What are the values of the following expressions, assuming that n is.pdf
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What are the values of the following expressions, assuming that n is 17 and m is 18? n / 10 + n % 10 = n % 2 + m % 2 = (m + n) / 2 = (m + n) / 2.0 = (int) (0.5 * (m + n)) = (int) Math.round(0.5 * (m + n)) = Solution Assuming the output required is in decimal and m and N are double. n / 10 + n % 10 = 8.7 n % 2 + m % 2 = 1 (m + n) / 2 = 17.5 (int) (0.5 * (m + n)) = 17 (int) Math.round(0.5 * (m + n)) = 18.
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The dividends received deduction
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apply to our crew. Then he will cover the two or three items that are relevant to our getting the
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2)they can implement the change in documentation
3)they can regenerate the code by using code generator
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package audioplay;
import javax.sound.sampled.UnsupportedAudioFileException;
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import javax.sound.sampled.LineUnavailableException;
import javax.sound.sampled.SourceDataLine;
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private static final int MAX_FILE_SIZE = 4096;
//Function to play audio file
void playMusic(String Path) {
File wavFile = new File(Path);
try {
AudioInputStream audioStream = AudioSystem.getAudioInputStream(wavFile);
AudioFormat format = audioStream.getFormat();
DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);
SourceDataLine al = (SourceDataLine) AudioSystem.getLine(info);
al.open(format);
al.start();
System.out.println(\"Staring the Music....!!\");
byte[] bytesBuffer = new byte[MAX_FILE_SIZE];
int bytesRead = -1;
while ((bytesRead = audioStream.read(bytesBuffer)) != -1) {
al.write(bytesBuffer, 0, bytesRead);
}
al.drain();
al.close();
audioStream.close();
System.out.println(\"Music stopped.!!\");
} catch (UnsupportedAudioFileException ex) {
System.out.println(\"The audio file not supported.Please try again with another audio file...!!!\");
ex.printStackTrace();
} catch (LineUnavailableException ex) {
System.out.println(\"Error Occurred\");
ex.printStackTrace();
} catch (IOException ex) {
System.out.println(\"Error..File may b corrupted.\");
ex.printStackTrace();
}
}
public static void main(String[] args) {
String audioFilePath = \"C:\\\\Users\\\\ttechnocraft\\\\Music\\\\adix.wav\";
AudioPlay player = new AudioPlay();
player.playMusic(audioFilePath);
}
}.
Identify the major provisions of the Affordable Care Act that will h.pdf
Identify the major provisions of the Affordable Care Act that will have an impact on individuals
and families.
Solution
The Patient Protection and Affordable Care Act ensures the access to quality, affordable health
care to all and create the transformation within the health care system necessary to contain costs.
It is designed to increase number of insured Americans, make healthcare & health insurance
affordable & improve performance of healthcare system in US.
Provisions of affordable care act:
1) Universal & continues access to health insurance: This act wants to accomplish the goal of
universal & continues access to healthcare & health insurance service to maximum US citizens.
Here there are two provisions
a) Insurance exchanges: The act wants to create online market place or health insurance
exchanges which will be helpful in reducing the prices of insurance plan & increasing
competition among service providers.
b) Expansion of coverage: The act wants to expand Medicaid coverage so that maximum
services like pediatric & dental care are also included in the plan.
2) Health insurance affordability: The law has specific subsidies for health insurance through tax
credits for lower income groups.
3) Hold insurance companies accountable (program integrity): The act includes numerous
provisions to terminate Medicaid that have been terminated in other programs. Suspending
payments for pending investigations or credible allegations of fraud & preventing inappropriate
claims.
4) Provider payments: States will receive 100% federal matching funds for the increase in
payments.
5) Children health insurance program: The act has provisions to extend funding for Children\'s
Health Insurance Program (CHIP).
Above provisions of this act will have an impact on individuals and families lives in a positive &
effective manner..
I have a simple TCP clientserver socket set up in Python. How can I.pdf
I have a simple TCP client/server socket set up in Python. How can I send specific sized
messages from the client to the server? Specifically 1, 100, 200, 400, 800 and 1000 bytes, and
also 1K, 2K, 4K, 8K, 16K and 32K bytes
I am using Python 2.7
Solution
You will use TCP to send and receive messages of size 1, 100, 200,
400, 800 and 1000 bytes. To measure throughput, you will use TCP to send and receive
messages of size 1K, 2K, 4K, 8K, 16K and 32K bytes. Note the difference in units for the
two sets of message sizes. For each measurement and for each message size, the client will
send at least ten probe messages to the server, which will echo back the messages.
Protocol phases: As in Part I, the client first needs to set up a TCP connection to
the server using the socket interface. The echo application will be extended, however, by
specifying the exact protocol interactions between the client and the server. This entails
specifying the exact message formats, as well as the different communication phases, as
outlined next.
1. Connection Setup Phase (CSP). This is the first phase in the protocol where the
client informs the server that it wants to conduct active network measurements in
order to compute the RTT and throughput of its path to the server. We will outline
the expected behavior from both the client and the server next.
CSP: Client. After setting up a TCP connection to the server, the client must send
a single message to the server having the following format:
\
• PROTOCOL PHASE: The protocol phase during the initial setup will be denoted
by the lower case character ’s’. This allows the server to differentiate between
the different protocol phases that the client can be operating at, as we will see.
• M-TYPE (MEASUREMENT TYPE): Allows the client to specify whether it
wants to compute the RTT, denoted by “rtt”, or the throughput, denoted by
“tput”.
• PROBES: Allows the client to specify the number of measurement probes that
the server should expect to receive. Once all the probe messages have been
echoed back and a sample measurement is taken for each one, the client should
compute an estimate of the mean (average) RTT or mean throughput, depending
on the type of measurement being performed. A detailed description of the probe
message’s format is provided in the description of the Measurement Phase.
MSG SIZE (MESSAGE SIZE): Specifies the number of bytes in the probe’s
payload.
• SERVER DELAY: Specifies the amount of time that the server should wait before
echoing the message back to the client. The default value should be 0. You
will vary this value later to emulate paths with longer propagation delays. Even
though increasing the server delay merely increases the processing time at the
server, it nevertheless causes the feedback delay, observed by the sender, to increase,
which has an effect somewhat similar to increasing the path’s propagation
delay.
• WS: A single white space to separate the different fields in the message. The
white .
Find the effective rate corresponding to the given nominal rate. (Ro.pdf
Find the effective rate corresponding to the given nominal rate. (Round your answer to the
nearest hundredth of a percentage point.)
7%/year, compounded monthly ?
Solution
I(t) = (1+(r/n))^(nt) Where I = annual rate, r = nominal rate, n = number of times
compounded per year, t = years I(t) = (1+ (.07/12))^(12*1)= 1.07229 Effective rate = 7.23%.
C++ AssignmentPlease read all the requirements and the overloading.pdf
C++ Assignment
Please read all the requirements and the overloading requiremets.
Here is what i have so far,
#include
using namespace std;
class ComplexNumbers
{
private:
double real;
double imaginary;
public:
void input ();
void output();
void setReal ( int );
int getReal ();
void setImaginary ( int );
int getImaginary();
ComplexNumbers();
ComplexNumbers(int);
ComplexNumbers(int, int);
friend bool operator == (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator + (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator - (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator * (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator - (ComplexNumbers);
};
void ComplexNumbers::setReal(int x)
{
real=x;
}
void ComplexNumbers::setImaginary(int y)
{
imaginary=y;
}
int ComplexNumbers::getReal()
{
return real;
}
int ComplexNumbers::getImaginary()
{
return imaginary;
}
//default constructor
ComplexNumbers::ComplexNumbers()
{
real=0;
imaginary=0;
}
//single parameterized constructor
ComplexNumbers:: ComplexNumbers(int z)
{
real=z;
}
//two parameterized constructor
ComplexNumbers::ComplexNumbers(int r,int u)
{
real = r;
imaginary= u;
}
void ComplexNumbers::input ()
{
int aReal(0),bImaginary(0);
cout<<\"enter real no\"<>aReal;
cout<<\"enter imaginary\"<>bImaginary;
real = aReal;
imaginary = bImaginary;
}
/* int real,imaginary;
cout<<\"enter real no\"<>real;
cout<<\"enter imaginary\"<>imaginary;*/
void ComplexNumbers::output()
{
cout<<\"the complex no is\"<
Solution
Your program is well-written. Apart from overloading the binary and unary operators. I have
modified your code and changed \"int\" to \"double\" in Constructors, and some methods.
FYI. Accessors are nothing but \"get\" methods in a class
Mutators are overloaded operators +,-,*. Means these operators are mutated over your
\"ComplexNumbers\" class.
The following is the modified program:
#include
using namespace std;
class ComplexNumbers
{
private:
double real;
double imaginary;
public:
void input ();
void output();
void setReal ( double );
double getReal ();
void setImaginary ( double );
double getImaginary();
ComplexNumbers();
ComplexNumbers(double);
ComplexNumbers(double,double);
bool operator == (ComplexNumbers&);
ComplexNumbers operator + (ComplexNumbers&);
ComplexNumbers operator - (ComplexNumbers&);
ComplexNumbers operator * (ComplexNumbers&);
ComplexNumbers operator - ();//unary minus
};
void ComplexNumbers::setReal(double x)
{
real=x;
}
void ComplexNumbers::setImaginary(double y)
{
imaginary=y;
}
double ComplexNumbers::getReal()
{
return real;
}
double ComplexNumbers::getImaginary()
{
return imaginary;
}
//default constructor
ComplexNumbers::ComplexNumbers()
{
real=0;
imaginary=0;
}
//single parameterized constructor
ComplexNumbers:: ComplexNumbers(double z)
{
real=z;
}
//two parameterized constructor
ComplexNumbers::ComplexNumbers(double r,double u)
{
real = r;
imaginary= u;
}
void ComplexNumbers::input ()
{
int aReal(0),bImaginary(0);
cout<<\.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
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True or False:
A budget variance is the difference between expected results and actual results at different levels
of operations?
Solution
Budget Variance: Budget Variance is calculated the difference between actual and expected
results. When the actual Revenue is greater than budgeted revenue than the variance is
favourabel and vice - Versa when the budgeted rvenue is more than the variance is unfavorable.
So Variance is the difference between the actual amount and budgeted amount.
So, Answer = True..
This level of measurement includes continuous measures A. Nominal; .pdf
This level of measurement includes continuous measures: A. Nominal; Interval B. Ordinal; Ratio
C. Interval; Nominal D. Interval; Ratio
Solution
Nomidal data type are group of people , name age etc. Ordinal data works for ranking orders.
They both deal with discrete measures.
the definition of interval includes the difference between quantityor event. it has arbitraty zero
point. The highest level of measurement is ratio level.It is comparative measurement, the
quotient of one quantity devided by another. It has an abosolute zero point. So, interval and ratio
levels deal with continuous measure.
Answer is D.
Throughout the middle ages European philosophers believed in…a sci.pdf
Throughout the middle ages European philosophers believed in…
a scientific and mythical view of the world.
an Aristotelian and Christian view of the world.
a Platonic and Christian view of the world.
a scientific and Christian view of the world.
a Platonic and mythical view of the world.A.
a scientific and mythical view of the world.B.
an Aristotelian and Christian view of the world.C.
a Platonic and Christian view of the world.D.
a scientific and Christian view of the world.E.
a Platonic and mythical view of the world.
Solution
Throughout the middle ages European philosophers believed in a Platonic and mythical view of
the world.E.
Throughout the middle ages European philosophers believed in a Platonic and mythical view of
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The term intangible assets is used in accounting to denotea. suc.pdf
The term \"intangible assets\" is used in accounting to denotea. such items as patents, copyrights,
and claims against customers which can be valued on a monetary basis.b. assets with lesser
economic significance because of the nature of such assets.c. properties without physical
characteristics that have long-term effects on a business enterprise.d. current or noncurrent
property items without physical characteristics.
Solution
As the intanbile asset has no physical characteristic but it can be used by the company for long
terms and have a monetary value
So correct answer is C as the properties without physical characteristics that have long effect on
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The methyl-accepting chemotaxis proteins of bacteriaa) integrate m.pdf
The methyl-accepting chemotaxis proteins of bacteria
a) integrate multiple signals through a two component phospho-relay system.
b) can only sense conditions that are favorable for bacterial growth.
c) are directly connected to the flagellar motor to guide the bacterium.
d) are sensor kinases tell the bacterium which way to go.
Solution
the methyl accepting chemotaxis protein of bacteria also called MCP THEY ARE SENSOR
KINASE TELL THE BACTERIA which way to go. the attractant for it is nutients and repeltant
are toxins. in E. coli Tar go towards aspartate and malate.
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so the all statement above are true..
The lagging strand is replicated with a series of Okazaki fragments .pdf
The lagging strand is replicated with a series of Okazaki fragments and that is why its synthesis
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The lagging strand is replicated with a series of Okazaki fragments and that is why its synthesis
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The files etcpasswd and etcshadow can be manually edited. Why wo.pdf
The files /etc/passwd and /etc/shadow can be manually edited. Why would you not want to do
this? What is the appropriate way to modify these files?
Solution
1. /etc/passwd is the file where the user information (Such as username, user ID, group ID,
location of home directory, login shell) is stored when a new user is created.
2. /etc/shadow is the file where important information (Such as encrypted form of the password
of a user, the day the password expires, whether or not the passwd has to be changed, the
minimum and maximum time between password changes, ...) is stored when a new user is
created.
So Both the file( passwd stores general user info and shadow stores user passwd info) store
sensitive data and information , if you manually modify and If you make a syntax error, you may
not be able to log in or become root anymore.
To modify these file use vipw and vigr command which will edit the files /etc/passwd and
/etc/group respectively.
The dividends received deduction Must exceed the applicable percentag.pdf
The dividends received deduction Must exceed the applicable percentage of the recipient
shareholder\'s taxable income Is affected by a requirement that the investor corporation must
own the investee\'s stock for a specified minimum holding period Is unaffected by the percentage
of the investee\'s stock owned by the investor corporation May be claimed by S corporations
Solution
The dividends received deduction
Is affected by a requirement that the investor corporation must own the investee\'s stock for a
specified minimum holding period.
Ans) B
Clarification:
DEDUCTIONS THAT CANNOT BE CLAIMED by S Corp,
dividends received deduction (allowed to C corporation).
us production deduction.
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Solution
The Chordae tendinae also known as heart string are chord like tendons that connects the
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Read the case Kenny An Effective Supervisor and in a 2-3 page p.pdf
Read the case \"Kenny: An Effective Supervisor\" and in a 2-3 page paper answer the following
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KENNY: AN EFFECTIVE SUPERVISOR
The most effective supervisor encountered by one of the authors of this textbook was named
Kenny, and he was maintenance supervisor in a chemical plant of an international corporation.*
The author was called in as a consultant because the plant was suffering from the results of the
ineffective, autocratic leadership of a former plant manager. Such leadership at the top adversely
affected all levels, resulting in low morale and losses from plant operations. In gathering data
about the plant through interviews, questionnaires, and observations, the consultant discovered
one main- tenance crew, unlike the rest of the departments in the plant, had very high morale and
productivity. Kenny was its supervisor. In the interview with Kenny, the consultant discov- ered
Kenny was a young man in his early thirties who had a two-year associate’s degree from a
community college. The consultant was impressed with his positive attitude, especially in view
of the overall low plant morale and productivity. Kenny said the plant was one of the finest
places he’d ever worked and the main- tenance people had more know-how than any other group
with which he had been associated. Kenny’s per- ception of his crew was they did twice as much
work as other crews, everyone worked together, and participa- tive management did work with
them. The consultant was curious about why pressure and criticism from the old, autocratic
manager seemed not to have had any effect on Kenny’s crew. The crew gave the consultant the
answer. They explained Kenny had the ability to act as a buffer between upper manage- ment
and the crew. He would get higher management’s primary objectives and points across without
upsetting his people. As one crew member described it: The maintenance supervisors will come
back from a “donkey barbecue” session with higher management where they are raising hell
about shoddy work, tak- ing too long at coffee breaks, etc. Other supervisors are shook up for a
week and give their staff hell. But Kenny is cool, calm, and collected. He will call us together
and report that nine items were discussed at the meeting, including shoddy work, but that doesn’t
apply to our crew. Then he will cover the two or three items that are relevant to our getting the
job done. Unfortunately, Kenny did have a real concern at the time of the consultant’s interview.
He was being trans- ferred from the highest-producing crew to the lowest producing one. In fact,
the latter was known as the “Hell’s Angels” crew. The crew members were a rene- gade group
who constantly fought with production people as well as with one another. The previous super-
visor had been terminated because he could not cope with them. After Kenny was assigned to the
new crew, he had to make a decision on the leadership strategies he would use in dealing with
them. His initial diagnosis.
Q4. How are automated tools used in the maintenance of information .pdf
Q4. How are automated tools used in the maintenance of information systems?
Solution
automated tools are used in the maintenance of information system in following ways
1)they are used in desinging of documentation
2)they can implement the change in documentation
3)they can regenerate the code by using code generator
4)they will automatically update the documentation during maintenance
5)they are used in reverse engineering
6)they are also used in reengineering.
PLEASE PLEASE use your OWN words, DO NOT copy and paste please expla.pdf
PLEASE PLEASE use your OWN words, DO NOT copy and paste please explain it with details
Explain how relational and object-oriented database management systems are used to construct
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Solution
The Relational Model
The Object-Oriented Model.
Nuclear Tree Focus on the relationships (monophyly, sister groups, .pdf
Nuclear Tree: Focus on the relationships (monophyly, sister groups, etc…). Address things that
look different group1 outg oatgroup2 0.0 A Fa T pol s all Shala u-ba. H our seya Stri SLxig
S.coy Snit
Solution
The phylogenetic tree is composed of tips, internal nodes and edges.
The tips (leaves) of the phylogenetic tree represent current day species, groups of descendant
taxa.
Two descendants that split from the same node are called sister groups, each one is the closest
relative of the other, they have a common ancestor that is unique to them, they have a lot of
evolutionary history in common and very little evolutionary history that is unique to each one.
In the given tree, so of the sis groups branches are not complete, or of different length, if they
have a difference in length then they are not sister groups.
Phylograms are phylogenetic trees in which the length of the edges represent evolutionary time;
therefore clades in phylograms represent true evolutionary history. in the given tree it ssems it is
a phylogram. but 0.02 represents length of branch that represents amount of genetic change. but
is not given for other branches.
Two differenet length of branches from sister groups are seen , which is unlikely to happen.
One branch shows a straight vertical line , and two groups are formed, this is not defined in thr
tree.
Each clade different branch length so can not be defined clearly.
Branch length of two outgroups are looks unusual..
Marine flowering plants produce. a sporophyte a gametophyte tetr.pdf
Marine flowering plants produce. a sporophyte a gametophyte tetra spores seeds
carpospores. Seed-producing plants first evolved. on land in freshwater in the ocean in
estuaries from red algae ancestors. Flowering plants that are salt tolerant are called algae.
conifers. halophytes. sporophytes. haptophytes. An example of a hydrophyte flowering plant
that can withstand and thrive under conditions of continued submergence is: cordgrass. pickle
weed. salt grass. turtle grass. blue grass. The horizontal stems of seagrasses are called:
rhizoids. rhizomes. stipes. blades. holdfasts. Which of the following is false about marine
plants? They are common in protected bays. They have xylem and phloem tissues. They are
found only in tropical areas. They produce seeds.
Solution
1)d)seeds because marine flowering plants produce seeds.
2)a)on land as the plants with seeds evolved on land.
3)d)halophytes as they are salt tolerant
4)d)turtle grass.
How important is Logistics Support Analysis to the ILS process Brie.pdf
How important is Logistics Support Analysis to the ILS process? Briefly critique.
Solution
ILS is an integrated and iterative process for developing material and a support strategy that
optimizes functional support, leverages existing resources, and guides the system engineering
process to quantify and lower life cycle cost and decrease the logistics footprint
readiness objectives, design for discard, maintenance task distribution, support equipment
,manpower and personnel requirements.
how to play .wav files in a java program on multiple devices!!! For .pdf
how to play .wav files in a java program on multiple devices!!! For instance when I send my
source files to my brother they need to be able to play on his computer as well as mine?
Solution
package audioplay;
import javax.sound.sampled.UnsupportedAudioFileException;
import java.io.IOException;
import javax.sound.sampled.DataLine;
import javax.sound.sampled.LineUnavailableException;
import javax.sound.sampled.SourceDataLine;
import java.io.File;
import javax.sound.sampled.AudioInputStream;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.AudioSystem;
public class AudioPlay {
// size required to read the audio stream
private static final int MAX_FILE_SIZE = 4096;
//Function to play audio file
void playMusic(String Path) {
File wavFile = new File(Path);
try {
AudioInputStream audioStream = AudioSystem.getAudioInputStream(wavFile);
AudioFormat format = audioStream.getFormat();
DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);
SourceDataLine al = (SourceDataLine) AudioSystem.getLine(info);
al.open(format);
al.start();
System.out.println(\"Staring the Music....!!\");
byte[] bytesBuffer = new byte[MAX_FILE_SIZE];
int bytesRead = -1;
while ((bytesRead = audioStream.read(bytesBuffer)) != -1) {
al.write(bytesBuffer, 0, bytesRead);
}
al.drain();
al.close();
audioStream.close();
System.out.println(\"Music stopped.!!\");
} catch (UnsupportedAudioFileException ex) {
System.out.println(\"The audio file not supported.Please try again with another audio file...!!!\");
ex.printStackTrace();
} catch (LineUnavailableException ex) {
System.out.println(\"Error Occurred\");
ex.printStackTrace();
} catch (IOException ex) {
System.out.println(\"Error..File may b corrupted.\");
ex.printStackTrace();
}
}
public static void main(String[] args) {
String audioFilePath = \"C:\\\\Users\\\\ttechnocraft\\\\Music\\\\adix.wav\";
AudioPlay player = new AudioPlay();
player.playMusic(audioFilePath);
}
}.
Identify the major provisions of the Affordable Care Act that will h.pdf
Identify the major provisions of the Affordable Care Act that will have an impact on individuals
and families.
Solution
The Patient Protection and Affordable Care Act ensures the access to quality, affordable health
care to all and create the transformation within the health care system necessary to contain costs.
It is designed to increase number of insured Americans, make healthcare & health insurance
affordable & improve performance of healthcare system in US.
Provisions of affordable care act:
1) Universal & continues access to health insurance: This act wants to accomplish the goal of
universal & continues access to healthcare & health insurance service to maximum US citizens.
Here there are two provisions
a) Insurance exchanges: The act wants to create online market place or health insurance
exchanges which will be helpful in reducing the prices of insurance plan & increasing
competition among service providers.
b) Expansion of coverage: The act wants to expand Medicaid coverage so that maximum
services like pediatric & dental care are also included in the plan.
2) Health insurance affordability: The law has specific subsidies for health insurance through tax
credits for lower income groups.
3) Hold insurance companies accountable (program integrity): The act includes numerous
provisions to terminate Medicaid that have been terminated in other programs. Suspending
payments for pending investigations or credible allegations of fraud & preventing inappropriate
claims.
4) Provider payments: States will receive 100% federal matching funds for the increase in
payments.
5) Children health insurance program: The act has provisions to extend funding for Children\'s
Health Insurance Program (CHIP).
Above provisions of this act will have an impact on individuals and families lives in a positive &
effective manner..
I have a simple TCP clientserver socket set up in Python. How can I.pdf
I have a simple TCP client/server socket set up in Python. How can I send specific sized
messages from the client to the server? Specifically 1, 100, 200, 400, 800 and 1000 bytes, and
also 1K, 2K, 4K, 8K, 16K and 32K bytes
I am using Python 2.7
Solution
You will use TCP to send and receive messages of size 1, 100, 200,
400, 800 and 1000 bytes. To measure throughput, you will use TCP to send and receive
messages of size 1K, 2K, 4K, 8K, 16K and 32K bytes. Note the difference in units for the
two sets of message sizes. For each measurement and for each message size, the client will
send at least ten probe messages to the server, which will echo back the messages.
Protocol phases: As in Part I, the client first needs to set up a TCP connection to
the server using the socket interface. The echo application will be extended, however, by
specifying the exact protocol interactions between the client and the server. This entails
specifying the exact message formats, as well as the different communication phases, as
outlined next.
1. Connection Setup Phase (CSP). This is the first phase in the protocol where the
client informs the server that it wants to conduct active network measurements in
order to compute the RTT and throughput of its path to the server. We will outline
the expected behavior from both the client and the server next.
CSP: Client. After setting up a TCP connection to the server, the client must send
a single message to the server having the following format:
\
• PROTOCOL PHASE: The protocol phase during the initial setup will be denoted
by the lower case character ’s’. This allows the server to differentiate between
the different protocol phases that the client can be operating at, as we will see.
• M-TYPE (MEASUREMENT TYPE): Allows the client to specify whether it
wants to compute the RTT, denoted by “rtt”, or the throughput, denoted by
“tput”.
• PROBES: Allows the client to specify the number of measurement probes that
the server should expect to receive. Once all the probe messages have been
echoed back and a sample measurement is taken for each one, the client should
compute an estimate of the mean (average) RTT or mean throughput, depending
on the type of measurement being performed. A detailed description of the probe
message’s format is provided in the description of the Measurement Phase.
MSG SIZE (MESSAGE SIZE): Specifies the number of bytes in the probe’s
payload.
• SERVER DELAY: Specifies the amount of time that the server should wait before
echoing the message back to the client. The default value should be 0. You
will vary this value later to emulate paths with longer propagation delays. Even
though increasing the server delay merely increases the processing time at the
server, it nevertheless causes the feedback delay, observed by the sender, to increase,
which has an effect somewhat similar to increasing the path’s propagation
delay.
• WS: A single white space to separate the different fields in the message. The
white .
Find the effective rate corresponding to the given nominal rate. (Ro.pdf
Find the effective rate corresponding to the given nominal rate. (Round your answer to the
nearest hundredth of a percentage point.)
7%/year, compounded monthly ?
Solution
I(t) = (1+(r/n))^(nt) Where I = annual rate, r = nominal rate, n = number of times
compounded per year, t = years I(t) = (1+ (.07/12))^(12*1)= 1.07229 Effective rate = 7.23%.
C++ AssignmentPlease read all the requirements and the overloading.pdf
C++ Assignment
Please read all the requirements and the overloading requiremets.
Here is what i have so far,
#include
using namespace std;
class ComplexNumbers
{
private:
double real;
double imaginary;
public:
void input ();
void output();
void setReal ( int );
int getReal ();
void setImaginary ( int );
int getImaginary();
ComplexNumbers();
ComplexNumbers(int);
ComplexNumbers(int, int);
friend bool operator == (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator + (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator - (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator * (ComplexNumbers , ComplexNumbers);
friend ComplexNumbers operator - (ComplexNumbers);
};
void ComplexNumbers::setReal(int x)
{
real=x;
}
void ComplexNumbers::setImaginary(int y)
{
imaginary=y;
}
int ComplexNumbers::getReal()
{
return real;
}
int ComplexNumbers::getImaginary()
{
return imaginary;
}
//default constructor
ComplexNumbers::ComplexNumbers()
{
real=0;
imaginary=0;
}
//single parameterized constructor
ComplexNumbers:: ComplexNumbers(int z)
{
real=z;
}
//two parameterized constructor
ComplexNumbers::ComplexNumbers(int r,int u)
{
real = r;
imaginary= u;
}
void ComplexNumbers::input ()
{
int aReal(0),bImaginary(0);
cout<<\"enter real no\"<>aReal;
cout<<\"enter imaginary\"<>bImaginary;
real = aReal;
imaginary = bImaginary;
}
/* int real,imaginary;
cout<<\"enter real no\"<>real;
cout<<\"enter imaginary\"<>imaginary;*/
void ComplexNumbers::output()
{
cout<<\"the complex no is\"<
Solution
Your program is well-written. Apart from overloading the binary and unary operators. I have
modified your code and changed \"int\" to \"double\" in Constructors, and some methods.
FYI. Accessors are nothing but \"get\" methods in a class
Mutators are overloaded operators +,-,*. Means these operators are mutated over your
\"ComplexNumbers\" class.
The following is the modified program:
#include
using namespace std;
class ComplexNumbers
{
private:
double real;
double imaginary;
public:
void input ();
void output();
void setReal ( double );
double getReal ();
void setImaginary ( double );
double getImaginary();
ComplexNumbers();
ComplexNumbers(double);
ComplexNumbers(double,double);
bool operator == (ComplexNumbers&);
ComplexNumbers operator + (ComplexNumbers&);
ComplexNumbers operator - (ComplexNumbers&);
ComplexNumbers operator * (ComplexNumbers&);
ComplexNumbers operator - ();//unary minus
};
void ComplexNumbers::setReal(double x)
{
real=x;
}
void ComplexNumbers::setImaginary(double y)
{
imaginary=y;
}
double ComplexNumbers::getReal()
{
return real;
}
double ComplexNumbers::getImaginary()
{
return imaginary;
}
//default constructor
ComplexNumbers::ComplexNumbers()
{
real=0;
imaginary=0;
}
//single parameterized constructor
ComplexNumbers:: ComplexNumbers(double z)
{
real=z;
}
//two parameterized constructor
ComplexNumbers::ComplexNumbers(double r,double u)
{
real = r;
imaginary= u;
}
void ComplexNumbers::input ()
{
int aReal(0),bImaginary(0);
cout<<\.
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