1. IP security mechanism can be not only used for security communica.pdf

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1. IP security mechanism can be not only used for security communications in Internet but also in local network or wide area network. a.True b.False 2. IP security is not transparent to the TCP applications because IP security is implemented in Network layer a.True b.False 3. The transport mode encapsulating security payload IPv4 can be only used as confidentiality purpose for the data. a.True b.False 4. In the following, which protocol(s) provide(s) IP security? a.TCP b.SSH c.AH d.ESP 5. A key idea that appears in both authentication and confidentiality mechanisms for IP is a.the association header. b.the IP header. c.the security association. d.the tunnel mode. Solution Answer-1: Option (A) True is correct. IP security mechanism used for security communications in Internet but also in local network or wide area network.IP security provides the capability to secure communications across a LAN, across private and public WANs, and across the Internet. Answer No-2 Obtion (B) False is correct. IP security is transparent to the TCP applications because IPSec is below the transport layer (TCP, UDP), so it is transparent to applications. Answer No-3 Option (B) False is correct. The transport mode encapsulating security payload IPv4 can not only used as confidentiality purpose for the data but also used for Authenticating Header. Note: The encapsulating security payload provides confidentiality service, including confidentiality of message contents and limited traffic flow confidentiality. And As an optional feature, ESP can also provide the same authentication services as Authentication Header. Answer-4 Option (C) AH is correct. The AH (authentication header) protocol provides support for data integrity and authentication of IP packets. Answer-5 Option (D) the tunnel mode is correct. Tunnel mode is the mode that appears both in authentication and confidentiality for IP i.e both AH(Authentication Header) and ESP( Encapsulating Security Payload) support two modes of use transport and tunnel mode. Tunnel mode encapsulates an entire IP packet within an IP packet to ensure that no part of the original packet is changed as it is moved through a network. Kindly Check and Verify Thanks...!!!.

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Note: The encapsulating security payload provides confidentiality service, including
confidentiality of message contents and limited traffic flow confidentiality. And As an optional
feature, ESP can also provide the same authentication services as Authentication Header.
Answer-4 Option (C) AH is correct.
The AH (authentication header) protocol provides support for data integrity and authentication of
IP packets.
Answer-5 Option (D) the tunnel mode is correct.
Tunnel mode is the mode that appears both in authentication and confidentiality for IP i.e both
AH(Authentication Header) and ESP( Encapsulating Security Payload) support two modes of
use transport and tunnel mode.
Tunnel mode encapsulates an entire IP packet within an IP packet to ensure that no part of the
original packet is changed as it is moved through a network.
Kindly Check and Verify Thanks...!!!

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IP Security One problem with Internet protocol (IP) is that it has no method for confirming the authenticity and security of data as it moves through the net. IP datagrams are typically routed between devices over disparate networks; as a result, information within these datagrams could be intercepted and altered. As use of the Internet for critical applications has increased, the need for enhancements to IP security became necessary. As a result, the Internet Engineering Task Force (IETF) created a set of protocols called IP Security, or IPsec, to support the secure exchange of packets over the Internet. IPsec is now a mandatory component of IPv6 and must be supported for any IPv6 implementation. IPsec is implemented in IPv6 using the authentication header (AH) and the encapsulating security payload (ESP) extension header. Answer the following questions in a 3- to 4-page, APA-formatted paper: 1 What is IPsec, and why is it necessary? How is IPsec used in VPN? 2 Which network layer currently suffers from attacks, and why? At which layers of the network stack architecture should a solution be attempted? Provide details. 3 How is IP security achieved? What is the basic authentication scheme? Which mechanisms are used? What are some of the application venues of IPsec? 4 How is a VPN implemented on a server so that its clients can connect to it? Remember to properly cite your sources according to APA guidelines. Solution IPSec IPsec also known as IP Security.Internet Protocol Security is a framework for a set of protocols that provide security for internet protocol. It can use cryptography to provide security. IPsec support network level data integrity, data confidentiality. As it is integrated at the internet layer (i.e. layer 3), it provides security for all the protocols in the TCP/IP. IPsec applied transparently to the applications, there is no need to configure separate security for each application the uses TCP/IP. IPsec provides security for IPsec provides two choices of security service: Authentication Header (AH), which essentially allows authentication of the sender of data, and Encapsulating Security Payload (ESP), which supports both authentication of the sender and encryption of data as well. The specific information associated with each of these services is inserted into the packet in a header that follows the IP packet header. Separate key protocols can be selected, such as the ISAKMP/Oakley protocol. IPsec is necessary for Earlier security approaches have inserted security at the Application layer of the communications model. IPsec is said to be especially useful for implementing virtual private networks and for remote user access through dial-up connection to private networks. A big advantage of IPsec is that security arrangements can be handled without requiring changes to individual user computers. Cisco has been a leader in proposing IPsec as a standard (or combination of standards and technologies) and has included support fo.

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1) The User Datagram Protocol (UDP) provides which of the following \"services\" to the user? A) Port addressing B) Duplication Protection C) Sequence Preservation D) Guaranteed Delivery 2) Which of the following is not a protocol architecture? A) OSI B) TCP/IP C) SNA D) All of the above 3) If a TCP/IP packet must first pass through a router on its way to the final destination, the IP address of the router will be found in: A) The user data B) The transport header C) The IP header D) The network header 4) The OSI Physical Layer specification includes the physical interface between devices, including all of the below except: A) Functions performed by individuals circuits of the physical interface B) Physical connector specs C) Error detection and control D) Representation of the bits (voltage, rate) E) Sequence of events in exchange 5) The OSI Data Link layer is concerned with reliability. Higher layers may assume error free transmission, provided systems are directly not connected. A) True B) False 6) T/F: Since, in the OSI model, each layer in the middle depends on lower layers for services and provides services to higher layers, a change in either the higher or lower layer wil require changes inside the layer in the middle? A) True B) False 7) True/False: TCP PDU is another name for a TCP segment A) True B) False Solution Please follow the data and description : 1) UDP : UDP provides two services not provided by the IP layer. It provides port numbers to help distinguish different user requests and, optionally, a checksum capability to verify that the data arrived intact. So the answer is OPTION A (Port addressing). 2) TCP/IP does not correspond to this model directly, as it either combines several OSI layers into a single layer, or does not use certain layers at all. So the answer is OPTION C (TCP/IP). 3) An IP header is header information at the beginning of an IP packet which contains information about IP version, source IP address, destination IP address, time-to-live, etc. So the answer is OPTION C (IP Header). 4) The specification Error detection and control is not included over the physical layer of the OSI reference model. So the answer is OPTION C (Error detection and control). 5) The Data Link Layer is primarily responsible for communications between adjacent network nodes. Network switches and hubs operate at this layer which may also correct errors generated in the Physical Layer. So the answer is OPTION A (True). 6) Yes, Since, in the OSI model, each layer in the middle depends on lower layers for services and provides services to higher layers, a change in either the higher or lower layer wil require changes inside the layer in the middle. So the answer is OPTION A (True). 7) The Layer 2 (Data Link Layer) PDU is the frame. The Layer 3 (Network Layer) PDU is the packet. The Layer 4 (Transport Layer) PDU is the segment for TCP or the datagram for UDP. So the answer is OPTION A (True). Hope this is helpful..

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Assignment is: \"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then \'paste\' in MS-Word)\" How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual Studios using the linked list below with what is being asked? Please need help Linked list : #include #include using namespace std; struct nodeType { int info; nodeType *link; }; void createList(nodeType*& first, nodeType*& last); void printList(nodeType*& first); void insertFront(nodeType*& first); void insertBack(nodeType*& last); void deleteFirst(nodeType*& first); void deleteLast(nodeType*& last, nodeType* first); int main() { nodeType *first, *last; int num; createList(first, last); int choice; while(true) { cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\ \"; cout<<\"Enter your choice: \"; cin>>choice; switch(choice) { case 1: insertFront(first); break; case 2: insertBack(last); break; case 3: deleteFirst(first); break; case 4: deleteLast(last, first); break; case 5: printList(first); break; case 6: return 0; default: cout<<\"Invalid menu option. Try again.\"<>number; while (number != -999) { newNode = new nodeType; // create new node newNode->info = number; newNode->link = NULL; if (first == NULL) { first = newNode; last = newNode; } else { last->link = newNode; last = newNode; } cout<<\"Enter an integer (-999 to stop): \"; cin>>number; } // end of while-loop } // end of build list function void deleteFirst(nodeType*& first) { nodeType *temp; temp= first; first= temp->link; delete temp; return; } void deleteLast(nodeType*& last, nodeType* current) { nodeType *temp; while(current->link != NULL) { temp=current; current=current->link; } temp=last; current->link=NULL; delete temp; last = current; return; } void insertFront(nodeType*& front) { int num; cout<<\"\ Enter the number to insert: \"; cin>>num; nodeType *newNode = new nodeType; newNode->info=num; newNode->link= front; front= newNode; return; } void insertBack(nodeType*& last) { int num; cout<<\"\ Enter the number to insert: \"; cin>>num; nodeType *newNode = new nodeType; newNode->info=num; newNode->link= NULL; last->link= newNode; last = newNode; return; } void printList(nodeType*& first) { cout<<\"Inside printList...printing linked list...\ \"<info << \" \"; current = current->link; } cout< #include using namespace std; struct nodeType { int info; nodeType *link; }; void createList(nodeType*& first, nodeType*& last); void printList(nodeType*& first); void insertFront(nodeType*& first); void insertBack(nodeType*& last); void deleteFirst(nodeType*& first); void deleteLast(nodeType*& last, nodeType* first); int main() { nodeType *firs.

Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf

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You maintain several virtual machines (VMs) in an offline state. How.pdf

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Which of the following statements is not TRUE?1)For a 64-bit computer, its memory address space is 264, not 642;2)A process is a program in execution;3)A child process has a different UID (User Identification) as its parent;4)DMA (Direct Memory Access) controls the flow of data between memory and some controller without constant CPU intervention. Solution Answer: 2 , 3 and 4 are true . 1 is false only process becomes program when it is been put from start state for processing. Every child has different UID with the parent. And we know that DMA controls the data flow..

Which of the following statements is not TRUE1)For a 64-bit compute.pdf

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You are running an ELISA on a sample to test for the presence of antigens from the influenza A virus. When it comes time to add the antibody, you realize that you only have antibodies for the influenza B virus. Would these antibodies give you accurate results? Why or why not? Solution Influenza virus is divided into three main types 1. Influenza A 2. Influenza B 3. Influenza C Influenza A virus is mostly observed in a wide variety of birds and mammalian species and Influenza B is mostly confined to humans. Influenza A and B have significant differences in antigens that are present in their membrane proteins. Antibodies are specific to antigens, as the antigens present on influenza A membrane are different from influenza B, the addition of influenza B antibodies instead of influenza A antibodies would result in inaccurate results..

You are running an ELISA on a sample to test for the presence of .pdf

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You are to write an efficient program that will read a dictionary of 100,000 words, and then check a document of 2,000,000 words. The program should insert into an array the position of each misspelled word. I have provided the driver program SpellRunner.cpp, and the necessary class stubs in speller.h, and speller.cpp. You may add any classes and/or code you wish to speller.cpp, and speller.h. You may also use, and alter any Weiss files. CreateDocs.out creates files used for testing. Doc-3-4-1 and Dict-3-4-1 are two testing files for small file. _____________________________________________________________________________ _________________ SpellRunner.cpp #include #include #include #include #include #include \"CPUTimer.h\" #include \"mynew.h\" #include \"speller.h\" // uncomment line below for DOS/Windows OS. // #define DOS #ifdef DOS #include #else #include #endif extern int maxRAM; extern int currentRAM; using namespace std; char* readDictionary(char *filename, char **dictionary, int dictSize) { int fileSize; char *s; struct stat statbuf; stat(filename, &statbuf); fileSize = statbuf.st_size; s = new char[fileSize + 1]; ifstream inf(filename); inf.read(s, fileSize); inf.close(); s[fileSize] = \'\\0\'; dictionary[0] = strtok(s, \"\ \"); for(int i = 1; i < dictSize; i++) dictionary[i] = strtok(NULL, \"\ \ \"); return s; } // readDictionary char* readDocument(char **document, int dictSize, int docSize, int seed) { char *s, filename[80]; int fileSize; struct stat statbuf; sprintf(filename, \"Doc-%d-%d-%d.txt\", dictSize, docSize, seed); stat(filename, &statbuf); fileSize = statbuf.st_size; s = new char[fileSize + 100]; ifstream inf(filename); inf.read(s, fileSize); inf.close(); s[fileSize] = \'\\0\'; document[0] = strtok(s, \"\ \"); for(int i = 1; i < docSize; i++) document[i] = strtok(NULL, \"\ \ \"); return s; } // readDocument() void readWrongs(int *wrongs, int dictSize, int docSize, int seed, int &wrongCount) { char filename[80]; wrongCount = 0; sprintf(filename, \"Wrongs-%d-%d-%d.txt\", dictSize, docSize, seed); ifstream inf(filename); while(inf >> wrongs[wrongCount++]); wrongCount--; } // readWrongs() void checkAnswers(int wrongs[], int wrongCount, int misspelled[], int misspelledCount) { for(int i = 0; i < wrongCount && i < misspelledCount; i++) if(wrongs[i] < misspelled[i]) { cout << \"Speller missed misspelled word # \" << wrongs[i] << endl; return; } else if(wrongs[i] > misspelled[i]) { cout << \"Speller thought correctly spelled word # \" << misspelled[i] << \" was wrong\ \"; return; } if(wrongCount != misspelledCount) cout << \"Speller found \" << misspelledCount << \" misspelled words when \" << \" there were really \" << wrongCount << \" misspelled words.\ \"; } // checkAnswers void cleanup(char **dictionary, char *docFilePtr, char **document, int *wrongs, int *misspelled) { delete [] dictionary; delete [] docFilePtr; delete [] document; delete [] wrongs; delete [] misspelled; } // cleanup() int main(int argc, char* argv[]) { char .

You are to write an efficient program that will read a dictionary of.pdf

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X = C B - B C D; Use Accumulator Register-Register (LoadSt.pdf

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