EBK COMPUTER SCIENCE ILLUMINATED
7th Edition
ISBN: 9781284174755
Author: Dale
Publisher: JONES+B CO
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Chapter 10, Problem 54E
Program Plan Intro
Given data:
It is given that the current value of the base register is “42993” and the current value of bounds register is “2031”.
Assumption:
Let take the starting address of application program as “A” and logical address as “L”.
- The physical address for the corresponding logical address “L” can be determined by adding the starting address “A” with the logical address “L”.
- The following is the diagrammatic representation of dynamic partition memory management system with base register and bounds register.
- In the above diagram, a particular program takes a partition size of “2031” for its accommodation in the main memory, which is provided as bounds register.
- Then, the little remaining space is allocated for other programs.
Formula to calculate the physical address:
- To calculate the physical address, the logical address (L) should be compared with bounds register value.
- If logical address is less than bounds register, physical address can be calculated.
- The physical address for the corresponding logical address is determined using the following formula,
- From the above formula, “L” represents logical address and “A” represents the starting address.
Explanation of Solution
b.
Calculation of physical address for logical address “1755”:
- Here, logical address “1755” is less than bounds register value “2031”. So, the physical address can be calculated.
- The physical address for the logical address “1755” is determined using the formula from Equation (1).
Substitute logical address “L” value as “1755” and starting address “A” value as “42993” in Equation (1)...
Explanation of Solution
c.
Calculation of physical address for logical address “3041”:
- Here, logical address “3041” is greater than bounds register value “2031”. So, the physical address cannot be calculated...
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// Method that calculates the square root of double variables
public static double myRoot(double number) {
double root;
root number/2;
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root old root;
root (root_old+number/root_old)/2;
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Chapter 10 Solutions
EBK COMPUTER SCIENCE ILLUMINATED
Ch. 10 - Prob. 1ECh. 10 - Prob. 2ECh. 10 - Prob. 3ECh. 10 - Prob. 4ECh. 10 - Prob. 5ECh. 10 - Prob. 6ECh. 10 - Prob. 7ECh. 10 - Prob. 8ECh. 10 - Prob. 9ECh. 10 - Prob. 10E
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