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CS101 Introduction to Computing Assignment No. 01 Semester Fall 25 october 2011 solution

Wednesday, October 26, 2011 Posted In Edit This
Assignment No. 01
Semester Fall 2011
Introduction to Computing-CS101
Total Marks: 15
Due Date: 31-10-2011

Q.1. There are multiple ways to enhance the performance of a computer system. Discuss what role a cache can play in boosting up the performance of a computer system? (5 Marks)

Q.2. Generally there is a concept that “Memory” and “Storage” are same terms. Do you agree? Justify your opinion with a real life example. (5 Marks)

Q.3. Convert the following Octal Number to Hexadecimal Number by writing each and every step of conversion process: (5 Marks)

(53324)8 => (?)16

Solution:

Q.1. There are multiple ways to enhance the performance of a computer system. Discuss what role a cache can play in boosting up the performance of a computer system? (5 Marks)

Answer:
 A CPU cache is a cache used by the central processing unit of a computer to reduce the average time to access memory it is different as compare to the main memory which is used to copy data from the secondary storage it store the data inside the page file and then take it from there as compare to cache memory it exists inside the cpu and directly gives the data to cpu to process the instructions lets see a figure which will make it more clear. The cache is a smaller, faster memory which stores copies of the data from the most frequently used main memory locations. As long as most memory accesses are cached memory locations, the average latency of memory accesses will be closer to the cache latency than to the latency of main memory. It is Volatile memory, also known as volatile storage, (vusolutions) is computer memory that requires power to maintain the stored information,When the processor needs to read from or write to a location in main memory, it first checks whether a copy of that data is in the cache. If so, the processor immediately reads from or writes to the cache, which is much faster than reading from or writing to main memory. Most modern desktop and server CPUs have at least three independent caches: an instruction cache to speed up executable instruction fetch, a data cache to speed up data fetch and store, and a translation [vu solutions] lookaside buffer (TLB) used to speed up virtual-to-physical address translation for both executable instructions and data. Data cache is usually organized as a hierarchy of more cache levels which are L1 and L2 etc.

Q.2. Generally there is a concept that “Memory” and “Storage” are same terms. Do you agree? Justify your opinion with a real life example. (5 Marks)

Answer:
AS we know nowadays these are very common terms which we use in our daily life so for that you called mostly the Ram as the memory of the system and the Storage device which is called the hard disk or Flash (Usb) memory or can be a memory card. So what’s the difference between them lets see it in real life that in Memory which is Volatile (vusolutions) and need powers to maintain if the electricity is off then it is going to wash each and everything. As compare to the Sorage Non-Volatile memory which doesn’t need the power all the time when the data is store to it and power went off next time you can see the data inside it. As we can see a new functions which start from the Xp Windows Os from Microsoft. Hibernate it store the data to the hard drive from the Memory You can say the Memory to storage when you hibernate the system it will open next time all the files and applications where you were working. So it’s an example from real life.

Q3).Convert the following Octal Number to Hexadecimal Number by writing each and every step of conversion process: (5 Marks) (53324)8 => (?)16

First we will convert this to binary to decimal

(53324)8 = 5*8^4+3*8^3+3*8^2+2*8^1+4*8^0
(53324)8 = 5*4096+3*512+3*64+2*8+4*1
(53324)8 = 20480+1536+192+16+4
(53324)8 = (22228)10

Now converting into hexadecimal
  
16
22228

16
1389
4
16
186
(13) D

5
6
  
So the hexa decimal equivalent is (56D4)16 

Another Solution:


Answer # 1

Cache is collection of duplicate data, that is used by central processing unit of a computer to reduce the average time to access memory it is different as compare to the main memory which is used to copy data from the secondary storage it store the data inside the page file and then take it from there. Cache is a smaller, faster memory which stores copies of the data from the most frequently used main memory locations. When microprocessor get any data he calls first cache chip if not available there then called same from the main memory. As long as most memory accesses are cached memory locations, the average latency of memory accesses will be closer to the cache latency than to the latency of main memory. It is volatile memory also known as volatile storage. Most modern desktop and server CPUs have at least three independent caches, an instruction cache to speed up executable instruction fetch, a data cache to speed up data fetch and store, and a translation look aside buffer used to speed up virtual to physical address translation for both executable instructions and data. Data cache is usually organized as a hierarchy of more cache levels. Cache has two types one work sepreatly and has slow speed, second not separate; work with microprocessor and faster than regular RAM. Cache chip in small size assist to microprocessor, but make short time to assess, resulting in a boost in performance. Accessing the cached copy make future access to data

Answer # 2

As we know nowadays these are very common terms which we use in or daily life so for that you called mostly the Ram as the memory of system and the storage device which is called hard disk or Flash (UBS) memory or can be memory card. Storage and memory are two different terms but play same role for receiving and transferring of data in computer. The difference between them is volatile and non volatile. When electricity goes off everything is wash or finish that is in volatile memory, as compare to the storage Non volatile memory which does not need the power all the time when the data store to it and power went off next time you see the data inside it. In the real life example we see a new function which starts from the Xp windows Os from Microsoft (Hibernate) it store the data to the hard drive from the memory we can say the memory to storage when we hibernate the system it will open next time all the files and application where we were working. So it is an example from real life.

Answer # 3

(53324) ^8

Firstly convert octal data to decimal

(53324)8 = 5*8^4 +3*8^3+3*8^2+2*8^1+4*8^0
(53324)8   = 5*4096+3*512+3*64+2*8+4*1
(53324)8 = 20480+1536+192+16+4
(53324)8 = (22228)10
Now converting into hexadecimal

16
22228

16
1389
4
16
186
(13) D

5
6


So the hexa decimal equivalent is (56D4)16

Eng001 Assignment No. 1 Fall 2011 solution

Wednesday, October 26, 2011 Posted In Edit This
ENG001 (Elementary English)

Fall 2011
Assignment # 01                                                                  
Total Marks: 15

Objective:
To access students’ knowledge about basic concepts of writing             
Instructions:

1. Late assignments will not be accepted.
2. If the file does not properly run, it will be marked zero.
3. Plagiarism will never be tolerated. Plagiarism occurs when a student uses work done by someone else as if it was his or her own.
4. If any assignment is found copied work, no marks will be awarded and the case may be referred to the academics for disciplinary action.
5. No assignment will be accepted through e-mail. 
6. The font color should be preferably black and font size should be 12 Times New Roman.

Q1: “Effective writing is much more a thinking process than a writing process.” Discuss. 5

Q2: Choose the best option: 5

Which of the following is an external source of information?
• Libraries
• Internet
• Resource persons
• All of the above (Correct)

Which of the following is NOT included in process of writing?
• Invention & Collection 
• Conciseness & Consideration (Correct)
• Organization & Drafting
• Revising & and Proofreading

Which of the following emphasizes to ‘put client first’?
• Creativity
• Conciseness
• Consideration
• Completeness

SMOG stands for __________;

• Simple Measure of Gobbledygook (Correct)
• Standard Measure of Gobbledygook
• Simple Management of Gobbledygook
• Standard Management of Gobbledygook

‘Students should simply open their minds to whatever pops into them’ refers to;

• Thinking
• Brainstorming
• Writing
• Listing

Q3: Recall the concept of Clustering and Listing to fill in the following blanks.    5



Advantages of Internet
1. Communication
2. Information
3. Entertainment
4. Downloading Software

CS601 Assignment Fall 2011 Solution

Wednesday, October 26, 2011 Posted In Edit This
for complete & in MS word file format, download now: CS601 Assignment solution


CS601 Solution.


Q.2 Idea ..
A packet and a frame are both packages of data moving through a network.


A packet exists at Layer 3 of the OSI Model, whereas a frame exists at Layer 2 of the OSI Model.


Layer 2 is the Data Link Layer. The best known Data Link Layer protocol is Ethernet.


Layer 3 is the Network Layer. The best know Network Layer protocol is IP (Internet Protocol).


To move through a network, a packet is encapsulated into one or more frames


Q.3


The numbers of cable links required for mesh topology are:
Number of devices = 11
Number of links = n (n-1)/2
=11 (11-1)/2 = 55 


Number of I/o ports = (n-1) = 11-1 = 10

Q.2 Idea ..
A packet and a frame are both packages of data moving through a network.


A packet exists at Layer 3 of the OSI Model, whereas a frame exists at Layer 2 of the OSI Model.


Layer 2 is the Data Link Layer. The best known Data Link Layer protocol is Ethernet.


Layer 3 is the Network Layer. The best know Network Layer protocol is IP (Internet Protocol).


To move through a network, a packet is encapsulated into one or more frames


Packet This term is consided by many to correctly refer to a message sent by protocols operating at the network layer of the OSI Reference Model. So you will commonly see people refer to packets . However, this termin is commonly also used to refer generically to any type of message, as i menitoned earlier.




Frame This term is most commonly associated with messages that travel at low levels of the OSI Reference Model. In particular, it is most commonly seen used in reference to data link layer messages. It is occasionally also used to refer to physical layer messages, when message formatting is performed by a layer 1 technology. A frame gets its name from the fact that it is created by taking higher-level packets or datagrams and "framing" them with additional header information needed at the lower level. 

STA301 Assignment No. 1 Fall 2011 solution

Wednesday, October 26, 2011 Posted In Edit This
Assignment No.1 (Course STA301)
Fall 2011 (Total Marks 30) 

Deadline
Your Assignment must be uploaded/ submitted before or on28th Oct, 2011STUDENTS ARE STRICTLY DIRECTED TO SUBMIT THEIR ASSIGNMENT BEFORE OR BY DUE DATE. NO ASSIGNMNENT AFTER DUE DATE WILL BE ACCEPTED VIA E.MAIL).

Rules for Marking
It should be clear that your Assignment will not get any credit IF:
·         The Assignment submitted, via email, after due date.
·         The submitted Assignment is not found as MS Word document file.
·         There will be unnecessary, extra or irrelevant material.
·         The Statistical notations/symbols are not well-written i.e., without using MathType software.
·         The Assignment will be copied from handouts, internet or from any other student’s file. Copied material (from handouts, any book or by any website) will be awarded ZERO MARKS. It isPLAGIARISM and an Academic Crime.
·         The medium of the course is English. Assignment in Urdu or Roman languages will not be accepted.
·         Assignment means Comprehensive yet precise accurate details about the given topic quoting different sources (books/articles/websites etc.). Do not rely only on handouts. You can take data/information from different authentic sources (like books, magazines, website etc) BUT express/organize all the collected material inYOUR OWN WORDS. Only then you will get good marks.

Objective(s) of this Assignment:

·         The assignment is being uploaded to strengthen the students’ BASIC concepts about statistics.
·         Practice for arranging and displaying already collected data into an organized form, using given statistical method.
·         Make students able to understand the concept of measures of central tendency.

Assignment.1 (Lessons 1-10)

Question 1:Marks: 5+5=10
i) From the table given below, Find
(a) Class boundaries
(b) How many men fall in the age of 50 or above? 
(c) How many men fall between the age group of 60 and 69?

Age Group (Men)
f
20-29
5
30-39
4
40-49
5
50-59
8
60-69
12
70-79
16
Total
50

(ii) Write down the following figure in the form of frequency table and also find the relative frequency.

Question 2:Marks: 5+5=10
(i) Calculate the mode from the following continuous distribution.

Height
Number (f)
57.5-60.0
6
60.0-62.5
26
62.5-65.0
190
65.0-67.5
281
67.5-70.0
412
70.0-72.5
127
72.5-75.0
38
Total
1080

(ii). Use the stem and leaf plot to answer these questions.

Statistics Test Scores
Stem
Leaf
6
1 1 4 6 7 8
7
2 3 5 7 9
8
1 3 5 6 6 7 7 8 9
9
0 0 3 4 6 8 9 9
10
0 0

a) What is the best test score?
b) How many students take the test?
c) How many students scored 90?
d) What is the lowest score?
e) Find the difference between the high and low scores.

Question 3:Marks: 3+3+4=10(i) Find the median for the following discrete frequency distribution.

Number of pupils per class
Number of Classes
20
1
21
0
22
1
23
3
24
6
25
9
26
8
27
10
28
7
Total
45


(ii) Calculate the weighted mean from the following data
Item
Expenditure
Weights
Food
290
7.5
Rent
54
2.0
Clothing
98
1.5
Fuel and light
75
1.0
Other items
75
0.5

(iii) A man gets a rise of 10% in salary at the end of his first year of service and further of 20 % and 25 % at the end of the second and third year respectively. The rise in each case is calculated on his salary at the beginning of the year. To what annual percentage increase is this equivalent?

Solution:






Class Boundaries f
19.5-29.5 5
29.5-39.5 4
39.5-49.5 5
49.5-59.5 8
59.5-69.5 12
59.5-79.5 16
total 50




(b) How many men fall in the age of 50 or above?


Men above 50
= 8+12+16=36
(c) men fall between the age group of 60 and 69=
=12
(ii) Write down the following figure in the form of frequency table and also find the relative frequency.


Class Boundaries f Relative frequncy
19.5-29.5 5 5/50=0.1
29.5-39.5 4 4/50=0.08
39.5-49.5 5 5/50=0.1
49.5-59.5 8 8/50=
59.5-69.5 12 12/5=
59.5-79.5 16 16/50=
total 50


Question 2:Marks: 5+5=10
(i) Calculate the mode from the following continuous distribution.
Height Number (f)
57.5-60.0 6
60.0-62.5 26
62.5-65.0 190
65.0-67.5 281----f1
l--67.5-70.0 412---model class
70.0-72.5 127-----f2
72.5-75.0 38
Total 1080


Ans=
Mode = l + fm-f1/(fm-f1)+(fm-f2) * h
Fm = frequency of model class = highest frequency =412 
H= difference between class boundries = 2.5


So…
Mode= 67.5+412-281/(412-281)+(412-127)*2.5
= 67.5+131/(131)+(285)*2.5
= 68.287


(ii). Use the stem and leaf plot to answer these questions.
Statistics Test Scores 
Stem Leaf
6 1 1 4 6 7 8 
7 2 3 5 7 9 
8 1 3 5 6 6 7 7 8 9 
9 0 0 3 4 6 8 9 9 
10 0 0 




a) What is the best test score?
Ans best score 100
b) How many students take the test?
Ans 30students
c) How many students scored 90?
Ans 02students made 90
d) What is the lowest score?
Ans lowest score= 61
e) Find the difference between the high and low scores.
Ans high score = 100 
Lowest = 61
Difference = 100- 61 =59


Question 3:Marks: 3+3+4=10(i) Find the median for the following discrete frequency distribution.Number of pupils per class Number of Classes
F cf 
20 1 1
21 0 1
22 1 2
23 3 5
24 6 11
25 9 20
26 8 28
27 10 38
28 7 45
Total 45


Median=n+1/2 for more detail page no61 of HO
45+1/2
median =23rd value is lying in 26the distribution so median = 26


(ii) Calculate the weighted mean from the following dataItem Expenditure Weights
Food 290 7.5
Rent 54 2.0
Clothing 98 1.5
Fuel and light 75 1.0 solve from page no58 of HO
Other items 75 0.5


(iii) A man gets a rise of 10% in salary at the end of his first year of service and further of 20 % and 25 % at the end of the second and third year respectively. The rise in each case is calculated on his salary at the beginning of the year. To what annual percentage increase is this equivalent?
Ans 
Annual percentage increase in his salary = 10%+20%+25%/3
= 55/3
= 18.33%






another solution: STA301 solution

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