I believe its "Thinking Universe"
The correct sequence is: 5,3,1,4,2
When you are installing software on your computer you need to first open the CD-ROM and then insert the installation CD into the CD-ROM. If the contents of the CD opens, click on the executable file to start the wizard, and then follow the prompts in the wizard. Finally, you often need to reboot your computer after installing a new program.
Answer:
a. 1 is a packet, 2 is data, 3 is a frame.
Explanation:
And what is not mentioned is segment which used TCP/UDP and is part of Transport layer. The packet carries the destination and sender IP address, and is part of the Network Layer. The frame has the Mac address of destination device and senders device and is part of data link layer.
Hence segment has no IP address, hence b. is not correct. Also, data cannot have the IP Address, and Frame has the MAC address, Hence, the above answer. And this arrangement is part of Data Encapsulation.
Also keep in mind data can be anything like a series of bits, or any and it can or not have a header.
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
