Question #1: Reliable Computing by Encoding Data and Address Bits 15 pts) The IB

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Question #1: Reliable Computing by Encoding Data and Address Bits 15 pts) The IBM 650 (built in 1953) computer used biquinary-coded decimal data and addresses to incorporate an error- checking scheme that ensured reliable operation. A biquinary code uses 7 bits comprising two bi bits, 0 and 5, and five quinary bits, 0, 1, 2, 3, and 4. Therefore, a biquinary code has the positional weights of 05-01234. For example, the decimal value of 6 is represented as 01-01000 and the biquinary code of decimal value of 3 is 10- 00010. Write the biquinary code for all 10 decimal values and show that there are exactly two 1s and five Os in the code, no matter what the decimal numeral is. Therefore, if any of the 7 bits inverts randomly due to noise or some other sources of error, the computer will be able to identify erroneous data. Question #2: Boolean Expressions for Word Problem (10 pts) Let variables S represent a package being small, H being heavy, and E being expensive. Let's consider a package that is not small as big, not heavy as light, and not expensive as inexpensive. Write a Boolean equation to represent the following: Your company specializes in delivering packages that are both small and inexpensive (a package must be small AND inexpensive for us to deliver it); you'll also deliver packages that are big but only if they are expensive. A particular truck can be loaded with packages only if the packages are small and light, small and heavy, or big and light. Simplify the equation. a. b.

Explanation / Answer

Solution:- As there are two separate questions, Please ask the second one by forming the other question. I'm going to answer the first question(WITH EXPLANATION) ofcourse,

The positional weightage of positional biquinary code is:-
05-01234, so, the numbers are represented as:-

0 -> 10-10000
1 -> 10-01000
2 -> 10-00100
3 -> 10-00010
4 -> 10-00001
5 -> 01-10000
6 -> 01-01000
7 -> 01-00100
8 -> 01-00010
9 -> 01-00001

As, we can see clearly that, there are exactly 2 1's and 5 0's in the representation of any number. Therefore any of the 7 bits inverts randomly, the computer will be able to identify erroneous data.

================Solution ends===============

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