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    Q. 1 Answer: Group client/server model The most suitable architecture model for this application is groupware client/server model. Because groupware server model is providing a set of application that enable communication among clients using text and images. Groupware are server based systems that support groups of people engaged in a common task and that provide an interface to a shared environment. Groupware client/server model is that which supports multiple users working on related tasks in local and remote networks. Group server is evolving concept that is more than just multiuser software which allows access to the same data. Groupware uses email, structured messages, agents , workflow, computer conferencing agents, file sharing systems among others. Groupware also makes problem-solving easier beacuse anyone in the team can make contributions. By analyzing the above mentioned system features and various system models we can choose groupware based system model. Because groupware server models work like team group. In the above case a company wants to develop an image scanner and editor applications for their office work. Office work like a team work. Ideas 2 Answer: Reading the above scenario, we can choose partitioning architecture technique. Because partitioning technique is a best suitable technique for distributing the responsibilities to different subsystems which will help in maintaining the software system. Partitioning can improve scalability, reduce contention, and optimize performance. It can also provide a mechanism for dividing data by usage pattern. For example, you can archive older data in cheaper data storage. Partitioning of architecture is an important concept. What we basically want to do is distribute the responsibilities to different subsystems so that we get a software system which is easy to maintain. Partitioning results in a system that suffers from fewer side effects. This ultimately means that we get a system that is easier to test and extend and hence is easier to maintain. A partition that divides a ship or plane into compartments. Screen. Partition consisting of a decorative frame or panel that serves to divide a space. Wall. An architectural partition with a height and length greater than its thickness; used to divide or enclose an area or to support another structure. Therefore, we can use the partitioning technique in the above Scenario. It is easy to use and maintain the system. Q. 2 Answer Idea class rightangletriangle{ constant int e = 3; // number of edges of the shape private: int a; // Length of Triangle height int b; // Length of triangle base public: float function1(){ // Method to calculate triangle area float z= (a*b)/2; // Formula for area calculation return z; } void function2(int x){ // method to set value for triangle height a = x; } void function3(int y){ // method to set value for triangle base b = x; } } ; Solution: Class TriangleRightAngle { constant int TOTAL_EDGES = ; private: int heightLengthMaximum; int baseLengthMaximum; public: float triangle AreaCalculate() { float wholeArea = (a*b)/2; return wholeArea; } void heightSetTriangle (int providedUserHeight){ heightLengthMaximum = providedUserheight; } void heightSetTriangle (int providedUserHeight){ baseLengthMaximum = providedUserbase; } };
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    This Assignment is related to Development phase and in this assignment you will focus only on the tasks given below: Consider the following ER Diagram and perform the given tasks; [image: U1DZCPN.png] Tasks: Solution Creating Database [image: uCnDRn1.png] [image: 9fUEMo3.png] Creating Tables 1. Client Table [image: vSyBdVc.png] [image: 8NS9wzr.png] 2. Event Table [image: RPCV0ml.png] [image: zhmOrVV.png] 3. Venue Table [image: RlTfCj7.png] [image: bKzN3l8.png] [image: VYwSfTf.png]
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    100% Solved: Q.1 Solution SELECT NAME FROM SALESRESP WHERE QUOTA >= (SELECT TARGET FROM OFFICES WHERE CITY = 'LAHORE') Q.2 Solution DECLARE eno salesresp.empl_num%type; commission salesresp.commision%type; update_commision number(10):=0; BEGIN Select empl_num, commision from SALESRESP where months_between(sysdate, joining_date) < 24 AND sal > 50000; update_commision:=commission-commision*0.01; update salesresp set commission = update_commision; dbms_output.put_line('Updated Commision ' || commision); END;
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    100% Solved: #include <iostream> #include <string.h> using namespace std; const int arrLength = 10; struct Student{ string userVUID; string userDetails; }std1={"NULL","NULL"}; class ArrQueue{ private: //Data Members Student arr[arrLength]; int front; int rear; public: //Constructor ArrQueue(){ for(int i=0; i<arrLength; i++) arr[i] = std1; front = -1; rear = -1; } //Member Functions void enQue(Student std){ if(isEmpty()){ arr[0] = std; front++; rear++; } else if(isFull()){ cout << " Queue is full."; } else{ arr[rear+1] = std; rear++; } } void deQue(){ if(isEmpty()){ cout<< " No Student In the Queue.\n"; } else if(rear == 0){ arr[front] = std1; front = -1; rear = -1; } else{ int tempfront = front; arr[front] = std1; for(int i=1; i<=rear; i++ ){ arr[front] = arr[i]; front++; } rear--; front = tempfront; } } int queLength(){ return rear+1; } bool isEmpty(){ if(front==-1 || rear==-1) return true; else return false; } bool isFull(){ if(rear == arrLength-1) return true; else return false; } void showQue(){ cout << "\n |Sr. VU ID Details |"; cout << "\n --- -------------------------------\n"; for(int i=front; i<=rear; i++){ cout<<" "<< i+1<< ". "<< arr[i].userVUID << " " << arr[i].userDetails <<"\n"; } } }; int main(){ /*Code For Even Id's Student std[] = {{"BC12345684","Bilal (BSCS)"}, {"BC12345685","Bilal (BSCS)"},{"BC12345686","Bilal (BSCS)"},{"BC12345687","Bilal (BSCS)"}, {"BC12345688","Bilal (BSCS)"}};*/ /*Code For Odd Id's */ Student std[] = {{"BC12345683","Bilal (BSCS)"}, {"BC12345684","Bilal (BSCS)"},{"BC12345685","Bilal (BSCS)"},{"BC12345686","Bilal (BSCS)"}, {"BC12345687","Bilal (BSCS)"}}; ArrQueue arrQue; cout << "\n -----------------------------------"; cout << "\n | Queue (After Adding Students) |"; cout << "\n -----------------------------------"; for(int i=0; i<=4; i++){ arrQue.enQue(std[i]); } arrQue.showQue(); cout << "\n -----------------------------------"; cout << "\n | Queue (After Removing Students) |"; cout << "\n -----------------------------------"; /* Code For Even Id's for(int i=0; i<=1; i++){ arrQue.deQue(); }*/ /*Code For Odd Id's*/ for(int i=0; i<1; i++){ arrQue.deQue(); } arrQue.showQue(); }
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    Solution Idea! #include <bits/stdc++.h> using namespace std; class HeapNode_Min { // Tree node of Huffman public: //Add data members here. char d; unsigned f; HeapNode_Min *l, *r; HeapNode_Min(char d, unsigned f = -1) { //Complete the body of HeapNode_Min function this->d = d; this->f = f ; this->l = NULL; this->r = NULL; } }; class Analyze { // two heap nodes comparison public: bool operator()(HeapNode_Min* l, HeapNode_Min* r) { //add return before statement and statement is completed. return (l->f > r->f); //Complete this statement } }; void display_Codes(HeapNode_Min* root, string s) // To print codes of huffman tree from the root. { if (!root) return; if (root->d != '$') cout root->d "\t: " s "\n"; display_Codes(root->l, s + "0"); display_Codes(root->r, s + "1"); //Complete this statement by passing arguments } void HCodes(char data[], int freq[], int s) // builds a Huffman Tree { HeapNode_Min *t,*r, *l ; // top, right, left priority_queue<HeapNode_Min*, vector<HeapNode_Min*>, Analyze> H_min; int a=0; while (a<s){H_min.push(new HeapNode_Min(data[a], freq[a])); ++a;} while (H_min.size() != 1) { l = H_min.top(); H_min.pop(); r = H_min.top(); H_min.pop(); t = new HeapNode_Min('$', r->f + l->f); t->r = r; t->l = l; H_min.push(t); } display_Codes(H_min.top(), ""); } int main() { int frequency[] = { 3, 6, 11, 14, 18, 25 }; char alphabet[] = { 'A', 'L', 'O', 'R', 'T', 'Y' }; //Complete this statement by passing data type to both sizeof operators int size_of = sizeof(alphabet) / sizeof(*alphabet); cout"Alphabet"":""Huffman Code\n"; cout"--------------------------------\n"; //Call Huffman_Codes function. HCodes(alphabet, frequency, size_of); return 0; }
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    @Muhammad-Arif https://cyberian.pk/topic/1116/eng101-assignment-2-solution-and-discussion/9
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    Q.1[image: AmOsgwE.png] Q.2 [image: tS465OP.png]
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    [image: dlR77xh.png] [image: DIUIHJx.png]
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    https://youtu.be/VSHrNQftyHQ
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    Answer to Q#1 Modern transponders can carry many different types of communications traffic. They can also receive signals from multiple ground stations, combining (multiplexing) or splitting (de-multiplexing) them for onwards transmission to other multiple ground stations. This method, by which many users share a common satellite resource, is called Multiple Access. There are several schemes for accomplishing this, each with its benefits and drawbacks. TDMA (Time Division Multiple Access) allocates a time slot to the user in a repetitive time frame. The signal is digitized and the data bits are stored in a buffer in a compressed time frame until their allocated time slot comes around when they are transmitted during their allocated time. At the receiver end of the link the bits are rearranged, spreading them out to reassemble the original digital signal and converted back to analogue form. The signal occupies the entire transponder bandwidth, but only during its allocated time slot. The rest of the time the bandwidth is available to other users. Digital signals typically have better noise immunity than analogue signals. FDMA (Frequency Division Multiple Access) shares the bandwidth between the users, with each user allocated a unique, narrower section of the available bandwidth. It works with analogue signals and all users have uninterrupted use of their own narrow frequency band or channel with all users occupying the available bandwidth simultaneously, each within their own narrow channel. The sender’s signal, called the baseband signal, is frequency shifted into the allocated frequency band for transmission and the receiver restores it back to the baseband. CDMA (Code Division Multiple Access) also known a Spread Spectrum, modulates the user’s signal with a pseudorandom code so that it occupies the full available spectrum, appearing as noise. The receiver uses the same pseudorandom code in an auto correlator device, which only recognizes a signal modulated with the same auto code and thus separates it from the noise. CDMA is more complex but has better noise immunity and provides greater security than the other two systems. Answer to Q#2 • Mobile Service Satellites (MSS) • Example: Satellite Phones
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    A
    @zaasmi thank you so much sir
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    Q. 1 Solution Idea Devices Visibility Consistency Constraints Affordance Keyboard Keyboard design is a good design all the Alphabet and numeric are in different section also the special function key are arranged in top of the keyboard, specially delete button is very good deployed at right top, which cannot be accidentally pressed during any typing. Keyboard has strict constraint i.e if we want to used special character which are on numeric buttons these cannot be activated without pressing the shift button. Or some button on keyboard especially on laptop is only work with pressing “fn” key. Atm machine ATM has good visibility because it have a limited amount of button on his left and right side and only some limited amount of processing screen. ATM has low consistence because there is no similar function to perform. ATM machine have strongly constraint because without the limited amount of action the individual can nothing to do. ATM Machine have no good affordance because naive users have facing a lot of difficulty while using the ATM. Digital watch Digital watch have no good visibility because its screen is not visible in full sun light and in darken area. It is not consistence device because every button has its function, we cannot perform similar task with different button or method. Digital watch has less affordance because its button have no label, so the naïve user can not configure the time or date. Digital camera It has consistence function because every time we press the button the same work will be done against mentioned button. Good constraint because a person cannot capture video except listed format. Also we cannot change function or setting of camera. Less affordance because the new user can used Digital camera with facing some difficulties. Q.2 Solution Idea: [image: pK2ZpIj.png]
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    Q1. Write a subroutine that will find the first even number from an array of your VU ID and calculate its factorial. (10 Marks) Note: Skip 0’s in your VU ID as shown below, VU ID: BC190206435 After skipping 0’s, array would be: Array: 1, 9, 2, 6, 4, 3, 5 The first even number is 2 in the array so its factorial will be calculated and saved in AX register. Solution: [org 0x100] jmp start data: dw 1, 9,2,6,4,3,5 ;initiliazation checkEven: mov dl,0002h mov ax,[data+bx] mov cx,[data+bx] xor ah,ah xor ch,ch div dl cmp ah,00h ;checking remainder je fact add bx,2 jmp checkEven fact: mov ax,0001 mov dx,0000 mult: mul cx loop mult ret start: mov si,00h mov bx,si call checkEven mov ax,0x4c00 int 0x21 Q2. Write a code in assembly language (using appropriate jumps) equivalent to this given code in C. (10 Marks) #include <stdio.h> int main() { int n1=1, n2=2; int largest; if( n1>n2) largest=n1; else largest = n2; return 0; } Solution: Assembly code: [org 0x100] jmp start n1 db 1 n2 db 2 largest db 0 start: mov ax, [n1] xor ah,ah mov bx, [n2] cmp ax,bx ja large mov [largest],bx large: mov [largest],ax mov ax, 0x4c00 int 21h
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