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Using nonalphanumeric characters makes any password much more difficult to crack, for two reasons First, adding nonalphanumeric characters forces the hacker to consider many more possible characters than just letters and numbers Second, most password crackers use combinations of common words and numbers to try to hack a password Because nonalphanumeric characters don t fit into common words or numbers, including a character such as an exclamation point will defeat these common-word hacks Not all serving systems let you use characters such as @, $, %, or \, however, so you need to experiment to see if a particular server will accept them Once you ve forced your users to choose strong passwords, you should make them change passwords at regular intervals While this concept sounds good on paper, and for the CompTIA Network+ exam you should remember that regular password changing is a good idea, in the real world it is a hard policy to maintain For starters, users tend to forget passwords when they change a lot One way to remember passwords if your organization forces you to change them is to use a numbering system I worked at a company that required me to change my password at the beginning of each month, so I did something simple I took a root password let s say it was m3y3rs5 and simply added a number to the end representing the current month So when June rolled around, for example, I would change my password to m3y3rs56 It worked pretty well! No matter how well your password implementation goes, using passwords always creates administrative problems First, users forget passwords and someone (usually you) has to access their account and reset their passwords Second, users will write passwords down, giving hackers an easy way into the network if those bits of paper fall into the wrong hands If you ve got the cash, there are two alternatives to passwords: smart devices and biometrics Smart devices are credit cards, USB keys, or other small devices that you insert into your PC in lieu of entering a password They work extremely well and are incredibly difficult to bypass The downside is that they might be lost or stolen If you want to go seriously space-age, then biometrics are the way to go Biometric devices scan fingerprints, retinas, or even the sound of the user s voice to provide a foolproof replacement for both passwords and smart devices Biometrics have been around for quite a while, but were relegated to extremely high-security networks due to their high cost (thousand of dollars per device) That price has dropped substantially, making biometrics worthy of consideration for some networks NOTE You can categorize ways to authenticate into three broad areas: ownership factors, knowledge factors, and inherent factors An ownership factor is something the user has, like an ID card or security token A knowledge factor is something the user knows, like a password or personal identification number (PIN) An inherent factor is something that is part of the user, like a fingerprint or retinal pattern You can use one or more of these factors to authenticate a user In fact, two-factor authentication is a fairly standard practice in secure facilities, like some government offices Clearly, passwords are not the only way anymore.

3 2 2 2 Figure 21 A plot of Re( f ) = ( x + xy + x ) /(x + y ) = u( x , y) The spike at the origin is a

Long before Janelle ever saved the Word document on her system when the systems were first installed someone who knew what they were doing set up and configured all the systems at MHTechEd to be part of a common network All this setup activity resulted in multiple layers of hardware and software that can work together behind the scenes to get that Word document from Janelle s system to Tiffany s Let s examine the different pieces of the network, and then return to the process of Tiffany grabbing that Word document

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Clearly the network needs a physical channel through which it can move bits of data between systems Most networks use a cable like the one shown in Figure 2-5 This cable, known in the networking industry as unshielded twisted pair (UTP), usually contains four pairs of wires that transmit data

10 5 0y

The theoretical maximum utilization (percentage of capacity used) of a segment goes down as the value of a (the length of the medium in relation to the frame size) goes up How does this work The media must be clear before another frame can be transmitted If a is large, then the stations must wait for the frame to be sent the length of the medium before it is clear It is important to remember that when data is on the wire, utilization is 100% When no data is on the wire, utilization is 0% and cuts into the overall utilization of the medium If, for example a = 1, then the maximum utilization of the medium can be at best only 50% As soon as the start of the frame hits the end of the wire, the end of the frame is being transmitted All devices must wait for the end of the frame to leave the wire Since the length of the wire in bits is the length of the frame, the same amount of time used to transmit the frame is now spent waiting for the frame to propagate all the way down the wire Thus the wire is only used to transmit data half of the time, so utilization is at best only 50% This is the best-case scenario; there is no overhead, no collisions, no errors, no lost data, and another frame is there to be sent at the exact point the first frame leaves the wire The following table illustrates several cases of a 10 Mbps segment that uses 5000-bit frames As the length of the cable increases, a goes up and the theoretical maximum utilization starts to go down dramatically You can see why it is important to keep a as low as possible.

Another key piece of hardware the network uses is a special box-like device called a hub (Figure 2-6), often tucked away in a closet or an equipment room Each system on the network has its own cable that runs to the hub Think of the hub as being like one of those old-time telephone switchboards, where operators created connections between persons who called in wanting to reach other telephone users

3 2 2 2 Figure 22 A plot of Im( f ) = ( y + x y y) /(x + y ) = v ( x , y)

Length (m)