3. The ALU supported set on less than (slt) using just the sign bit of the adder. Lets try a set on less than operationusing the values -710 and 610. To make it simpler to follow the example, lets limit the binary representations to 1bits: 10012 and 01102. Then we get 10012 – 01102 = 10012+10102 = 00112. This result would suggest that –7 > 6,which is clearly wrong. Hence, we must factor in overflow in the decision.AinvertOperationBinvertCarrylnResultb.2Less3SetOverflowOverflowdetectionThere are 4 possible cases for the Set and Overflow detection bits generated by the ALU, shown in the truth tablebelow. Fill in the desired value for slt for each of these cases, and then explain how use a logical operation to combinethe Set and Overflow detection bits to get this desired outcome. Draw the changes you need to make to the ALU toget the correct slt behavior on the diagram above.Set Overflow SLT1111

Answered: Image /qna-images/answer/e513e980-6d45-4afd-bf1f-1cfa6c2d9444.jpg

3. The ALU supported set on less than (slt) using just the sign bit of the adder. Lets try a set on less than operation using the values -710 and 610. To make it simpler to follow the example, lets limit the binary representations to 1 bits: 10012 and 01102. Then we get 10012 – 01102 = 10012 + 10102 = 00112. This result would suggest that -7 > 6, which is clearly wrong. Hence, we must factor in overflow in the decision.

3. The ALU supported set on less than (slt) using just the sign bit of the adder. Lets try a set on less than operation using the values -710 and 610. To make it simpler to follow the example, lets limit the binary representations to 1 bits: 10012 and 01102. Then we get 10012 – 01102 = 10012 + 10102 = 00112. This result would suggest that -7 > 6, which is clearly wrong. Hence, we must factor in overflow in the decision.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

3. The ALU supported set on less than (slt) using just the sign bit of the adder. Lets try a set on less than operation
using the values -710 and 610. To make it simpler to follow the example, lets limit the binary representations to 1
bits: 10012 and 01102. Then we get 10012 – 01102 = 10012+10102 = 00112. This result would suggest that –7 > 6,
which is clearly wrong. Hence, we must factor in overflow in the decision.
Ainvert
Operation
Binvert
Carryln
Result
b.
2
Less
3
Set
Overflow
Overflow
detection
There are 4 possible cases for the Set and Overflow detection bits generated by the ALU, shown in the truth table
below. Fill in the desired value for slt for each of these cases, and then explain how use a logical operation to combine
the Set and Overflow detection bits to get this desired outcome. Draw the changes you need to make to the ALU to
get the correct slt behavior on the diagram above.
Set Overflow SLT
1
1
1
1

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