How long does the session between A and B last, using connection-orientedcommunication? Explain.?
150568
untill the termination phase\n\nThe connection termination phase uses a four-way handshake, with each side of the connection terminating independently. When an endpoint wishes to stop its half of the connection, it transmits a FIN packet, which the other end acknowledges with an ACK. Therefore, a typical teardown requires a pair of FIN and ACK segments from each TCP endpoint.\n\nOr\n\nA connection can be "half-open", in which case one side has terminated its end, but the other has not. The side which has terminated can no longer send any data into the connection, but the other side can.
166191
In general synchronous connection would be better suited for short distance, because of a more reliable connection. \nSynchronous is easier to use by software on each side.\n\nAsynchronous uses a fire and forget mechanism which targets less reliable transmissions and needs extra mechanisms to make sure transmitted data is consistent.
138858
[quote]\nCan I collaborate these 2 connections into one, and would that theoretically mean I could double the connection speed to 54Mbps x 2 = 108Mbps?\n[end quote]\n\nNo, i don't think it works that way. What i presume is that only one connection is active at a time since you have no control on what connection a program work. And btw, the speed of one wireless connection is far higher than what you get by your ISP.
573465
Go in for multisessions
483901
Thats a good question!\nI know you are doing Netwrok Technology CS2004N ;)\n well,\nYou should first assume the distance between rocket comples and the lecture room, say 150 meters.\nthe we have different kinds of delays to be calculated.\nThe first one is, transmission delay:\n total sie of data = * number of slides * 8(bits)\nthis would be: 10 * 1000 * 50 = 4MB\n\nNow we know the average data transmission rate, which is 1 Mbps.\n\nTransmission delay would be:\n4MB / 1 Mpbs = 4seconds\n\nNow propagation Delay would be:\nDistanse / Speed of Sound\nwhich would be 150/343 = 0.43 seconds\n\nProccessing Delay:\n50 * 10 * 1ms = 500 milli seconds = 0.5 seconds\n\nTotal Delay = 0.5 + 0.43 + 4 = 4.93 Seconds\nThis would give you the delay, BUT remeber this is using UDP, NOT TCP, if you want to have TCP you need 2 more delays, Ack Delay and Handshake Delay.(I think its ok to use UDP) If you want to use TCP ACK delay equals to propagation delay, and handshake delay is 2 * propagation delay.\nGood Luck \nMRB
8837
session\n\nIn communications, the active connection between a user and a computer or between two computers.\n\nUsing an application program (period between starting up and quitting).\n\nOne or more tracks of audio or data that were recorded at one time on a compact disc. See multisession.
828268
You're probably thinking of TCP-IP and wikipedia has an interesting article on its history, and comparisons between the two. http://en.wikipedia.org/wiki/TCPIP  \n(It's far better than me boring you with explanations of models and standards)
287213
The same way you transfer any kind of UDP packets; dial-up makes no difference.  Usually users are not concerned with the underlying transport mechanisms, TCP or UDP.  UDP packets may be transferred programatically using standard methods.\n\nWhat are you trying to accomplish?
321301
It depends on your ISP.  Some will just charge you more for having 2 at the same time.  Others will break the first connection when the second gets on, but most ISPs do something to discourage the practice because it's "freeloading."
32674
A standard form of communication between multiple objects.
963015
C:&#92;>ping yahoo.com  <--what you are pinging\n\nPinging yahoo.com [216.109.112.135] with 32 bytes of data: <--the address was found at 216.109.112.135, now let's start pinging\n\nReply from 216.109.112.135: bytes=32 time=17ms TTL=45  <--The website replied from that IP address.  You sent the site 32bytes of data to test with.  The reply took 17ms (milliseconds) to get back to you.  TTL isn't important here.\nReply from 216.109.112.135: bytes=32 time=18ms TTL=45\nReply from 216.109.112.135: bytes=32 time=19ms TTL=45\nReply from 216.109.112.135: bytes=32 time=18ms TTL=45\n\nPing statistics for 216.109.112.135:\n    Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),  <-- you sent 4 packets and all were answered successfully.\nApproximate round trip times in milli-seconds:\n    Minimum = 17ms, Maximum = 19ms, Average = 18ms  <---self explanatory.\n\nIf you want to ping a site "forever", use:  ping whatever.com -t\n\n...where whatever.com is the real site you want to ping.
45930
Follow our steps and you'll have your PCs talking to each other in no time.\n\n\nhttp://www.pcworld.com/howto/article/0,aid,102461,00.asp
424294
Depends on your internet connection.\n\nAnywhere from 45 mins on a 56k modem to seconds on a T1 connection.
533325
In the TCP/IP protocol suite, data is exchanged in the form of packets. Because of the way TCP/IP performs routing, the path taken by packets can change dynamically throughout a session. This means that if Host A and B are communicating and Host A sends two packets, they could potentially take different paths to Host B, and could arrive out of order. Host B must therefore have a way to re-assemble the packets (containing the data in octets) in correct order. \n\nTCP/IP provides a way to do this using sequence numbers. Starting from an initial sequence number (ISN) that each host provides as part of the session setup, each octet is assigned a sequence number that is one larger than that of its predecessor. For example, if Host B provides 123 as the initial sequence number for its session with Host A, Host A's first octet is numbered 123, the second is numbered 124, the third is numbered 125, etc. \n\nIf someone attempts to launch an attack against another computer and the attacker can predict how a host selects ISNs, it's possible to conduct two types of attacks, known as IP address spoofing and session hijacking. The more random the ISNs are, the more difficult it is to carry out these attacks. \n\nFor related information on this problem, please visit the following Microsoft Web site: \nhttp://www.microsoft.com/technet/security/bulletin/MS99-046.asp (http://www.microsoft.com/technet/security/bulletin/MS99-046.asp)\nFor additional security-related information about Microsoft products, please visit the following Microsoft Web site: \nhttp://www.microsoft.com/technet/security/
324995
if you are on dialup, your connection might be slow and u might be getting disconnected from time to time.
405325
I'll answer the question generally, since individual products may, as you suggest do a little looking around without telling you. You might want to watch the activity lights when you connect, and try to correlate them with the connection sequence. \n\nI'm assuming here that you have the commonest sort of arrangement, a router (connected to DSL, cable ? etc) and a laptop with a wireless client radio inside. \n\nEvery 'access point' (one type of radio operating mode) must broadcast a beacon signal to tell possible client radios in range something about itself. For 802.11 b/g (and probably n, but I haven't checked) the beacon is broadcast on the only channel that is legal everywhere, channel 10. It's also broadcast with the lowest possible legal speed. This means that any client 802.11 b/g radio in range will be able to hear and understand the beacon. \n\nOnce the client radio hears all the local beacons, it chooses one (usually based on signal strength) that it can interact with. So an 802.11b client radio, which knows nothing about 802.11g format signal types (ie, 'modulation schemes') won't be able to work with an 802.11g access point and both had better be willing to find the highest performance connection possible to both, in this case one of the 802.11b schemes, and not any of the 802.11g expanded schemes. \n\nThis takes a bit of time. but once both ends have discovered the highest performance configuration they can both support, the radios can actually tell the software (drives, operating system, ...) what's going one. At that point, Ip addresses have to be assigned to allow any traffic to and from the Net. If the Operating system is configured to use DHCP, it must contact the DHCP server it's been told about, and 'lease' an IP address from the block of addresses the  ISP has had assigned to it. Next time on the ISP, the IP address assigned might be entirely different. Or the Internet software using the client radio might have been given a permanent IP address (called 'static' address assignment in contrast to the 'dynamic' assignment under DHCP). \n\nAll of this takes time. Which further depends on the connection scheme (speed, modulation, etc) the radios have agreed on. So a marginal 802.11b connection (distant location, poor antenna placement, bad polarization match, ...) might only be able to manage the barest minimum, 1Mbps. A really good 802.11g connection, could use the highest speed possible. which will of course never be as fast as the ads claim. there is a good deal of handshaking and  timeout settings, and error announcement sent back to the other radio when found. \n\nAnd there is a backoff_and_try_a_lower_speed_connection negotiation when there are more than x errors (varies by manufacturer default, but is sometimes changeable by the user for a particular installation in the case of some utilities and software drivers). And an occasional try_to_use_a_better_speed_ connection if we haven't had errors in the most recent y seconds. Either requires another negotiation sequence, which again takes time. \n\nAnd, if the range is marginal (almost too far away) there will not only be automatic up regulation attempts, and perhaps some down regulation attempts, but more a few entire reestablish_the_connection restarts as signal drops below the radio's sensitivity threshold (all real world equipment has them). \n\nAnd some operating systems take their own sweet time to recognize a new device (in Windows, you may have to run the add_new_hardware wizard in the control panel manually. Sometimes the whole add new hardware routine more than once to succeed.  \n\nFinally, on initial start up, most browsers depend not only on hardware initialization, connection negotiation, DHCP negotiation, and so on. But also must do a DNS lookup, and download the current versions of a 'home page' (which may be larded with images from many other sites). It can
378111
Don't know how you can do this, but you can get the same information from Task Manager/Networking tab.  Click on View and Select Columns and check Bytes Sent and Byte Received.  This only works for the time period that you keep Task Manager running.