1 Homework #5. (Due Nov 30) Name .docx

1
Homework #5. (Due: Nov 30)
Name: (ID: )
* You must hand-write your solutions.
* You can add assumptions to solve the questions conveniently
(explicitly write your assumptions).
1. Florida Corp. currently has an EPS of $4.04, and the
benchmark PE for the company is 21 (the benchmark
PE can be the PE ratio of comparable companies). Earnings are
expected to grow at 5.5% per year.
a) What is your estimate of the current stock price?
b) What is the target stock price (i.e., forecasted stock price) in
one year?

c) Assuming the company pays no dividend, what is the implied
return on the company’s stock over the
next year? What does this tell you about the implicit stock
return using PE valuation?
2
2. Most corporations pay quarterly dividends on their common
stock rather than annual dividends. Barring
any unusual circumstances during the year, the board raises,
lowers, or maintains the current dividend

once a year and then pays this dividend out in equal quarterly
installments to its shareholders.
a) Suppose a company currently pays an annual dividend of
$3.60 on its common stock in a single annual
installment, and management plans on raising this dividend by
3.8% per year indefinitely. If the required
return on this stock is 10.5%, what is the current share price?
b) Now suppose the company in (a) actually pays its annual
dividend in equal quarterly installments; thus,
the company has just paid a dividend of $ 0.90 per share, as it
has for the previous three quarters. What
is your value for the current share price now? (Hint: find the
equivalent annual end-of-year dividend for
each year.) Comment on whether you think this model of stock
valuation appropriate. Assume that the
quarterly dividends are reinvested at the required return.

3
3. Panther Corp. has a bond issue with a face value of $1,000
that is coming due in one year. The value of
Panther’s assets is currently $1,090. Rick Grimes, the CEO,
believes that the assets in the firm will be worth
either $920 or $1,380 in a year. T-bill rate is 4.8%.
a) What is the value of Panther’s equity? The value of the debt?
Use no arbitrage concept. (Hint: you can
consider equity as call option.)
b) Suppose Panther can reconfigure its existing assets in such a
way that the value in a year will be $800
or $1,600. If the current value of assets is unchanged, will the
stock holders favor such a move? Why or

why not?
4. A call option with an exercise price of $45 and four months
to expiration has a price of $3.80. The stock
is currently priced at $42.75, and the risk free rate is 5% per
year. What is the price of a put option with
the same exercise price?
4
5. Devise a portfolio using only call options and shares of stock
with following value (payoff) at option
expiration date. You need to specify the strike prices of call
options that you employ.

6. You are considering purchasing a put option on a stock with a
current price of $33. The exercise price
is $35, and the price of the corresponding call option is $2.25.
According to the put-call parity theorem, if
the risk-free rate of interest is 4% and there are 90 days until
expiration, the value of the put should be
____________.

0
50
payoff
S(T) 60 50 110
5
7. Suppose that today is January 1 of the first year. Miami Corp.
expects that the EBIT in this year will be
$300. During the same period, depreciation costs will be $14
and amortization will be $6. Capital
expenditures are $60, and the planned increase in net working
capital is $30. The tax rate is 0.35. The
debts of Miami Corp. are $250. The weighted average cost of
capitals is 10%.
(a) What is the free cash flow to the firm (FCFF)?
(b) If the firm pays $20 as interest expense and there is no

change in the debts, what is the free cash flow
to the equity (FCFE)?
(c) Suppose that the FCFF of the firm will increase at 10%
during year 2 and 8% during year 3. After year
3, the growth rate will be 5%. What is the value of the equity?
Use FCFF for (c).
(d) Suppose that the FCFE of the firm increases at 7% per year.
What is the cost of equity of Miami Corp.?
© Dmitriy Shironosov/ShutterStock, Inc.

LABORATORY
■■
Study■how■networks■route■packets■to■various■destination■h
osts.
■■ Learn■how■networks■ensure■reliable■delivery.
■Software■needed:
■ 1)■ Apps■from■the■Lab■Manual■website:
a)■ TCP/IP■(TCP■IP.jar)
b)■ Network■Routing■(Network■Routing.jar)
19
OBJECTIVES
REFERENCES
Networking
© Jones & Bartlett Learning, LLC
NOT FOR SALE OR DISTRIBUTION
208 | Laboratory 19
BACKGROUND
This lab explores the basic concepts of networking, elaborating
on two specific aspects of
computer networking. Review these topics from your textbook,
lecture notes, or online
resources:
■ Computer networks, packet switching, routers, network
topologies: ring, star

■ Network protocols, TCP/IP, IP addresses
ACTIVITY
❯ Part 1
Though we tend to take networks for granted (except when
they’re down!), they are
surprisingly complicated. Setting one up involves much more
than just running a
wire between two computers. Not only must the machines agree
on a whole series of
protocols and identifying conventions before any
communication can happen, but the
wire itself is subject to seemingly malevolent forces working to
corrupt the fragile data
traveling through it!
The TCP/IP protocol suite is a whole system of software,
protocols, and management
decisions. At its heart are several protocols that break messages
into packets and send
them from source computers to destination computers. In this
activity, we will see a
simplified form of TCP (Transmission Control Protocol) as it
reliably sends a message to
a destination computer.
Start the TCP/IP app. This app simulates a reliable connection,
which means that
a big message is correctly sent in its entirety from source to
destination. If anything
goes wrong, the software makes heroic attempts to recover the
information. While it is
impossible to absolutely guarantee that the message will arrive
intact, the networking
software can make it very likely that it will be delivered

correctly.
In this app, there are only two hosts, or nodes, numbered 0 and
1. Host 0 is a
computer that is trying to send your message to host 1. You will
play the role of one of
those malevolent forces of nature, damaging and even
destroying data packets. Will the
software recover the data properly and save the day? Let’s hope
so!
Networking | 209
Click on the Example button, which inserts a message into the
Your message text
area. Click Run, and then Send a message. Host 0 moves the
message to the To be sent
area and proceeds to “packetize” it into 10-character packets.
Here you can see the first
10 characters—A very lon (the two spaces are part of the
count)—moving slowly in a
packet across the wire.
What exactly does the DAT 1 0 144 in the packet mean? This is
the packet’s header
and contains crucial information. First, DAT identifies the type
of packet being sent.
There are three types of packets: DAT for “data,” ACK for
“acknowledgment,” and NAK for
“negative acknowledgment.” We will talk more about these in a
minute.

Next in the header comes the destination address, which is 1.
Though this seems a
bit silly in a two-node network, bigger networks obviously
require a destination address.
The number 0, which appears next, is the sequence number. As
messages are broken into
packets, each packet is assigned a number: 0 in the first packet,
1 in the second packet,
and so forth. If any packet were to arrive out of order, the
destination computer could
examine these sequence numbers and reassemble the message in
the correct order. After
all, if part of the message were scrambled, the meaning could
change completely. (“You
owe us $1000” means something quite different than “You owe
us 000$1.”)
Finally, 144 is the checksum, which alerts the destination
computer that the packet
was damaged. This app uses an extremely simple checksum
algorithm, merely taking
the ASCII value of each character in the packet, adding these
together, and performing
modulo 256 (keep the remainder after dividing by 256) so that
the checksum is always
between 0 and 255, inclusive. This checksum algorithm was
actually used in some early
protocols, but it is prone to serious problems. Today more
advanced ones are used, most
notably the cyclic redundancy code. We’d tell you how it works,
but that would change
our lab into an advanced mathematics class!

210 | Laboratory 19
After the first packet arrives at host 1, it sends back an ACK
packet to acknowledge
that the data arrived successfully. The type of the packet is
ACK; the destination is 0
(host 0), and the sequence number that it sends back is 1,
meaning that it expects host
0 to send it a packet with the sequence number 1 next. Finally,
there is no real data in
an ACK packet, so the checksum is 0. This explains why the
header is ACK 0 1 0.
Networks are always contending with the forces of chaos that
can corrupt the data
being sent: noise on the line, equipment failures, and other sorts
of mayhem. So let’s
wreak havoc on our poor little network and see what happens.
Select Damage packets that
are touched from the pull-down menu. You can do this while the
app is sending packets.
As a packet comes sliding along the wire, click on it. This
should somehow corrupt the
data. For example, the s below turned into the Japanese yen
symbol!
What does host 1 do when this packet arrives? How does it even
know that the packet
is damaged? It recomputes the checksum using the data it
receives, and it matches the
checksum in the header with the recomputed one. They don’t
match this time, so host

1 sends host 0 a NAK packet with the same sequence number.
This tells host 0 that the
packet was corrupted and needs to be resent.
When host 0 gets the NAK, it resends the packet. Various status
messages appear in the
large text area at the bottom of the app, explaining what is
going on. © Jones & Bartlett Learning, LLC
Networking | 211
Now let’s experiment with more ways to make life difficult.
Select Delete packets
that are touched from the pull-down menu. When another packet
comes along, click on
it with your mouse and it should disappear.
Packets that do not even arrive pose a bigger challenge than
ones that are damaged. How
does host 1 ever know that a packet is missing? Actually, it
doesn’t, but host 0 is looking
for an ACK or NAK packet in response to each DAT packet that
it sends. So if one never
arrives, host 0 assumes the packet is lost, and it sends the
packet again.
The host won’t wait forever to receive the ACK or NAK packet.
It sets a timer on each
packet and if nothing arrives before the timer counts down to 0,
it times out and resends
the packet.

Of course, all of this assumes that everything, including the
hosts and the software, is
working correctly to catch any errors that may occur. As you
can imagine, there are about
a million things that can happen to foul up this pretty picture,
and we will explore some
of them in the exercise. But suppose the packets are temporarily
held up somehow. They
aren’t lost, but the timer goes off anyway. What does host 1 do
when it gets a duplicate
packet? And does it send another ACK? Yikes! This networking
is rapidly spinning out of
control! Scientists who designed the Internet had to think about
all these possibilities and
decide how to handle them. Fortunately, they found reliable
solutions so today we can sit
in the comfort of our homes and travel the entire world with our
fingertips.
212 | Laboratory 19
❯ Part 2
The second app simulates how computers route packets to their
final destinations.
Routing is enormously complicated because there are so many
variables. We will see
some of this unexpected complexity as we watch this app.
Start the Network Routing app. Let’s go for the most complex
example right away.
Select Example 5 from the pull-down menu.

Each blue disk represents a node, or a computer in the network.
The lines between
nodes are telecommunication lines, perhaps telephone wires or
Ethernet cables. Each
node has its own IP address, a 32-bit number that is always
represented as four octets
separated by dots.
In this app, the lines are “full duplex,” which means that
packets can flow in either
direction between the two connected hosts. This isn’t always the
case in the real world,
but many lines are full duplex.
You can move the nodes around on the screen by dragging them.
Their attached
lines move with them. You can even add new nodes. To make a
new node, double-click on
any open space and the app places a blue dot there with the
unusable address of 0.0.0.0.
You are expected to change that right away!
To change the characteristics of a node, double-click on the
node and a new
window appears:
Networking | 213
This window displays status information in the right column,
showing packets that this

node has received and some statistics. At the top are three fields
that you can change,
the first being the IP address of the node. You can also
customize the message you want
to send, and you can add the address of the recipient. If the
Destination node field is
blank, no messages will be sent by this node. More than one
node can send and receive
messages at the same time when this app runs.
The connection and routing tables below these fields are the
focus of this app’s study.
The connection table lists the IP addresses of nodes that directly
connect to this node.
Most networks assume, quite logically, that two nodes can find
each other and send
packets if they are directly connected. It is when nodes are not
directly connected to their
destinations that things get interesting!
The purpose of the routing table is to tell the node what to do
when it is not directly
connected to the destination node. For example, if node X wants
to send a packet to node
Y, which is not directly connected, it must send the packet to
node Z instead, and node Z
then assumes the responsibility of getting it to node Y.
Hopefully, Z will not mistakenly
send it back to X! If so, a routing loop would occur and the
packet would travel the wires
endlessly, like the Flying Dutchman. This prospect so worried
early developers of TCP/IP
that they put a timer on each packet so that it would just
evaporate after going through
too many nodes.

The routing table has two columns. The first column lists the
destination, and the
second lists the next hop. The next hop must be a directly
connected node. There are
also a number of special cases that can shorten routing tables. If
the word “DEFAULT”
appears in the routing table, then the node merely sends packets
to the first node in the
directly connected table. This works nicely for leaf nodes,
which connect to only one
other node, but won’t work for a complex topology, or the
configuration of the network.
Most of the time, nodes with two or more connections need to
explicitly list all
distant nodes, along with the next hop. However, this quickly
gets tedious and requires
too much work to update, as many new nodes are added. One
possible alternative is
to list some of the destination/next hop pairs, and leave the
others unspecified. The
Network Routing app lets you do this by putting an asterisk in
the destination column, as
shown below for the ring network (Example 3):
214 | Laboratory 19
Finally, what can we do if the network is under attack and the
routing table is now
obsolete due to out-of-commission routers? Early TCP/IP
developers worried about this,

too, and came up with the “hot potato algorithm.” This app
allows you to type “HOT
POTATO” in the routing table. Nothing else is needed. Hot
potato works this way: When
a packet comes in, the node makes a random guess as to which
directly connected node
will get the packet to its final destination, and it sends it to that
next hop. This may not
work and it may result in routing loops, but it just might do the
trick, too. Imagine
a bunch of people on the beach around a campfire passing
around a hot sweet potato
wrapped in aluminum foil. As they energetically toss it to one
another, there is a good
chance that it will end up in the hands (or lap) of someone who
wants to eat it!
There are many other routing algorithms that were designed
with various emergencies
or unusual conditions in mind. One such algorithm is called
“flooding,” in which every
packet is copied to every outgoing wire, not just to one random
one, as in hot potato.
What happens in the Internet, which is built on TCP/IP? What
routing algorithms
are used? The local area networks that attach to the Internet
often use specialized
routing algorithms. For example, Ethernet is like a telephone
where everyone can hear
everyone else’s calls. So when the phone rings, or a new packet
comes in, each computer
listens to the beginning of the packet to see if it is meant for
them.
The backbone network that manages the long-distance, high-

volume traffic on the
Internet uses several dynamic routing protocols. In such
systems, nodes change their
routing tables from time to time as they are forwarding packets.
Unlike the “Network
Routing” app, in which you, the network administrator, decide
once and for all what the
routing table for node X is, dynamic routing systems measure
the characteristics of the
traffic and the attached lines, and adjust their routing tables so
that packets will flow
most efficiently through the entire system. This sounds
daunting, and it is, kind of like
continuously adjusting the timing on all traffic lights so that
cars on the streets get to
their destinations in the least amount of time. While the system
isn’t perfect, it’s worked
on the Internet for over 40 years.
CS1150 Introduction to Computer Science
Lab #9 – Networking, 40 points
Objective:
To study how networks route packets to various destination
hosts. Learn how networks ensure reliable delivery.
Instructions:
Make sure to do the Activity as outlined in the LabCH19.pdf
file. The activity is actually a tutorial that will help you to
solve the problems in the exercises. It is expected that you will
complete the Activity before you begin the exercises.
When you are finished, save and rename the completed

document as “firstname_lastname_lab9.docx”. Then, submit
the saved document to Pilot.
EXERCISE 1
1) Start the “TCP/IP” applet.
2) Copy and paste the following text into the “Your Message”
box:
Computer networking is essential in our world today.
Note that this message is exactly 52 ASCII characters long,
counting spaces and the period at the end.
It should look like this:
Then, press the button “Send a message” and watch.
3) Count how many DAT and ACK packets were sent. (Don’t
rely on the “Status” box, actually count the packets yourself.)

Write your answers here:
DEFINE DAT AND GIVE NUMBER OF DAT PACKETS:
DEFINE ACK AND GIVE NUMBER OF ACK PACKETS:
4) Determine the steps for calculating how many characters
were sent during the entire exchange.
Assuming that:
· Every DAT or ACK packet has a packet header (example:
DAT 1 0 221)
· Besides the packet header, every DAT packet carries AT
MOST 10 characters of data.
· The last packet carried less than 10 characters of data. (52
data characters in the message, remember? 52 doesn’t divide
nicely into 10)
· ACK packets have no data, just a packet header.
· Every blank space, even in a packet header, counts as a
character.
· Assume that every packet header is 9 characters long
We calculated that exactly 171 characters were sent back and
forth in total during the entire process. How did we get that
number?
Show the mathematical steps to calculate the number 171 here,
below this line.:
5) Calculate the overhead, expressed as a percentage. Here is
the process:
· There are 52 characters of actual data in the message,

including blanks and punctuation.
· Subtract the number of characters of actual data (52
characters) from the total number of characters sent in both
directions in all the packets (from earlier in the lab). This gives
you the number of characters sent that were not actually data,
just part of the cost/overhead of sending the actual data.
· Divide the number of overhead characters by the total number
of characters to calculate the overhead expressed as a
percentage of the total.
Write the overhead percentage here, below this line:
6) Imagine you have a million-character message to send.
Calculate how many packets will be needed and how many
characters will be sent in total for the entire process to move the
message from node 0 to node 1:
HOW MANY PACKETS TOTAL:
HOW MANY CHARACTERS TOTAL:
7) What would be an obvious way to decrease the overhead
percentage? Why might this solution backfire? Under what
conditions would that occur?
OBVIOUS WAY TO DECREASE OVERHEAD PERCENTAGE:
HOW THAT COULD BACKFIRE:
UNDER WHAT CONDITIONS:

8) What is the purpose of IP address? What is the purpose of
Routers?
EXERCISE 2
1) Start or restart the “TCP/IP” applet again. Once again, copy
and paste the following text into the “Your Message” box:
Computer networking is essential in our world today.
2) This time, change the selection away from Leave packets
undamaged to Delete packets that are touched.

3) Delete some data packets by clicking on them as they move
along the wire, and watch the re-transmission after timeout.
4) What happens if you delete the re-transmitted packet? Does
the TCP/IP applet need to take any special action?
Write what happens here, below this line:
Is special action needed if re-transmitted packet is deleted?
Write your answer here, below this line:
5) What happens if you delete ACK or NAK packets?
EXERCISE 3
1) Start the “Network router” applet.
Select Example 3, the ring network.

2) Double-click on the node 37.61.25.46. List the nodes it is
directly connected to (the nodes in its connection table).
Write your list of nodes here, below this line:
3) If 37.61.25.46 wants to send packets to a node that is not
directly connected, to which node will it first send the packets?
(Check the routing table by double clicking on the node.)
Write the address of the next node here, below this line:
4) Run the applet for a while, letting it generate packets
continuously. Double-click on 37.61.25.46 again and look at its
statistics. How many packets were sent? Received? Forwarded?
PACKETS SENT:
PACKETS RECEIVED:
PACKETS FORWARDED:
5) Which nodes are sending packets out, and to whom?
NODES THAT ARE SENDING PACKETS:
NODES THAT ARE RECEIVING PACKETS:
6) Click on 138.92.6.17. Write down its statistics.
PACKETS SENT:
PACKETS RECEIVED:
PACKETS FORWARDED:

EXERCISE 4
1) Start the “Network router” applet. Select Example 4, the star
network.
2) Look at the routing and connection tables for the center node
and several other nodes. Describe any pattern you can see in
these tables.
Write your description here, below this line:
3) How is the connection table for the center node different
from the other nodes?
4) Select Generate when I click on a node from the pull-down
menu.
5) If you double-click on 159.121.2.13, you will see that its
destination node is 126.14.5.46. Run the applet, click on
159.121.2.13, and watch the packets go. What color does the
sending computer turn briefly? What color does the destination
computer turn? What does it mean if a node flashes green?
WHAT COLOR (SENDING COMPUTER):
WHAT COLOR (DESTINATION COMPUTER):

WHAT DOES IT MEAN IF A NODE FLASHES GREEN:
Try it again with 8.10.20.25 to confirm.
6) Many early computer networks used a star topology like this
example. What would happen if the center node in this type of
network dies?

Exercise 5
Create a new topology using 4 nodes.
1) Create 4 nodes with following IP addresses
Node 1 = 130.108.7.11
Node 2 = 130.108.7.22
Node 3 = 130.108.7.33
Node 4 = 130.108.7.44
2) Make Node 3 as a forwarding node and connect it with all the
other remaining nodes.
3) Let the sender receiver pair be as following:
Sender
Receiver
Node 1
Node 2
Node 2
Node 4
Node 4
Node 1
4) Your topology should look like as given below:
5) Paste your screenshot below this line:
Rubric (40 pts possible):

Exercise
Points
1
Question 1
0
Question 2
0
Question 3
1
Question 4
1
Question 5
2
Question 6
1
Question 7
1
Question 8
2
2
Question 1
0
Question 2
1
Question 3
1

Question 4
3
Question 5
3
3
Question 1
1
Question 2
1
Question 3
1
Question 4
1
Question 5
2
Question 6
2
4
Question 1
1
Question 2
1
Question 3
1
Question 4
1

Question 5
2
Question 6
2
5
Question 1
2
Question 2
2
Question 3
2
Question 5
2
Total
40
Lab_9_Applets/Network Routing.jar
META-INF/MANIFEST.MF
Manifest-Version: 1.0
Created-By: 1.6.0_38 (Sun Microsystems Inc.)
Main-Class: Network

Network$1.classsynchronizedclass Network$1 extends
java.awt.event.WindowAdapter {
void Network$1(Network);
public void windowClosing(java.awt.event.WindowEvent);
}
Network$2.classsynchronizedclass Network$2 extends Thread {
void Network$2(Network);
public void run();
}
Network.classpublicsynchronizedclass Network extends
java.awt.Frame implements java.awt.event.ActionListener,
java.awt.event.ComponentListener,
java.awt.event.MouseListener,
java.awt.event.MouseMotionListener,
java.awt.event.ItemListener {
java.awt.Button runB;
java.awt.Button stopB;
java.awt.Button helpB;
java.awt.Button newB;
java.awt.Button editB;
java.awt.Choice exampleCH;
java.awt.Choice methodCH;
java.awt.Color acolor;
Node[] nodes;
staticfinal int MAXNODES = 25;
boolean mustStop;
boolean running;
int x;
int y;
int buttonHeight;
int topnode;

int nextNodeNumber;
java.awt.Image buffer;
java.awt.Graphics gg;
int current;
boolean dragging;
boolean genContinuously;
publicstatic void main(String[]);
public void Network();
public void actionPerformed(java.awt.event.ActionEvent);
public void itemStateChanged(java.awt.event.ItemEvent);
public void runme();
public void stop();
public void
componentResized(java.awt.event.ComponentEvent);
public void
componentHidden(java.awt.event.ComponentEvent);
public void
componentMoved(java.awt.event.ComponentEvent);
public void
componentShown(java.awt.event.ComponentEvent);
public void mouseMoved(java.awt.event.MouseEvent);
public void mouseDragged(java.awt.event.MouseEvent);
public void mouseEntered(java.awt.event.MouseEvent);
public void mouseExited(java.awt.event.MouseEvent);
public void mousePressed(java.awt.event.MouseEvent);
public void mouseReleased(java.awt.event.MouseEvent);
public void mouseClicked(java.awt.event.MouseEvent);
public void doubleClick(java.awt.event.MouseEvent);
public void newNode(int, int);
public void grow();
public void singleClick(java.awt.event.MouseEvent);
public void paint(java.awt.Graphics);
public void update(java.awt.Graphics);
public void runaux();
private void delay(int);
private void help();

private void loadExample();
public Node find(String);
publicstatic int convert(String);
publicstatic String convert(int);
}
Node.classpublicsynchronizedclass Node {
String id;
Network parent;
public int x;
public int y;
String messageToSend;
StringList connectionTable;
StringList routingTable;
StringList packetsReceived;
StringList info;
String dest;
Queue outgoing;
java.awt.Color mycolor;
int numericalID;
boolean mustGenerate;
staticfinal int SLEEPTIME = 250;
int numpacketsSent;
int numpacketsRecd;
int numpacketsDropt;
int numpacketsForwarded;
staticfinal int diameter = 25;
public void Node(Network, String, int, int);
public void connect(String);
public void clearRoutes();
public void addRoute(String, String);
public boolean within(int, int);
publicstatic String getDest(String);
publicstatic String getSource(String);

publicstatic String getMessage(String);
public void send(String, String);
public void deliver(String);
public void route();
public String findNextHop(String);
public void generatePacket(String);
public float xgetRandom();
public int getRandom();
public int getRandom(int);
public String status();
}
NodeWindow$1.classsynchronizedclass NodeWindow$1 extends
void NodeWindow$1(NodeWindow);
}
NodeWindow.classpublicsynchronizedclass NodeWindow
extends java.awt.Frame implements
java.awt.event.ActionListener,
java.awt.event.ComponentListener,
java.awt.event.MouseListener, java.awt.event.ItemListener {
java.awt.Choice freqCH;
java.awt.TextArea conTA;
java.awt.TextArea routeTA;
java.awt.TextArea recvdTA;
java.awt.TextArea infoTA;
java.awt.TextField ipTF;
java.awt.TextField sendTF;
java.awt.TextField destTF;
java.awt.Label Label1;

java.awt.Label label5;
java.awt.Button okB;
java.awt.Button cancelB;
staticfinal int X = 0;
staticfinal int Y = 0;
Node node;
Network parent;
public void NodeWindow(Node, Network);
public void
public void
public void
public void
}
Popup$1.classsynchronizedclass Popup$1 extends
void Popup$1(Popup);

}
Popup.classsynchronizedclass Popup extends java.awt.Frame {
java.awt.TextArea ta;
public void Popup(String);
public void Popup(String, java.awt.Color);
public void Popup(String, int, int);
public void Popup(String, int, int, java.awt.Color);
public void Popup(String, int, int, int, int, java.awt.Color);
public void add(String);
public void setFont(java.awt.Font);
}
Queue.classpublicsynchronizedclass Queue implements
java.io.Serializable, Cloneable {
java.util.Vector theQueue;
public void Queue();
public boolean isEmpty();
public void enqueue(Object);
public Object dequeue();
public Object top();
public int length();
public Object clone();
}
StringList.classpublicsynchronizedclass StringList {
String[] contents;
int size;
public void StringList();
public void StringList(String, String);
public void StringList(String);

public int length();
public StringList shift(int);
public StringList copy();
public int find(String);
public String get(int);
public void put(String, int);
public String toString();
public String toString(char);
public void substitute(String[], String[]);
private void grow();
private void grow(int);
}
U.classpublicsynchronizedclass U {
public void U();
publicstatic int atoi(String);
publicstatic long atol(String);
publicstatic double atod(String);
publicstatic String[] copy(String[]);
publicstatic String[] tokenize(String);
publicstatic String[] tokenize(String, String);
publicstatic String detokenize(String[]);
publicstatic boolean equals(String[], String[]);
publicstatic void sleep(long);
publicstatic int power(int, int);
publicstatic String convert(int, int);
publicstatic boolean isint(String);
publicstatic String dec2bin(int);
publicstatic int bin2dec(String);
publicstatic String twoscomplement(String);
publicstatic String padout(String, char, int);
publicstatic String squish(String, char);
publicstatic java.awt.Color translateColor(String);

publicstatic String getField(String, int);
publicstatic String getField(String, int, char);
}
Lab_9_Applets/TCP IP.jar
META-INF/MANIFEST.MF
Manifest-Version: 1.0
Created-By: 1.6.0_38 (Sun Microsystems Inc.)
Main-Class: Tcpip
Input.classsynchronizedclass Input extends java.awt.Frame
implements java.awt.event.ActionListener {
boolean iamready;
java.awt.Button okB;
java.awt.Button cancelB;
public void Input(String);
public void Input(String, int, int, int, int, java.awt.Color);
public boolean ready();
public String get();
}
Node$1.classsynchronizedclass Node$1 extends Thread {
void Node$1(Node);
public void run();

}
Node$2.classsynchronizedclass Node$2 extends Thread {
void Node$2(Node);
public void run();
}
Node.classsynchronizedclass Node {
int nextSeqNum;
int nextAckExpected;
String messageToSend;
String remainingMessage;
String outgoingData;
int id;
Tcpip parent;
int x;
int y;
int diameter;
Packet currentPkt;
Packet inbox;
int direction;
int timer;
publicstaticfinal int TRANSITDELAY = 90;
Packet p;
boolean waitingAck;
public void Node(int, int, Tcpip, int, int, int);
public boolean inject(String);
public void deliver(Packet);
public void start();
private void processAck();
private void processData();
public void draw(java.awt.Graphics);
public void clear();
}

Packet.classsynchronizedclass Packet {
String data;
String type;
int dest;
int seq;
int chksum;
int x;
int y;
int width;
int height;
publicstaticfinal int XINCREMENT = 10;
Tcpip parent;
boolean delivered;
public void Packet(String, String, int, int, Tcpip, int, int);
public String toString();
public void advance(int);
public void draw(java.awt.Graphics);
publicstatic int computeChecksum(String);
public boolean within(int, int);
public void damage();
}
Popup$1.classsynchronizedclass Popup$1 extends
void Popup$1(Popup);
}
Popup.classsynchronizedclass Popup extends java.awt.Frame {
public void Popup(String);
public void Popup(String, java.awt.Color);

public void Popup(String, int, int);
public void Popup(String, int, int, java.awt.Color);
public void Popup(String, int, int, int, int, java.awt.Color,
String);
public void setFont(java.awt.Font);
}
Tcpip$1.classsynchronizedclass Tcpip$1 extends
void Tcpip$1(Tcpip);
}
Tcpip.classpublicsynchronizedclass Tcpip extends
java.awt.Frame implements java.awt.event.ActionListener,
java.awt.event.ComponentListener, java.awt.event.ItemListener,
java.awt.event.MouseListener {
java.awt.TextArea status;
java.awt.TextField inputTF;
java.awt.TextField tf1;
java.awt.TextField tf2;
java.awt.Label lab;
java.awt.Button runB;
java.awt.Button sendB;
java.awt.Button stopB;
java.awt.Button helpB;
java.awt.Button exampleB;
Node[] nodes;
boolean mustStop;
int x;
int y;
int buttonHeight;

java.awt.Choice damageCH;
boolean doDamage;
public void Tcpip();
public void run();
public void stop();
public void
public void
public void
public void
public void singleClick(java.awt.event.MouseEvent);
public void msg(String);
private void delay(int);
private void help();
private void loadExample();
}
U.classpublicsynchronizedclass U {
public void U();

publicstatic int atoi(String);
publicstatic long atol(String);
publicstatic double atod(String);
publicstatic String[] copy(String[]);
publicstatic String[] tokenize(String);
publicstatic String[] tokenize(String, String);
publicstatic String detokenize(String[]);
publicstatic boolean equals(String[], String[]);
publicstatic void sleep(long);
publicstatic int power(int, int);
publicstatic String convert(int, int);
publicstatic boolean isint(String);
publicstatic String dec2bin(int);
publicstatic int bin2dec(String);
publicstatic String twoscomplement(String);
publicstatic String padout(String, char, int);
publicstatic String squish(String, char);
publicstatic java.awt.Color translateColor(String);
}

1 Homework #5. (Due Nov 30) Name .docx

Recommended

Recommended

More Related Content

Similar to 1 Homework #5. (Due Nov 30) Name .docx

Similar to 1 Homework #5. (Due Nov 30) Name .docx (18)

More from oswald1horne84988

More from oswald1horne84988 (20)

Recently uploaded

Recently uploaded (20)

1 Homework #5. (Due Nov 30) Name .docx