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Q10/the minimum sum- of- products for the following function isF(A,B,C,D,E)=[m(0,3,5,7,9)+d(2,4,6,10,15) using Quine - McClusky method : A'BC'D'E+A'B'E'+A'B'D+A'B'C O Option 6 A'B'E'+A'B'D+A'B'C+A'CDE O None of them O A'B'E+A'B'E'+A'B'D+A'B'C O A'CDE+A'BC'D'E+A'B'E'+A'B'D+A'B'C

Q10/the minimum sum- of- products for the following function isF(A,B,C,D,E)=[m(0,3,5,7,9)+d(2,4,6,10,15) using Quine - McClusky method : A'BC'D'E+A'B'E'+A'B'D+A'B'C O Option 6 A'B'E'+A'B'D+A'B'C+A'CDE O None of them O A'B'E+A'B'E'+A'B'D+A'B'C O A'CDE+A'BC'D'E+A'B'E'+A'B'D+A'B'C

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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Q10/the minimum sum- of- products for the following function isF(A,B,C,D,E)=[m(0,3,5,7,9)+d(2,4,6,10,15) using Quine - McClusky method : A'BC'D'E+A'B'E'+A'B'D+A'B'C O Option 6 A'B'E'+A'B'D+A'B'C+A'CDE O None of them O A'B'E+A'B'E'+A'B'D+A'B'C O A'CDE+A'BC'D'E+A'B'E'+A'B'D+A'B'C
Transcribed Image Text:Q10/the minimum sum- of- products for the following function isF(A,B,C,D,E)=[m(0,3,5,7,9)+d(2,4,6,10,15) using Quine - McClusky method : A'BC'D'E+A'B'E'+A'B'D+A'B'C O Option 6 A'B'E'+A'B'D+A'B'C+A'CDE O None of them O A'B'E+A'B'E'+A'B'D+A'B'C O A'CDE+A'BC'D'E+A'B'E'+A'B'D+A'B'C
Expert Solution
Step 1

A'CDE+A'BC'D'E+A'B'C'+A'B'C+A'B'C.


Explanation:

Approach to solving the question:

 

1) First, we use the Quine - Mcclusky method to find the minimum sum- of- products. 

 

2) Next, we apply the distributive law to simplify the expression. 

 

3) Finally, we apply the associative law to simplify the expression.

 

 

Detailed explanation:


 

1) The minimum sum- of- products for the function

isF(A,B,C,D,E)=Em(0,3,5,7,9)+Ed(2,4,6,10,15) using Quine - Mcclusky

method is A'CDE+A'BC'D'E+A'B'C'+A'B'C+A'B'C.

 

2) Applying the distributive law, we have

 

A'CDE+A'BC'D'E+A'B'C'+A'B'C+A'B'C

=A'(CDE+BC'D'E+B'C')+A'B'C+A'B'C 

 

 

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