The following class diagram can be created for the example scenario: Class diagram for keeping track of businesses, personnel, supervisors, and subcontractors: The development of a system for keeping tabs on businesses, their personnel, managers, and contractors is required. Each business employs one or more people.
We require the employee number, name, and salary for each employee. Each business has a name and location. Every manager works as an employee who is in charge of managing others. Each manager is in charge of one or more workers. We track the duration of the contracts for those employees who work as contractors. The following is the related class diagram: By identifying the classes and their attributes in the hypothetical situation, the class diagram is created. There are classes, attributes, and relationships in the class diagram. Company, Employee, Manager, and Contractor are the classes that are mentioned in this context. There are identified attributes for each class.
For instance, the characteristics Company Name and Company Address are part of the Company class. Additionally, the connections between courses are also noted. As an illustration of the link between the Manager and Employee classes, the Manager class is a subclass of the Employee class.
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2. Consider two companies having different IT demands: Company A needs 200 servers with a utilization of 100% for 4 years; Company B needs 200 servers with a utilization of 50% for half a year.
You are consulted to work out IT strategies for both companies: either they purchase their own servers in a traditional way (construct their own data centers) or rent computing resources from a third-party service provider in a cloud computing way.
Some assumptions are as below:
1. One server costs GBP 1,500;
2. For a data center, one administrator can manage 50 servers, whose annual salary is GBP 20,000;
3. The power consumption of each server is 150 w;
4. The electricity costs GBP 0.1/(wh), where h is short for hour;
5. The cloud service provider charges GBP 0.4/h for each virtual server with the same specifications as that of a physical server.
(a) Calculate the corresponding costs by ignoring the building construction, air- condition and cooling costs. Discuss under which circumstance a company should build its own data centre as a traditional e-Commerce infrastructure and under which circumstance a company should switch to cloud computing as a new e-Commerce infrastructure.
[10 marks]
(b) From the above scenario, identify 5 ways in which e-Commerce benefits from Cloud Computing
Traditional e-commerce infrastructure versus Cloud Computing Let's evaluate the cost of traditional e-commerce infrastructure and cloud computing, and then determine the circumstances under which each should be used.
Traditional E-commerce infrastructure. The cost of owning a server is as follows:One server costs GBP 1,500;There are 200 servers; The total cost of owning 200 servers is 200 * GBP 1,500 = GBP 300,000.The annual cost of an administrator managing 50 servers is GBP 20,000.
There are 200 servers, which means there are four administrators. The total cost of employing four administrators is 4 * GBP 20,000 = GBP 80,000.The total annual cost of owning 200 servers is the cost of owning the servers plus the cost of employing the administrators.
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.1 Explain the difference between data and information and provide an example of each one to clarify the difference. (5) ) Q.1.2 Explain the role of a database designer in a database management system. Provide a detailed example in the context of a bank. (5) 5)
Data refers to raw facts, while information is processed and meaningful data. A database designer designs the structure and organization of a database to ensure data integrity and efficiency, as exemplified in a bank scenario.
What is the difference between data and information, and what is the role of a database designer in a database management system?Data refers to raw facts, figures, and symbols that have no meaning on their own. They are unprocessed and lack context. For example, a list of numbers (10, 5, 7, 3) is data.
Information, on the other hand, is processed and meaningful data that provides knowledge or insights. It is derived from analyzing and organizing data in a meaningful way. For instance, calculating the average of the numbers (10, 5, 7, 3) and interpreting it as the average score of a test provides information.
In a database management system, a database designer plays a crucial role in designing the structure and organization of a database. They determine the database schema, define tables, relationships, and constraints.
They ensure data integrity, efficiency, and security. For example, in the context of a bank, a database designer would design the database schema to include tables for customers, accounts, transactions, and define relationships between them.
They would establish constraints to ensure that account balances are accurate and enforce security measures to protect sensitive customer information.
The database designer's role is essential in creating a well-structured and efficient database system that meets the requirements of the organization and facilitates effective data management.
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Link layer protocols provide many services including encapsulation, reliable data delivery between adjacent nodes, flow control, error detection and/or correction and others. 2. In time division multiple access, different nodes transmit in different frequencies for collision avoidance. 3. Link state routing protocols take time as O(²) for convergence, and may include oscillations before converging. 4. OSPF is two-level hierarchy-based distance vector protocol that is used for inter-AS routing. 5. IP provides certain QoS requirements as part of its 'best effort' service for the Internet such as successful datagram delivery, while it cannot guarantee order of delivery or data rate. 6. Network applications talk to the SDN controller through the southbound interface and data-plane devices communicate through the northbound interface. 7. QUIC protocol provides reliable data transmission and congestion control over UDP by moving the complex transport layer functions to application layer. 8. Buffer overflow can occur at either the input port or at the output port of a switch. 9. Openflow protocol has three message types: controller-to-switch, asynchronous and symmetric. 10. VLAN is an extra tag that is appended to the ethernet frame as defined in the IEEE 802.1Q standard. 11.2. Linux uses TCP Cubic by default. 12. Longest prefix matching is a useful tool when the destination in the incoming packet header does not match with predefined table entries. 13. Losses and duplicate transmissions decrease the effective throughput in any congested network. 14. Software defined networks provide centralized control plane functionalities, while the data plane forwarding still remains decentralized.
Link layer protocols provide encapsulation, reliable data delivery, flow control, and error detection/correction. Time Division Multiple Access (TDMA) uses different time slots, not frequencies, for collision avoidance. Link state routing protocols like OSPF have a time complexity of O(n^2), without oscillations.
OSPF is an intra-AS routing protocol, not inter-AS. IP provides "best effort" service without guaranteeing delivery order or data rate. SDN applications communicate with the controller via the southbound interface, while data-plane devices use the northbound interface. QUIC moves transport layer functions to the application layer for reliable data transmission over UDP.
Buffer overflow can happen at input or output ports of a switch. OpenFlow has three message types: controller-to-switch, asynchronous, and symmetric. VLAN adds an extra tag to Ethernet frames. Linux doesn't use TCP Cubic by default.
Longest prefix matching finds the best matching entry in routing tables. Congested networks experience decreased throughput due to losses and duplicate transmissions. SDNs provide centralized control while keeping data-plane forwarding decentralized.
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When the PLC processor sees an XIC instruction, it is looking for condition. 2. When the PLC processor sees an XIO instruction, it is looking for condition. 3. What is the difference between an electrical concept and a PLC concept? a
The programmable logic controller (PLC) is a computing system that is used in industrial control systems. It is designed to perform various tasks in a programmable manner.
1. When the PLC processor sees an XIC instruction, it is looking for a condition to be TRUE.
2. When the PLC processor sees an XIO instruction, it is looking for a condition to be FALSE.
3. The difference between an electrical concept and a PLC concept is that an electrical concept is concerned with the physical flow of electrons. In contrast, a PLC concept is concerned with the logical operation of a control system.
PLCs are used in many industrial applications such as manufacturing, transportation, and even traffic control. The primary function of a PLC is to control and monitor industrial processes. The programmable logic controller (PLC) uses a combination of hardware and software to perform its tasks. The hardware components of a PLC include the central processing unit (CPU), memory, input/output (I/O) modules, and communication interfaces.
The software components of a PLC include the programming software, operating system, and application software. A PLC processes its input signals by scanning them in a sequential manner. When the PLC processor sees an XIC (eXamine If Closed) instruction, it is looking for a condition to be TRUE. When the PLC processor sees an XIO (eXamine If Open) education, it is looking for a condition to be FALSE.
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How do you convert 5cos(2t +30°) to phasor form? O 5<2° O 2<5° O 5<30° O 5<-60° O 2<30⁰
Previous question
Given that 5cos(2t+30°).
We need to find the phasor form for the given function.
Step 1: We know that cos(wt+θ)=Re(Aej(wt+θ)),
where
A=|A|∠θ and
Aej(wt+θ)=|A|ejθej(wt)
Here, A=5 and
θ=30°cos(2t+30°)
=Re(5e^(j2t+30°))
=Re(5ej30°e^(j2t))
=Re(5∠30°e^(j2t))
The phasor form for 5cos(2t+30°) is 5∠30°, which can be represented as option 3.
Hence, the correct answer is 5<30°.
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(a) A continuous time signal x(t) was sampled to obtain the discrete time signal x[n] = 4e for n 20, and x[n] -0 otherwise, Write an expression for the continuous time signal x(t) and find the sampling frequency in Hz that was used to obtain the discrete sequence x[n]. [4 marks] (b) Suppose the signal x(t) from vibration measurement by a transducer device can be described by the equation: X(t) = 1 + cos(2000 ml) + sin(6000xt) + cos(12000nt) V. Assume ADC device performs sampling of the signal at a rate of 0.2 ms and uses 12 bits for each sample representation. 0) With the support of a simple diagram, briefly explain how the sampling may be accomplished. [4 marks) (ii) Find the frequencies that are contained in the signal x/t). [3 marks] (ii) What is the Nyquist rate that must be used for sampling the signal x(t) for it to be adequately recovered from the samples. [3 marks) (iv) Find the number of samples that would be obtained after the ADC has sampled 60 seconds duration of the signal. Will you consider this number of samples the same if the Nyquist rate is used? Show this. [6 marks) (1) Find the discrete signal x[n] produced after the ADC sampling operation of x(t) (in simplest form or in terms of principal frequency). What is the discrete angular frequency of x[n). [10 marks] (vi) Find the analog signal y(t) that is reconstructed from x[n) using an ideal interpolation, i.e., y(t) = y[nFs) - x[nFs). [4 marks) (vi) Find the frequencies contained in the reconstructed signal y(t). [3 marks) (vii) Is the signal y(t) the same as the original signal x(t)? Give reasons why for your answer (as to why it is the same or not the same). [4 marks) (viii) How much memory space (MB) will be required to store the sampled data after the ADC sampling operation. [3 marks) (ix) Explain the difference between zero-order hold and first order hold method of signal reconstruction. Use diagram to support your answer.
(a) Detailed explanation of Continuous time signal x(t) was sampled to obtain the discrete time signal x[n] = 4e for n 20, and x[n] -0 otherwise:The given sequence can be written as x[n] = 4e n=20, and x[n] = 0 for other n. We can write the expression for continuous time signal x(t) using the given sequence as shown below:x(t) = 4e for 20T ≤ t ≤ 21T= 0 otherwise. The sampling frequency in Hz that was used to obtain the discrete sequence x[n] is given by Fs = 1/T, where T is the time interval between two samples.
b) Detailed explanation of the given information:A diagram to explain how the sampling may be accomplished is shown below:Sampling can be explained as a process of converting a continuous-time signal into a discrete-time signal. This can be done using an analog-to-digital converter (ADC) device. The given signal x(t) can be expressed asX(t) = 1 + cos(2000πt) + sin(6000πt) + cos(12000πt)V.
Frequencies that are contained in the signal x(t) can be found as follows:Frequency spectrum of x(t) can be represented as shown below:Frequency spectrum of x(t) contains three frequencies, which are 2 kHz, 6 kHz, and 12 kHz. Nyquist rate that must be used for sampling the signal x(t) for it to be adequately recovered from the samples is given by the formula fN = 2fM, where fM is the highest frequency in the signal. The highest frequency in the signal x(t) is 12 kHz.fN = 2fM= 2 × 12 kHz= 24 kHz.The number of samples that would be obtained after the ADC has sampled 60 seconds duration of the signal can be calculated as shown below:N = (60 s)/(0.2 ms/sample)= 300000 samples.The number of samples would be the same if the Nyquist rate is used as it is greater than the sampling rate, which is 5 kHz.The discrete signal x[n] produced after the ADC sampling operation of x(t) is given by the formula x[n] = x(nTs) = X(n/Fs), where Ts = 1/Fs.
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Find the transfer function H(z) for the following digital systems y(n)=x(n) +0.75x(n-1) +0.5x(n-2) ii. y(n)= x(n)-0.5y(n-1) 111. y(n)=0.5 x(n)-0.25x(n-2) + 0.5 x(n-3) - 0.1y(n-1) - 0.28y(n-2) d. Write a Matlab code to solve the previous question
Scripts lack input and output arguments, making them the most basic type of code file.
Thus, They can be used to automate a number of MATLAB instructions, such as calculations that must be carried out frequently from the command line or a list of commands that must be referred to.
A new script can be made using the following methods. Right-click on any highlighted commands in the Command History and choose Create Script. Simply select the New Script button under the Home menu.
Use the editing feature. For instance, the command edit creates and opens the file new file name if it doesn't already exist. MATLAB creates a new file called Untitled if new file name is left blank.
Thus, Scripts lack input and output arguments, making them the most basic type of code file.
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1. What happens when a std::unique_ptr is passed by value to a function? In the following code. Please explain in detail
#include auto f(std::unique_ptr ptr) { *ptr = 42; return ptr; } int main() { auto ptr = std::make_unique(); ptr = f(ptr); } 2. Problem: set intersection
Input: An array of size n and an array of size m
Output: true if two sets are disjoint, false if otherwise
Write your algorithm to solve this problem. Analyze the worst-case complexity in terms of m and n, considering the case where m << n (m is a lot smaller than n).
1. When a std::unique_ptr is passed by value to a function, ownership is transferred and the function can modify the pointed object, as demonstrated in the code.
2. The algorithm checks if two sets are disjoint by using a hash set or binary search tree, with a worst-case complexity of O(n + m log n) for a tree or O(n + m) for a hash set when m << n.
1. What happens when a std::unique_ptr is passed by value to a function, as shown in the provided code? Explain in detail.2. Write an algorithm to determine if two sets are disjoint given an array of size n and an array of size m. Analyze the worst-case complexity in terms of m and n, considering the case where m << n.1. When a std::unique_ptr is passed by value to a function, ownership of the dynamically allocated object is transferred to the function parameter. In the given code, the function `f` takes a std::unique_ptr `ptr` as a parameter and assigns the value 42 to the object pointed to by `ptr`. The modified `ptr` is then returned from the function.
In the main function, a std::unique_ptr `ptr` is created using std::make_unique(). It is then passed as an argument to the function `f(ptr)`. The function `f` modifies the value of the object pointed to by `ptr` to 42 and returns the modified pointer. However, the returned pointer is assigned back to `ptr`, effectively replacing the original pointer.
2. Algorithm for set intersection:
- Initialize a hash set or a balanced binary search tree to store elements from the first array.
- Iterate through each element in the second array and check if it exists in the set or tree. If found, return false as the sets are not disjoint.
- If the loop completes without finding any common elements, return true as the sets are disjoint.
Worst-case complexity analysis:
- Constructing the set or tree takes O(n) time and space complexity.
- Iterating through the second array takes O(m) time.
- Checking if each element exists in the set or tree takes O(log n) time for a balanced binary search tree or O(1) time on average for a hash set.
- Therefore, the worst-case complexity for the disjoint set algorithm in terms of m and n, considering m << n, is O(n + m log n) for a balanced binary search tree or O(n + m) for a hash set.
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a Problem Description Implement a recursive function named order that receives as arguments an array named a and an integer named n. After the function executes, the elements in the array must become in ascending order without using global or static variables. Examples Before After [40,70,80, 60,40] [40, 40, 60, 70, 80) Write a C program that performs the following: o Asks the user to input an integer n. • Creates an n-element 1-D integer array named random • Fills each element in the array by random multiples of 10 between 10 and 100 Inclusive. • prints the array. passes the array to the function order, then prints the array again. Organize the output to appear as shown in the sample output below Enter number of elements 5 The array before sorting: 40 70 80 60 40 The array after sorting: 40 60 70 40 80
Below is the C program that asks the user to input an integer n, creates an n-element 1-D integer array named random, fills each element in the array by random multiples of 10 between 10 and 100 Inclusive, prints the array, passes the array to the function order, and then prints the array again.```
#include
#include
#include
#define MAX 100
void order(int *a, int n) {
int i, j, temp;
for(i = 0; i < n-1; i++) {
for(j = i+1; j < n; j++) {
if(*(a+i) > *(a+j)) {
temp = *(a+i);
*(a+i) = *(a+j);
*(a+j) = temp;
}
}
}
}
int main() {
int i, n, a[MAX];
printf("Enter number of elements: ");
scanf("%d", &n);
srand(time(NULL));
for(i = 0; i < n; i++)
*(a+i) = rand() % 10 * 10 + 10;
printf("\nThe array before sorting: ");
for(i = 0; i < n; i++)
printf("%d ", *(a+i));
order(a, n);
printf("\n\nThe array after sorting: ");
for(i = 0; i < n; i++)
printf("%d ", *(a+i));
return 0;
}
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A file has 7-20,00ey bytes) sa622Aybek). ADENT records of fixed length. Each record has the following fields. Name (30 (40 bytes), PHONE (10 (8 Sabyte). Majar_dept_code (4 bytes), Minar_dept code (4 bytes), Class_code (4 bytes, integer), and Degree program (3 bytes) An additional byte is used as a deletion marker. Suppose only 80% of the STUDENT records have a value for PHONE, 85% for MAJORDEPTCODE, 15% for MINORDEPTCODE, and 90% for DEGREEPROGRAM, and we use a variable-length record file. Each record has a 1-byte field type for each field occurring in the record(NAME, SSN, ADDRESS, BIRTHDATE, SEX, CLASSCODE), plus the 1-byte deletion marker and a 1-byte end-of-record marker. Suppose we use a spanned record organization, where each block has a 5-byte pointer to the next block (this space is not used for record storage). (a) Calculate the average record length Rin bytes. (5 marks) (b) Calculate the number of blocks needed for the file. (5 marks) A file has r-20,000 STUDENT records of fixed length. Each record has the following fields. Name (30 bytes), San (ykeh). Address (40 bytes), PHONE (10 bytes), Birth_date (8 bytes). SEX byte), Major_dept_code (4 bytes). Minor_dept_code (4 bytes), Class_code (4 bytes, integer), and Degree program (3 bytes). An additional Byte is used as a deletion marker. Suppose only 80% of the STUDENT records have a value for PHONE, 85% for MAJORDEPTCODE, 15% for MINORDEPTCODE, and 90% for DEGREEPROGRAM, and we use a variable-length record file. Each record has a 1-byte field type for each field occurring in the record(NAME, SSN, ADDRESS, BIRTHDATE, SEX, CLASSCODE), plus the 1-byte deletion marker and a 1-byte end-of-record marker. Suppose we use a spanned record organization, where each block has a 5-byte pointer to the next block (this space is not used for record storage). (5 marks) (a) Calculate the average record length Rin bytes. (b) Calculate the number of blocks needed for the file. (5 marks)
The average record length is approximately 7.13 bytes.
Approximately 18,548 blocks are needed for the file.
(a) To calculate the average record length (R) in bytes, we need to sum up the lengths of all fields in a record and divide it by the number of fields.
Record length = Length of Name + Length of PHONE + Length of Major_dept_code + Length of Minor_dept_code + Length of Class_code + Length of Degree_program + Length of deletion marker + Length of end-of-record marker
Record length = 30 + 10 + 4 + 4 + 4 + 3 + 1 + 1 = 57 bytes
The average record length (R) can be calculated as follows:
R = Record length / Number of fields
R = 57 / 8 ≈ 7.13 bytes
(b) To calculate the number of blocks needed for the file, we need to consider the number of records and the block size.
Number of records = 20,000
Block size = Record length + Pointer size
Assuming the pointer size is 5 bytes, we can calculate the total size of a block:
Block size = Record length + 5
Block size = 57 + 5 = 62 bytes
Number of blocks needed = Number of records / (Block size / Record length)
Number of blocks needed = 20,000 / (62 / 57)
Number of blocks needed ≈ 18,548 blocks
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A faulty casino jackpot machine is exhibiting signs of abnormalities every day. 6 am: The sound system, screen display and 'cash in' alert display are all turned OFF. 8 am: The sound system will be turned ON only. 12 noon: The screen display will be turned ON only. 6pm: The sound system and screen display will be turned ON only. 11 pm: The 'cash in' screen display will be turned ON only. After which, the faulty casino jackpot machine will return to the same state at 6 am the following day. Note that all unused states will return to the same stage at 6 am daily. (a) Sketch the state transition diagram for the faulty casino jackpot machine and explain if it is a self-correcting one. (7 marks) (b) Using only J-K flip-flops, state the simplified Boolean expression for each input of the flip-flop. (8 marks) (c) Recall the operations of J-K flip-flops and derive the state transition table.
The operation of J-K flip-flop can be described with the help of the following state transition table:
Clock (Clk)J inputK inputState before the clock pulseQnState after the clock pulse
Qn+16 AMDonDon0 AMDonDoff0 AMDonDoff8 AMDonDon8 AMDonDoff8 AMDoffDon8 AMDoffDoff8 AMDoffDon12 NoonDoffDon8 AMDoffDon6 PMDoffDon14 PMDoffDon11 PMDonDoff14 PMDonDoff11 PMDonDoff6 AMDonDon0 AM
a) The state transition diagram for the faulty casino jackpot machine is given below.
The faulty casino jackpot machine is self-correcting because all unused states will return to the same state at 6 am the following day. It is a cyclic machine because the machine moves from state to state and ultimately returns to the initial state. The self-correcting characteristic of the machine makes it a good state machine.
b) The simplified Boolean expression for each input of the flip-flop with J-K flip-flops is given below:
J-K Flip-Flop
The simplified Boolean Expression for each input
J1 (Q, Q') = (K1'Q' + J1'Q)Q1' + (K1'Q)Q1Q1 (Q, Q')
= (J1'Q' + K1'Q)Q1' + (J1'Q)Q1J2 (Q1, Q1')
= (K2'Q1' + J2'Q1)Q2' + (K2'Q1)Q2Q2 (Q1, Q1')
= (J2'Q1' + K2'Q1)Q2' + (J2'Q1)Q2
The above simplified Boolean expressions for each input of the flip-flop using J-K flip-flops.
c) J-K flip-flop can be built by cross-coupling of the NOT gates, AND gates, and OR gates. The operation of J-K flip-flop can be described with the help of the following state transition table:
Clock (Clk)J inputK inputState before the clock pulseQnState after the clock pulse
Qn+16 AMDonDon0 AMDonDoff0 AMDonDoff8 AMDonDon8 AMDonDoff8 AMDoffDon8 AMDoffDoff8 AMDoffDon12 NoonDoffDon8 AMDoffDon6 PMDoffDon14 PMDoffDon11 PMDonDoff14 PMDonDoff11 PMDonDoff6 AMDonDon0 AM
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While this week's work focuses on three primary types of data storage, flat text, JSON, and Python's pickle, there is a wide world of meaningful/ accessible data types avaialble to Python practictioners. Pick another data format (different from the three we used in class) and discuss the various strengths, weaknesses, and "gotchas" of that data format and how it compares to the three we used in class and any other comparisons you might feel are apt. In particular, describe potential use cases for the data format and why you feel it is a more appropriate format than any other for these use-cases. Finally, provide us a simple implementation of reading and writing to that particular data format using Python.
The XML file format is another data format that is commonly used besides the ones already discussed in class (flat text, JSON, and Python's pickle).XML stands for eXtensible Markup Language. It is a markup language like HTML, but its main purpose is to store and transport data instead of displaying data.
The following are the various strengths, weaknesses, and "gotchas" of the XML data format:
Strengths:• XML is self-describing, which means that it contains metadata that describes the data it holds. This means that there is no need to maintain a separate data dictionary.• XML is a vendor-neutral standard, meaning that it can be used across different platforms and operating systems.• XML can be used for many purposes, including data exchange, configuration files, and web services.• XML can be parsed easily using many programming languages.
Weaknesses:• XML is a verbose format, which means that it can be bulky and take up more space than other formats.• XML can be complicated to learn and use. This means that it can take longer to develop an XML-based system than a system that uses a simpler format.• XML can be sensitive to the order of its elements.Gotchas:• XML is case-sensitive, so it is important to ensure that the tags are spelled correctly.• XML can be vulnerable to injection attacks, so it is important to ensure that the data is properly sanitized before it is stored or processed.
The following are the potential use cases for the XML data format:• Configuration files• Data exchange• Web services• Storing large amounts of dataFinally,
here's an example of reading and writing to an XML file using Python: Example Code:import xml.etree.ElementTree as ET # Reading from an XML file tree = ET.parse('file.xml') root = tree.getroot() for child in root: print(child.tag, child.attrib) # Writing to an XML file root = ET.Element('data') item = ET.SubElement(root, 'item') item.set('name', 'item1') item.set('value', '10') tree = ET.ElementTree(root) tree.write('file.xml').
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From chapter 8 of Zelle, modify his program average7.py to work with a file that contains positive and negative numbers. For this problem, please use readline( ). Your job is to compute and print two values:
The average of the positive numbers
The average of the negative numbers any 0 values should be treated as positive.
In your solution only call the float function ONCE for each number processed. For your convenience, Zelle's file can be found in average7.py
The `float` function is only called once for each number processed, as required. Also note that this program assumes that the file contains only numbers, one per line. If the file contains other data, the program will fail.
In order to modify the program average7.py in chapter 8 of Zelle to work with a file containing positive and negative numbers, follow the steps below:First, open the file that contains the numbers using the built-in `open` function. Next, create two counters - one for positive numbers and one for negative numbers.
def main():
fileName = input("Enter the name of the data file: ")
infile = open(fileName, 'r')
posCount = 0
negCount = 0
posSum = 0.0
negSum = 0.0
line = infile.readline()
while line != "":
num = float(line)
if num >= 0:
posCount += 1
posSum += num
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Describe what cach of the four Maxwell's equations mean. 0.D = b. 1.B = 0 aB VXE = al ad VII = J + d. al
James Clerk Maxwell was a Scottish physicist who formulated a set of equations that united the concepts of electricity and magnetism in the mid-19th century.
Each of the four Maxwell's equations has a different meaning, which is as follows:
Gauss's Law for Electric Fields (Equation 1): It relates electric field lines to their sources, which are electric charges. This equation describes how electric charges generate electric fields, which then influence other electric charges.Gauss's
Law for Magnetic Fields (Equation 2): It states that magnetic field lines do not have sources. Instead, they either originate or terminate at the magnetic poles of an object.
Faraday's Law of Electromagnetic Induction (Equation 3): It states that any changes in magnetic flux that cut across a conductor generate an electromotive force (EMF) in that conductor. This equation describes how electric fields can be induced by magnetic fields, which is the principle behind electric generators.
Ampere's Law with Maxwell's Addition (Equation 4): It relates magnetic fields to the flow of electric charges. It explains how magnetic fields are generated by electric currents and how changing electric fields can generate magnetic fields.
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The two most common types of schedules used in construction are the bar chart and a) Flow chart b) Network diagram Progress diagram d) Timing program Leave blank Close ✓End the Exam
The correct answer is: b) Network diagram. Network diagrams, also known as activity-on-node (AON) diagrams or precedence diagrams, are widely used in construction scheduling.
The two most common types of schedules used in construction are the bar chart and the network diagram.
a) Flow chart: Flow charts are used to represent the sequence of activities in a process but are not typically used as primary scheduling tools in construction projects.
b) Network diagram: Network diagrams, also known as activity-on-node (AON) diagrams or precedence diagrams, are widely used in construction scheduling. They depict the relationships between activities, including dependencies and durations, using nodes and arrows.
c) Progress diagram: Progress diagrams are not a common scheduling tool in construction. They are often used to visually represent the progress or status of activities within a project but are not focused on scheduling activities.
d) Timing program: Timing programs refer to scheduling software or tools that help in creating, managing, and updating construction schedules. While timing programs can utilize bar charts and network diagrams, they are not a distinct type of schedule themselves.
Therefore, the correct answer is: b) Network diagram.
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Write Problem Definition, Overview and Implementation
for the following
1. Compare and contrast MANET and VANET
MANET and VANET are two network technologies that have been designed to provide mobile wireless communications in different settings.
Overview: MANET (Mobile Ad Hoc Network) is a self-configuring network that is composed of mobile nodes that can be connected in a wireless manner. It is designed for use in scenarios where there is no fixed infrastructure, such as in emergency situations, military operations, or in areas where there is no pre-existing network infrastructure.
VANET (Vehicular Ad Hoc Network), on the other hand, is a type of MANET that is designed specifically for use in vehicular communications. It is used in vehicles to provide wireless connectivity between them, and to enable communication between vehicles and the surrounding infrastructure.
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In vCenter Server High Availability, the ______ node is used to maintain quorum.
a. passive
b. primary
c. active
d. secondary
e. witness
If you use VMware Host Client to communicate directly with an ESXi host, communications go directly to the ______ process and the vCenter
Server database is not updated.
a. vpxa
b. hosta
c. vpxd
d. hostd
e. vpx
In vCenter Server High Availability, the witness node is used to maintain quorum.
The witness node in vCenter Server High Availability plays a crucial role in maintaining quorum. Quorum refers to the majority of nodes in a cluster agreeing on the state of the system. In this case, the witness node acts as a tiebreaker and helps determine the availability and health of the vCenter Server. It does not actively participate in the processing or execution of virtual machines, but it contributes to the decision-making process in case of failovers or cluster reconfiguration. By having the witness node, the cluster can achieve a majority decision, ensuring the continuity of services and maintaining the high availability of the vCenter Server environment.
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Which of the following valves is better for on/ off control? a) Butterfly valve b) Ball valve c) Plug valve d) Knife valve In globe valves, the flow rate control is determined by a) Size of the opening b) Lift of the valve plug c) Pressure difference d) Gravity A--------is a connector showing the relationship between the representative shapes. A. Line B. arrow C. Process D. box
The ball valve is better for on/off control. This is due to the fact that it has a simple design, is less prone to clogging, and is capable of tight sealing.
In globe valves, the flow rate control is determined by the lift of the valve plug.
A line is a connector showing the relationship between the representative shapes.
Thus,
In first statement option b is correct. In second statement option b is correct. In third statement option A is correct.Learn more about valve:
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Complete the following code to toggle B4 every 1 ms. Prescaling =8, and the frequency of the oscillator is 8 MHz. DDRB, loop: SBI 4 CALL delay
Given the following code, one must complete it to toggle B4 every 1ms. Prescaling = 8 and the oscillator frequency is 8 MHz:DDRB, loop: SBI 4 CALL delayTo toggle the B4 in a delay of 1ms we need to calculate the time in which the signal needs to be high and low.
To calculate the time, we use the formula given below:Time Period = 1 / Frequency where the frequency is equal to the oscillator frequency divided by the prescaling.To calculate the frequency, we use the formula given below:Frequency = Oscillator Frequency / PrescalingThus, the frequency will be equal to:8MHz / 8 = 1 MHzWe want to toggle the B4 pin every 1ms.
Therefore, we need to calculate the ON time and OFF time for the signal. Let us suppose that we need to turn ON the signal for half the time and OFF for half the time. Thus, the ON time and OFF time will be equal to:On time = Off time = 0.5 msNow, we need to calculate the number of clock cycles needed to produce the delay of 0.5ms.
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i have an HTML code for a web page and i want to add a weather API to it how can I do that ?
p.s: I want a step by step explanation please
The weather widget can be added anywhere on your web page using an HTML tag. Here is an example: This is just a basic example, and you can customize the widget to suit your needs. You can also add more features, such as the current temperature, the forecast for the next few days, and more.
To add a weather API to an HTML code for a web page, you can follow these steps:-
Step 1: Sign Up for a Weather API Service. There are several weather API services out there, including Open Weather Map, Weather Underground, and Accu Weather. Choose one of them and sign up for their service to obtain an API key, which is necessary to access the weather data.
Step 2: Obtain the API URL Once you have signed up for the service and obtained your API key, you will need to obtain the API URL, which is the address that your web page will use to access the weather data.
Step 3: Add the API URL to Your HTML CodeNow you need to add the API URL to your HTML code by creating a script tag in the head section of your HTML document. The script tag should include the API URL and the API key, which you obtained earlier.
Step 4: Add the Weather Widget, once you have added the script tag to your HTML code, you can now add the weather widget to your web page.
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List two Turkey-based ERP software and describe their main features.
When it comes to Turkey-based ERP software, there are a variety of solutions available. Some of the most popular ERP software options available in Turkey are:
1. Logo Tiger:
Logo Tiger is an ERP software solution developed by Logo Software, which is based in Turkey. The software includes a range of features and tools, including accounting, inventory management, customer management, and more. Some of the main features of Logo Tiger include:
Inventory Management: Allows you to manage inventory and track stock levels, including real-time updates and alerts when inventory levels are low.
CRM: The software includes customer relationship management tools, which allow you to track customer information, monitor customer interactions, and provide personalized customer service.
Accounting: Logo Tiger includes a range of accounting features, such as financial reporting, general ledger, and accounts payable/receivable.
2. Netsis ERP:
Netsis ERP is another Turkey-based ERP software solution that is widely used by businesses of all sizes. The software includes a variety of features, including financial management, inventory management, and customer relationship management. Some of the key features of Netsis ERP include:
Financial Management: The software includes a range of financial management tools, such as general ledger, accounts payable/receivable, and financial reporting.
Inventory Management: Netsis ERP includes advanced inventory management tools, such as real-time tracking and alerts, purchase order management, and more.
CRM: The software includes customer relationship management tools that allow you to manage customer interactions, monitor customer data, and provide personalized customer service.
In conclusion, these two Turkey-based ERP software are Logo Tiger and Netsis ERP. Logo Tiger provides inventory management, CRM, and accounting, while Netsis ERP provides financial management, inventory management, and CRM as well.
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Match the following: Binding is done at execution time Where a process is accessing/updating shared data [Choose ] Demand-paged virtual memory Best fit critical section Belady's anomaly Atomic instruction Race condition Dynamic storage-allocation algorithm which results in the smallest leftover hole Translation look-aside buffer Page-fault rate may increase as the number of allocated frames increases Nonvolatile memory Dynamically linked library Optimal page replacement Thrashing Enhanced second chance algorithm Results when several threads try to access and modify the same data concurrently Executes as a single, uninterruptible uni [Choose ] A small, fast-lookup hardware cache for [Choose]
What is the purpose of a cache memory in a computer system.
What are the primary components of a central processing unit (CPU)?Binding is done at execution time - Dynamic linking
Where a process is accessing/updating shared data - Critical section
Demand-paged virtual memory - Page-fault rate may increase as the number of allocated frames increases
Best fit - Dynamic storage-allocation algorithm which results in the smallest leftover hole
Belady's anomaly - Optimal page replacement
Atomic instruction - Executes as a single, uninterruptible unit
Race condition - Results when several threads try to access and modify the same data concurrently
Translation look-aside buffer - A small, fast-lookup hardware cache for virtual-to-physical address translation
Nonvolatile memory - Thrashing
Dynamically linked library - Enhanced second chance algorithm
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7-50 Stereo FM transmission was studied in Sec. 5-7. At the transmitter, the left-channel audio, m(t), and the right-channel audio, mp(t), are each preemphasized by an f₁ = 2.1-kHz network. These preemphasized audio signals are then converted into the composite baseband modulating signal m(t), as shown in Fig. 5-17. At the receiver, the FM detector outputs the composite baseband signal that has been corrupted by noise. (Assume that the noise comes from a white Gaussian noise channel.) This corrupted composite baseband signal is demulti- plexed into corrupted left- and right-channel audio signals, m(t) and mp(t), each having been deemphasized by a 2.1-kHz filter. The noise on these outputs arises from the noise at the output of the FM detector that occurs in the 0- to 15-kHz and 23- to 53-kHz bands. The subcarrier frequency is 38 kHz. Assuming that the input SNR of the FM receiver is large, show that the stereo FM system is 22.2 dB more noisy than the corresponding monaural FM system.
A stereo FM transmission (7-50) was analyzed in Section 5-7. The left-channel audio, m(t), and the right-channel audio, mp(t), are preemphasized at the transmitter using an f₁=2.1-kHz network. These preemphasized audio signals are converted into the composite baseband modulating signal m(t) as shown in Fig. 5-17.
At the receiver, the FM detector yields the composite baseband signal that has been contaminated by noise. Assume that the noise originates from a white Gaussian noise channel. The corrupted composite baseband signal is demultiplexed into two corrupted left and right audio signals, m(t) and mp (t), respectively, each of which has been subjected to a 2.1-kHz filter and is therefore demphasized.
The noise in these outputs is caused by noise at the output of the FM detector in the 0- to 15-kHz and 23- to 53-kHz bands. The subcarrier frequency is 38 kHz. Assume that the FM receiver's input SNR is large. Now show that the stereo FM system is 22.2 dB more noisy than the corresponding monaural FM system.
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Q.1.3 Create a hierarchy chart that accurately represents the logic in the scenario below: Scenario: The application for an online store allows for an order to be created, amended, and processed. Each of the functionalities represent a module. Before an order can be amended though, the order needs to be retrieved.
Here is a hierarchy chart that accurately represents the logic in the given scenario:
```
Application
|
- Create order
|
- Retrieve order
|
- Amend order
|
- Process order
```
In the above hierarchy chart, the Application is the main module. The Application has four sub-modules - Create Order, Retrieve Order, Amend Order, and Process Order. The Create Order, Amend Order, and Process Order modules can be accessed directly. However, before the Amend Order module can be accessed, the Retrieve Order module needs to be accessed first.
According to the scenario given, before an order can be amended, it must first be retrieved. Therefore, the flow of the program would have to go through the retrieval module before going to the amendment module . The order processing module has a control flow arrow pointing to it, showing that an order can only be processed after it has been amended to the customer's satisfaction.
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When configuring a router, when would you choose basic management setup rather than extended setup? 2 marks
What do square brackets [ ] in a prompt for input indicate? 2 marks
Why is it important to configure an enable or enable secret password on a router? 1 mark
If this is a test can you provide the answer choices? Or are you just asking
Explanation:
create a linked list with 10 elements.convert the linked list into array list.write a Java program so that the elements with odd and even indices will be retrieved seperately.
In this program, we first create a linked list and add 10 elements to it. Then, we convert the linked list to an array list using the ArrayList constructor.
Here's a Java program that creates a linked list with 10 elements, converts it into an array list, and retrieves the elements with odd and even indices separately:
import java.util.ArrayList;
import java.util.LinkedList;
public class LinkedListToArray {
public static void main(String[] args) {
// Create a linked list
LinkedList<Integer> linkedList = new LinkedList<>();
// Add elements to the linked list
for (int i = 1; i <= 10; i++) {
linkedList.add(i);
}
// Convert linked list to array list
ArrayList<Integer> arrayList = new ArrayList<>(linkedList);
// Retrieve elements with odd indices
System.out.println("Elements with odd indices:");
for (int i = 0; i < arrayList.size(); i++) {
if (i % 2 != 0) {
System.out.println(arrayList.get(i));
}
}
// Retrieve elements with even indices
System.out.println("Elements with even indices:");
for (int i = 0; i < arrayList.size(); i++) {
if (i % 2 == 0) {
System.out.println(arrayList.get(i));
}
}
}
}
Finally, we retrieve the elements with odd indices and even indices separately by iterating over the array list and checking the index using the modulus operator.
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d²y dy +4 + 10y = f(t) dt² dt a) Assume y(0) = 2 and y'(0) = 0. Find the corresponding transfer function H(s) = Y(s)/F(s) b) from Y(s), identify the zero-input response and the zero-state response.
The differential equation is d²y/dt² + 4(dy/dt) + 10y = f(t).Part a:To find the transfer function H(s) = Y(s)/F(s), where F(s) and Y(s) are the Laplace transforms of f(t) and y(t), respectively, we can first take the Laplace transform of the differential equation.
L{d²y/dt²} + 4L{dy/dt} + 10L{y} = L{f(t)}s²Y(s) - sy(0) - y'(0) + 4sY(s) + 2y(0) + 10Y(s) = F(s)Substituting y(0) = 2 and y'(0) = 0 and solving for Y(s), we get:Y(s) = F(s) / [s² + 4s + 10]Taking the Laplace transform of the output y(t) = h(t) * f(t), where h(t) is the unit impulse response, we get Y(s) = H(s)F(s).
Equating the two expressions for Y(s), we get:H(s) = 1 / [s² + 4s + 10]Part b:The zero-input response is the solution to the homogeneous equation d²y/dt² + 4(dy/dt) + 10y = 0. The characteristic equation is s² + 4s + 10 = 0, which has complex conjugate roots s = -2 ± i√6. The solution is:y(z) = e^{-2t}[c_1cos(√6t) + c_2sin(√6t)]The zero-state response is the solution to the inhomogeneous equation d²y/dt² + 4(dy/dt) + 10y = f(t) with zero initial conditions.
We can use the convolution theorem to find the output:Y(s) = H(s)F(s) = F(s) / [s² + 4s + 10]y(t) = L^{-1}[Y(s)] = L^{-1}[F(s) / (s² + 4s + 10)] = e^{-2t}[sin(√6t)/√6] * u(t)where u(t) is the unit step function.
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CPS310- COMPUTER ORGANIZATION II Lab #4 ARC Subroutines Submission instruction: Labs will be done individually. Please complete and submit this lab on D2L by the submission deadline according to the details provided by your TA. Each student should submit a pdf file that includes all their written work and screenshots of the results of the simulations. Please note that Lab 4 will be graded based on correct completion of the following questions in addition to the answer to TA's questions. Notes: Make sure you store and restore the registers used by the subroutine. Pass all the parameters via stack data structure between caller and callee unless specified otherwise. PART A - myFact Write a program that calls a subroutine called myFact by passing a positive integer value via stack. The subroutine myFact calculates factorial of the positive integer value and returns the result in register %r3. Then the caller program stores the result in the memory location 4048. PART B-SQRT Write a program that calls a function called SQRT by passing a positive integer value via stack. SORT performs a square root of the positive integer number and returns the result in register %r3. Then caller program stores the result in the memory location 6048. Note: SQRT calculates the closest integer value the square root of the positive number. A few examples are provided below. sqrt(13) returns 4 sqrt(12) returns 3 sqrt(10) returns 3 sqrt(9) returns 3 sqrt(8) returns 3 sqrt(5) returns 2 sqrt(4) returns 2 sqrt(3) returns 2
myFactThe program that calls a subroutine called myFact by passing a positive integer value via the stack and stores the result in the memory location 4048. The subroutine myFact calculates the factorial of the positive integer value and returns the result in register %r3. For the correct completion of this question, the following code needs to be written.```.section ".data"myFactResult: .word 0 #This is where the result is stored. We will put this in 4048myInt: .word 3 #The value we want to pass into the subroutine.stack: .skip 4 #Used to hold the value we want to pass into the subroutine. .
section ".text" .globl main #Make main visible to the linker. main: #Start of code. sub sp, sp, #16! str lr, [sp, #12]! ldr r1, =myInt ldr r0, =stack str r1, [r0, #0]! #Push parameter onto stack bl myFact #Call subroutine ldr r0, =myFactResult #Where to put the result. ldr r1, =myFact str r3, [r0] #Store result. ldr lr, [sp, #12]! add sp, sp, #16! bx lr #End of main myFact: sub sp, sp, #12! str lr, [sp, #8]! mov r3, #1 #The starting point for the factorial. mov r2, sp #Where to store the next value. ldr r1, [sp, #4] #Load the integer into r1. myFact_loop: cmp r1, #0 beq myFact_done str r1, [r2, #0]! #Store current value. sub r1, r1, #1 #Get the next value. b myFact_loop myFact_done: ldr r0, [sp, #0]! #Restore LR. add sp, sp, #12! bx lr ```Part B: SQRTWrite a program that calls a function called SQRT by passing a positive integer value via the stack. SQRT calculates the closest integer value to the square root of the positive number and returns the result in register %r3.
Then, the caller program stores the result in the memory location 6048. To complete this question, the following code needs to be written.```.section ".data" sqrtResult: .word 0 #The location where the result is stored. We will put this in 6048. intValue: .word 13 #The value we want to pass into the function. stack: .skip 4 #Used to hold the value we want to pass into the subroutine. .section ".text" .globl main #Make main visible to the linker. main: #Start of code. sub sp, sp, #16! str lr, [sp, #12]! ldr r1, =intValue ldr r0, =stack str r1, [r0, #0]! #Push parameter onto stack bl SQRT #Call the function. ldr r0, =sqrtResult #Where to put the result. ldr r1, =SQRT str r3, [r0] #Store the result. ldr lr, [sp, #12]! add sp, sp, #16! bx lr #End of main SQRT: sub sp, sp, #12! str lr, [sp, #8]! mov r3, #0 #The value we will return. mov r2, sp #The location to store the current value. ldr r1, [sp, #4] #The integer to take the square root of. SQRT_loop: cmp r1, #0 beq SQRT_done sub r1, r1, #1 str r1, [r2, #0]! #Store the current value. add r3, r3, #1 #Add 1 to the result. cmp r1, #0 bne SQRT_loop #Repeat until we are done. SQRT_done: ldr r0, [sp, #0]! #Restore LR. add sp, sp, #12! bx lr ```Explanation:In the first part of the question, we had to create a program that calls a subroutine called myFact and calculate the factorial of the positive integer value and returns the result in register %r3. Then, the caller program stores the result in the memory location 4048. We did this by writing a code that first pushed the value onto the stack and then called the subroutine to calculate the factorial. The value was then stored in the memory location 4048 by the caller program.In the second part of the question, we had to create a program that calls a function called SQRT by passing a positive integer value via the stack and stores the result in the memory location 6048. SQRT calculates the closest integer value to the square root of the positive number and returns the result in register %r3. We did this by writing a code that pushed the value onto the stack and then called the function to calculate the square root. The value was then stored in the memory location 6048 by the caller program.
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On Matlab/command window Draw the step response of the system whose transfer function is given by y(s) = R(S) 9 for 5
We can see here that to draw the step response of a system in MATLAB, you can use the step function. Assuming the transfer function of the system is given by G(s) = Y(s) / U(s) = 9 / (s + 5).
What are the steps to plot the step response?We can see here that the steps are:
1. Define the transfer function:
num = 9; % Numerator coefficients of the transfer function
den = [1 5]; % Denominator coefficients of the transfer function
G = t f(num, den); % Create the transfer function object
2. Plot the step response:
step(G); % Plot the step response
The complete code will be:
num = 9; % Numerator coefficients of the transfer function
den = [1 5]; % Denominator coefficients of the transfer function
G = t f(num, den); % Create the transfer function object
step(G); % Plot the step response
When you run this code in the Matlab command window, it will generate a plot showing the step response of the system.
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Create a flowchart and a java program of the problem stated as follows. Create user-defined methods that will perform 4 functionalities of an Automated Teller Machine (Withdraw, Deposit, Transfer, Balance Check) 1. Withdraw (method name: withdrawl * input the amount to be withdrawn. Compute for the remaining balance after the operation (balance = balance - withdrawn) 2. Deposit (method name: deposit) * input the amount to be deposited. Compute for the updated balance after the operation (balance = balance + deposit) 3. Transfer (method name: transter] *Input the account number and the amount to be transferred. Compute for the updated balance after the operation (balance = balance - moneySent) 4. Balance Check (method name: balanceCheck) * Display the remaining balance. 113 erty of and for the exclusive Use Of SLU. Reproduction, sloning in a retrieval system, distributing, uploading or posting online, or transmiting in any form or by any ns, electronic, mechanical, photocopying, recording, or otherwise of any part of this document without the prior written permission of SLU, is strictly prohibited. Case es Sample Input/Output: Depicted below are sample outputs when the program is executed (the items in bold characters are input from the user, while the items in bold italic are calculated and printed by the program): ATM- ATM- 1. Withdraw 1. Withdraw 2. Deposit 2. Deposit 3. Transfer 3. Transfer 4. Check Balance 4. Check Balance 5. EXIT 5. EXIT Choose operation: 1 Choose operation: 1 Enter amount to be withdrawn: 1000 Enter amount to be withdrawn: 6000 Collect your money Insufficient Balance Check Balance? (Y/N): Y Check Balance? (Y/N): Y Balance: 4000 Balance: 4500 ATM- ATM 1. Withdraw 1. Withdraw 2. Deposit 2. Deposit 3. Transfer 3. Transfer 4. Check Balance 4. Check Balance 5. EXIT 5. EXIT Choose operation: 2 Choose operation: 3 Enter amount you want to deposit: 500 Enter Account Number:0566323454 Check Balance! (Y/N): N Enter Transfer Amount: 300 Check Balance? (Y/N): Y Balance: 3200 *Note: The initial balance is 0. The user is given an option to check the balance or not. The program will run until the exit option is chosen. Required: The flowchart, the java file (FamilyName_ATM.java) containing the code and 4 for more) image files (Samplel, Sample2, Sample3, and Sample4) containing different sample input/output of the program.
Flowchart and Java Program for Automated Teller Machine (ATM)Here is the flowchart and Java program for Automated Teller Machine (ATM).
We have used a switch case statement to perform different operations based on user input. We have also defined 4 methods for performing the 4 functionalities of the ATM as required.// Java program for Automated Teller Machine (ATM)
import java.util.Scanner;
public class ATM {
static int balance = 0;
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int choice = 0;
do {
System.out.println("ATM-");
System.out.println("1. Withdraw");
System.out.println("2. Deposit");
System.out.println("3. Transfer");
System.out.println("4. Balance Check");
System.out.println("5. EXIT");
System.out.print("Choose operation: ");
choice = sc.nextInt();
switch (choice) {
case 1:
withdraw();
break;
case 2:
deposit();
break;
case 3:
transfer();
break;
case 4:
balanceCheck();
break;
case 5:
System.out.println("Thank you for using this ATM, have a nice day!");
break;
default:
System.out.println("Invalid choice! Please choose a valid operation.");
}
} while (choice != 5);
}
static void withdraw() {
System.out.print("Enter amount to be withdrawn: ");
int withdrawn = sc.nextInt();
if (balance >= withdrawn) {
balance = balance - withdrawn;
System.out.println("Collect your money");
} else {
System.out.println("Insufficient balance!");
}
}
static void deposit() {
System.out.print("Enter amount you want to deposit: ");
int deposit = sc.nextInt();
balance = balance + deposit;
System.out.println("Deposit successful!");
}
static void transfer() {
System.out.print("Enter Account Number: ");
long accountNumber = sc.nextLong();
System.out.print("Enter Transfer Amount: ");
int moneySent = sc.nextInt();
if (balance >= moneySent) {
balance = balance - moneySent;
System.out.println("Transfer successful!");
} else {
System.out.println("Insufficient balance!");
}
}
static void balanceCheck() {
System.out.print("Check Balance? (Y/N): ");
char choice = sc.next().charAt(0);
if (choice == 'Y' || choice == 'y') {
System.out.println("Balance: " + balance);
}
}
}
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