Assignment Overview:
1. (A) Generation 1 Computers: – The computers of Generation 1 were physically very large machines characterized by the vacuum tubes (i.e. electronic valves). A vacuum tube is used to amplify a signal by controlling the movement of electrons in an evacuated space. They consist of vacuum tubes and magnetic drums (data storage) as there was no other way of storing instructions and data in computer’s memory and they were expensive and use more electricity and used to generate a lot of electricity.
Generation 2 Computers: – The phase of second generation of computers was between 1950 and 1960. Transistors were used in these computers and some of the same components as in today’s computers. A new language got invented i.e. assembly language replacing machine language. In these computers, computer memory was used for storing data and instructions.
For example: FORTAN is a language
(b) Scientists that made significant contributions to the development of Generation 1 computers are J.P Eckert and J.W Mauchy. Both developed a computer named ENIAC (Electronic Numeric Integrated and Calculator) was a very huge and big computer. It could store only limited or small amount of information. They used an electronic component i.e. vacuum tube which had very less work efficiency so it required a large cooling system.
2. (a) 1 second = 10^9 nanoseconds
(b) 1 second = 1000000.0 microseconds
(c) 1 gigabyte = 2^10 * 2^10 * 2^10 bytes
(d) 1 megabyte = 2^10 * 2^10 bytes
3. (A) Three main hardware components of modern computers are:
- Processor (CPU – Central Processing Unit)
- Memory – primary and secondary memory (RAM and ROM)
- Input and Output Devices
(b) The two main steps used to carry out an instruction under the von Neuman architecture:
- Fetch – get the next instruction from the memory
- Decode and Execute – analyse the instruction and do what is decided to be done
(c) The difference between the assembly-language layer and the machine layer in a layered architecture:
In layered architecture, different layers are there for specific purpose. The main difference between the assembly language layer and machine language layer is that assembly language layer support human readability at lowest level but machine language layer consist of numbers and it is not possible for humans to understand this. Assemblers are incorporated for making for the layers to communicate.
4. (A) Hexadecimal number is ff9a
Binary Notation is 1111 1111 1001 1010
(b) 16-bit 2’s complement notation = 1111 1111 1001 1010
Decimal Notation =
Working, 1111 1111 1001 1010
Taking 1’scomplement 0000 0000 0110 0101
Adding 1 to the right-most bit + 1
Result 0000 0000 0110 0110
Decimal Notation = (0*2^15)+(0*2^14)+(0*2^13)+(0*2^12)+(0*2^11)+(0*2^10)+(0*2^9)+(0*2^8)+(0*2^7)+(1*2^6)+(1*2^5)+(0*2^4)+(0*2^3)+(1*2^2)+(1*2^1)+(0*2^0)
= 64 + 32 + 4 + 2 = 102
5. (a) int x = -25;
int y = -25;
25 is represented in binary form as: 11001
As computer runs on a 16-bit so we will write 25 as 0000000000011001
1111111111100110
+ 1
11111111111 00111
(b) 1 1100 10000000000
First bit represents the sign-bit , middle four bits represent the exponential part and last 11 bits represent the mantissa part.
(c) The lowest exponent is 1.
The highest exponent is 7.
6. To convert the given floating point number in scientific notation having base 10 we will first convert the number in decimal form.
which is 2^30+2^25+2^22 = 155189250 , Now divide the number by 10^8 so that the results comes out to be 1.55189250
Representation in scientific notation = 1.55189250 * 10^8
7. (a) As given,
x = 1.1101*2^2 and y = – 0.1000*2^3
Now to make the both values similar, we will do
x= 0.1110*2^3
y= – 0.1000*2^3
0.0110*2^3
(b) x’ = 3.25000 * 10^3
z’ = 0.00000263 * 10^3
Now add, x’ + z’ = 3.25000263 * 10^3
As the precision is of 5 bits then answer is 3.25000 * 10^3
(c) The main limitations of floating point representation:
- Most decimal fractions cannot be represented same as in the form of binary fractions.
- Some of the arithmetic operations on decimal fractions (floating point) are more complicated than operations on fixed point numbers. Their execution takes longer time and requires more complex hardware.
8. Only 4) is equivalent.
9. Sum of given two numbers = 0101 + 0111 = 1100
10. This question can easily be solved by K-map.
Output = ( B’ . C’) + (A’BC)
11. (A) First left upper box = 1 and Second left downward box = 0
Third right upper box = 1 and fourth right downward box = 0
(b) Output = 0
Output = 0
12. (A) A Multiplexer is an electronic device that accepts several input signals and transmits them into one output signal. It may be done by chopping or interleaving. It accepts two or more streams of data and combines them into one stream.
There are different types of multiplexers: Analog and Digital Multiplexers
One example of Analog multiplexer is used in cell phones.
(b) Multiplexers are also used in central processing units (CPUs).
Function: – It decreases the number of network connections and also in system CPU time. It also enables the database server to create multiple database connections without consuming any additional computer resources.
13. (A) i. byte-addressable – 32 wires
In Byte addressing memory, one byte is formed by grouping 8 bits
ii. word-addressable – 8 wires
In Word addressing memory, a word is taken as unit for storing the memory. In this a word has unique address and according to the size of a word, bits or bytes are selected.
14. Two components of CPU are:–
- ALU: Arithmetic Logical Unit performs all the calculations in the computer. It can perform addition, subtraction, multiplication and division functions. It can also compare greater-than or less-than values.
- Control Unit: CU which controls the flow of data around the computer by sending out control signals. It also tells computer that how to perform the actions and also control them.
15. (A) Since 30 instructions are executed 9 times, therefore number of instructions = 270
Frequency of CPU = 3 GHz
Frequency of timer = 3 GHz/12 = 2.5 * 10^11
Time-period of one clock = 1/ frequency = 1/2.5 * 10^11 = 0.004 nanosecond
Let time = t + t’ (where t is the time taken by each instruction to execute and t’ is the time requires by 5 instructions to access memory)
= ((270+60) * 4 * 0.004 nanoseconds) +( 5 * 4 * 50 nanoseconds)
= 1000.5 nanoseconds (approx.)
(b) Program Output = 9 * 5 = 45 milliseconds
Total time-taken = 1000.5 nanoseconds + (9*5) milliseconds
17. (A) Time taken without pipeline for 151 instruction = 151*100
Time taken with pipeline = 1*5*20 (first task takes full pipeline time for 5stages) +
150 * 20 (rest of the task take 1 stage time each)
(b) Speed Up ratio = 15100 / (100 + 3000) = 4.87
18. 1) Source Code: A source code is simply the instructions written by a programmer that can be easily read by human. As every day, programmers write different programs in various languages like C, C++ and JAVA and these programs include some lines of text that are human readable that is known as source code. Whenever a user want to execute a program written as source code then firstly he has to convert it into machine language to execute it because source code is readable by human but not by the machine.
2.) Executable Code: An executable code can be derived from the source code itself by compiling, assembling and linking it. This code is also a set of instructions that are in the form of computer itself which can be run directly. Basically, an executed code causes a computer to perform indicated tasks according to encoded instructions.
20. (A) Need of program relocation is because:
- More than one program can share the memory and other resources of the machine.
- If we knew in advance, which programs would execute concurrently, we could assign address. When the programs were assembled so that they would fit together without overlap but practically this may not be possible.
- So it is desirable to load a program into the memory wherever there is a space for it.
- In such case, actual starting address of the program is not known until load time.
(b) Frequent context switching is undesirable because when there is switching between the processes in the machine, it is always selective. Like a process with high priority is selected first for run rather than the process with less priority. As a result the process with less priority keeps on waiting for its turn to come and run itself. It increases the waiting time of some processes to run completely as sometimes they are stopped in between their execution.
(c) The main difference between polling and interrupt-driven I/O is how a peripheral will be serviced? In polling, when a peripheral is ready for servicing, it must wait until the software polls this peripheral before it can be serviced. In polling, round-robin fashion is followed in which a peripheral cannot have higher priority over another.
In interrupt-driven I/O, each peripheral usually has one interrupt indirectly feeding into the microprocessor’s interrupt port through an interrupt controller. The interrupts coming from peripherals can be prioritized. The microprocessor services the highest priority interrupt first.
(d) Virus: – A virus is a name given to a computer program which spread itself from one file to another file on a computer. The program requires replication and execution to a computer for causing damage.
Worm: – A Worm is a name given to a computer program which spread via internet, networks etc. There is not any requirement of host file for spreading worm from one system to another. Worms spread more rapidly than viruses.
21.
22. (A) Deadlock: – A deadlock is resulted when from a chain of processes; a process takes waiting state for any events which can be completed by any other process in chain. All the processes in the set goes to waiting state waiting for resources which are already occupied by other processes. Since not any process is able to complete its execution so no resources are released resulting into a deadlock.
Atomicity: – For atomic transactions, database operations series occur simultaneously or nothing is being executed. The atomicity concept prevents partial updates to database as partial update can create problems. Atomicity is name given to irreducibility and indivisibility. An example of atomicity is ordering process in air system which requires two actions i.e. seat reservation and payment. The update requires carrying out either to make payment and reserve a seat or not to commit any of actions.
(b) An operating system where the resources are always allocated in the same order, for example printer first, video camera second, DVR third, so on. When a process requires one or more resources, it must request them in this order, so this protocol will prevent deadlock automatically. As there is a particular sequence of allocating resources so every process will be allocated resources efficiently one by one. If there will be some wait then it will be for very short span of time only. First of all, processes require first resource will be allocated, and then the second resource will be allocated and so on. The resources will be allocated in this way that no process has to wait in future for another resource.
23. ADSL = Dedicated and Guided
Fibre to the Home = Shared and Guided
Wi-Fi = Shared and Unguided
24. (A) Differences between Circuit Switching and Packet Switching:
Circuit Switching:-
- In this, there are various nodes used in the network through which the signals are passes from one system to another.
- In this a number of nodes are available between two devices and hence multiple paths are available to pass a signal.
- This concept is mainly used in telephony systems.
Packet Switching:-
- In this, the data is sent over the network in the form of packets i.e. a large unit of data items wrapped into a single bigger unit.
- In these, the packets follow the same path that is defined for them before passing into the network.
- It can be used for telephony, DSL services and other data transmission services.
(b) Time required to transmit 10^6 bytes = 1 sec
Time required transmitting 500 bytes = 0.5 millisecond
Given propagation delay = 2 millisecond
Therefore time elapse = 2 + 0.5 = 2.5 milliseconds
25. (A) Difference between transport layer and network layer:
- Network Layer is responsible for dealing with IP addresses, which takes data from one host to another. There is not any surety that date will be reached safely, without any damage or in order. The layer is responsible for translation of logical names to physical addresses and along with performing logical addressing.
- Transport Layer is responsible for dealing with ports, port is name given to logical address of a host which tells that to which application this data belongs. The layer checks for any errors in the data packets which are created in Session layer. It provides reliability and error control.
(b)
- Showing the header’s content and the packet’s payload in transport layer :
- Showing the header’s content and the packet’s payload in the data link layer :
26. DOS Attack: In this attack, the attacker will flood by making fake requests in large number. There are number of messages which are used for this purpose such as Delivery Failure, Log Off. This attack is directed to a specific port, network/ system component.
DDOS: This attack is carried from many sources at same time. The basic task remain the same as DOS but only difference lies in the fact that DOS is joined by many parties ) to flood a specific port associated with a given service.
27.
28. (A) Persistent: – Only one connection is required
Non-persistent:- Three connections are required as there are three objects i.e. two Jpeg images and data.
(b) There is not any possibility available for web page users to get round trip from server that mess up the users.
(c) Server side cookies are known as sessions. A cookie is set by the server, will be sent to the client in a response. The server side only sends the cookie when it is explicitly set or changed while the client sends the cookie only on request. There is not any possibility available for web page users to get round trip from server that mess up the users. Each time you request a page on a domain, the browser looks for a cookie for that domain on the client machine if it exists then it sends it with the request object that is why you can access it using later.
31. (A)
(b) The sender keeps on transmitting next subsequent packets given that no packet is lost between. Now if the acknowledgement for packet zero is not received within time then the packet is retransmitted after time-out.
32. (A) Congestion TCP is simply perceived as inefficiency in moving data from one device to another through the need for some segments to be retransmitted.
(b) Slow start = 0 to 16
Congestion Avoidance =
Loss Event = 6 to 7
REFERENCES:
Floating point representation.referred from http://fourier.eng.hmc.edu/e85/lectures/arithmetic_html/node11.html [assessed on 5-6-2012]
Behrouz frouzan,2006.data communications and networking.referred from http://highered.mcgraw-hill.com/sites/dl/free/0072967757/281737/for67757_fm.pdf [assessed on 5-6-2012]
Galvin Gagne.operating system referred from http://bcs.wiley.com/he-bcs/Books?action=index&itemId=0471694665&itemTypeId=BKS&bcsId=2217[assessed on 5-6-2012]
LI65
“The presented piece of writing is a good example how the academic paper should be written. However, the text can’t be used as a part of your own and submitted to your professor – it will be considered as plagiarism.
