K & G has a huge sales network. To control the sales, it is divided into the regions. Each region has multiple zones. Each zone has different cities. Each sales person is allocated to different cities. The objective is to track Sales figure at different granularity levels of Region. Also to count number of products sold. Products are categorized as High end and Low end products. Identify the following: Dimensions and Facts Design a suitable schema (Please make schema on paper or using any online tool and attach its image)

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
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  1. K & G has a huge sales network. To control the sales, it is divided into the regions. Each region has multiple zones. Each zone has different cities. Each sales person is allocated to different cities. The objective is to track Sales figure at different granularity levels of Region. Also to count number of products sold. Products are categorized as High end and Low end products.

Identify the following:

  1. Dimensions and Facts
  2. Design a suitable schema (Please make schema on paper or using any online tool and attach its image)

 

  1. United Garments is looking to implement BI to improve the decision making capability within the organization. To which decision making level you will recommend United Garments to implement BI and why?
Given-
We need to satisfy the given data is right for the given diagram. The condition is that the one
output will be HIGH and other two are LOW.
The given values of A and Bare-
A3=1
A2=1
A1=0
A0=1
B3=1
B2=1
B1=1
BO=0
Solution-
To get the solution we start from the starting and calculate the values -
For first 2-bit comparator-
A=11 and B=11
So the outputs A=8 is HIGH(1) and other two are LOW(0).
For second 2-bit comparator-
A=01 and B=10
So the outputs A<B8 is HIGH(1) and other two are LOW(0).
So now we take HIGH outputs because all operations performs on LOW bit is always gives
LOW(0) in output so the all outputs if them is going to 0.
So by taking the HIGH(1) bits -
The two high bits we get on A=B and A-B, so these values are meet on third "AND" gate.
So the output of third AND gate is 1 and this is then gone on second OR gate. And we know that
it the one bit is high then the output of the OR gate will be HIGH(1) always.
So the final output of second OR gate A<B is HIGH(1) and other two gates output will be
LOW(0).
Transcribed Image Text:Given- We need to satisfy the given data is right for the given diagram. The condition is that the one output will be HIGH and other two are LOW. The given values of A and Bare- A3=1 A2=1 A1=0 A0=1 B3=1 B2=1 B1=1 BO=0 Solution- To get the solution we start from the starting and calculate the values - For first 2-bit comparator- A=11 and B=11 So the outputs A=8 is HIGH(1) and other two are LOW(0). For second 2-bit comparator- A=01 and B=10 So the outputs A<B8 is HIGH(1) and other two are LOW(0). So now we take HIGH outputs because all operations performs on LOW bit is always gives LOW(0) in output so the all outputs if them is going to 0. So by taking the HIGH(1) bits - The two high bits we get on A=B and A-B, so these values are meet on third "AND" gate. So the output of third AND gate is 1 and this is then gone on second OR gate. And we know that it the one bit is high then the output of the OR gate will be HIGH(1) always. So the final output of second OR gate A<B is HIGH(1) and other two gates output will be LOW(0).
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