms

1byte = 8 bits and 1ms = 10⁻³ s

Throughput = ( 1090 × 8 )bits / ( 44.2 × 10⁻³ )s

Throughput = 8720 bits / ( 44.2 × 10⁻³ s )

Throughput = 197.285 × 10³ bps

Now Utilization will be;

Utilization = Throughput / Bandwidth

we substitute

Utilization = ( 197.285 × 10³ bps ) / ( 26 × 10⁶ bps )

Utilization = 0.0076

Utilization is percentage will be ( 0.0076 × 100)% = 0.76%

∴ Over all utilization for the 5 segments will be;

⇒ 5 × 0.76% = 3.8%

Therefore, for the 5 segments, the utilization is 3.8%

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3. In a 32-bit CPU, the largest integer that we can store is 2147483647. Verify this with a
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Answer:

No, it can't be verified with a pseudocode.

Explanation:

We can not verify this with a pseudocode because the largest integer that we can store in 32-bit integer goes by the formula 2^32 - 1 = 4, 294, 967,295 and this means that it has 32 ones. Or it may be 2^31 - 1 = 2, 147, 483,647 which is a two complement signed integer.

Despite the fact that it can not be verified by using pseudocode, we can do the Verification by writting programs Through some programming language or in plain English code.

In a 32-bit CPU, the largest integer that we can store is 2147483647 because we can store integer as 2^31 and - (2^31 + 1).

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