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Identify the value of 'Y' that will give the output of disp(X) equal to 5 using the code given below.

Identify the value of 'Y' that will give the output of disp(X) equal to 5 using the code given below.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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**Problem Statement:** Identify the value of 'Y' that will give the output of `disp(X)` equal to 5 using the code given below. ```plaintext for k = 1:3 X = 5; X = k + Y; end disp(X) ``` **Solution Options:** - Y = 4 (Incorrect) - Y = 3 - Y = 2 (Correct) - Y = 1 **Explanation:** To solve the problem, we need to identify the value of Y that results in the variable X equaling 5 at the conclusion of the loop. The code runs a loop with `k` ranging from 1 to 3, and within the loop, X is set to 5 initially, and then reassigned to the value of `k + Y`. To ensure `disp(X)` outputs the value of 5, after the loop where `k` is 3, the equation `3 + Y = 5` must hold true. Solving for Y gives: ``` Y = 5 - 3 Y = 2 ``` Detailed Debugging: 1. **Initialization**: Begin with `X = 5`. 2. **Loop Iteration**: - For `k = 1`, `X = 1 + Y` - For `k = 2`, `X = 2 + Y` - For `k = 3`, `X = 3 + Y` 3. **Final Evaluation**: `X = 3 + 2 = 5` Thus, the correct answer is Y = 2.
Transcribed Image Text:**Problem Statement:** Identify the value of 'Y' that will give the output of `disp(X)` equal to 5 using the code given below. ```plaintext for k = 1:3 X = 5; X = k + Y; end disp(X) ``` **Solution Options:** - Y = 4 (Incorrect) - Y = 3 - Y = 2 (Correct) - Y = 1 **Explanation:** To solve the problem, we need to identify the value of Y that results in the variable X equaling 5 at the conclusion of the loop. The code runs a loop with `k` ranging from 1 to 3, and within the loop, X is set to 5 initially, and then reassigned to the value of `k + Y`. To ensure `disp(X)` outputs the value of 5, after the loop where `k` is 3, the equation `3 + Y = 5` must hold true. Solving for Y gives: ``` Y = 5 - 3 Y = 2 ``` Detailed Debugging: 1. **Initialization**: Begin with `X = 5`. 2. **Loop Iteration**: - For `k = 1`, `X = 1 + Y` - For `k = 2`, `X = 2 + Y` - For `k = 3`, `X = 3 + Y` 3. **Final Evaluation**: `X = 3 + 2 = 5` Thus, the correct answer is Y = 2.
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