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Embedded Advanced Sampling Environment (EASE). The philosophy of EASE, a sampling technology for networks, is that there is no need to deploy expensive RMON technology when low-cost EASE devices perform a similar function. The RMON standard states that every packet must be analyzed. This means that fast hardware must be deployed to cope with the ever-increasing network speed and traffic volume. The potential exists for this to be prohibitively expensive at higher transmission rates (100 Mbps and higher). Currently there are no RMON probes that can handle FDDI utilization above 40 percent because it s just too expensive to produce the hardware. The idea behind EASE is that looking at every packet is not necessary. Sampling already is used in many other areas, such as industrial quality control and financial auditing. By sampling only 2 percent of frames, the cost of the monitoring hardware that needs to operate only at 2 percent of the media speed is greatly reduced. Using echo test. None of the standards heretofore discussed actually allows a network manager to directly monitor the latency in the network. Latency is the amount of time it takes a source message sent on the network to reach its destination, and for the source to be notified of receipt. A simple echo test allows network managers to measure the latency in the network. Figure 29.12 is an example of an echo test between London and Boston. The data from the results can be graphed, revealing potential performance bottlenecks on a network. Echo tests can be configured to work with alarms to notify network staff automatically when latency exceeds acceptable limits.
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Breaking long lines in this fashion is often used to make programs more readable. It can also help prevent excessively long lines from wrapping.
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As explained, in C# all variables must be declared. Normally, a declaration includes the type of the variable, such as int or bool, followed by the name of the variable. However, beginning with C# 3.0, it became possible to let the compiler determine the type of a local variable based on the value used to initialize it. This is called an implicitly typed variable. An implicitly typed variable is declared using the keyword var, and it must be initialized. The compiler uses the type of the initializer to determine the type of the variable. Here is an example:
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2. Making and Using Graphs According to your predicted boiling point, in what state
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In this chapter we study power in circuits in more detail. In particular, our focus will be power in circuits with sinusoidal sources. We begin by considering maximum power transfer.
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I want to nd out what I really want to do rather than always focusing on helping others gure out what they want to do.
If we know the voltage in an inductor, we can nd the current by integrating. The result is i(t) = i(0) + 1 L
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In general, an object reference variable can refer only to objects of its type. There is, however, an important exception to C# s strict type enforcement. A reference variable of a base class can be assigned a reference to an object of any class derived from that base class. This is legal because an instance of a derived type encapsulates an instance of the base type. Thus, a base class reference can refer to it. Here is an example:
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The equation is D(PV) 5 P DV V DP The trick in solving this problem is knowing what value to use for the constant pressure or constant volume on the right side of the equation. Since both pressure and volume are changing, the appropriate value to use for the constant pressure and constant volume is the average value of the pressure or volume. Here we will make an assumption that the average value is equal to the midway point between the initial and final values. This assumption holds true under certain conditions, for example, if the pressure and volume change at a constant rate. If we write each of the changes (denoted with D, the Greek letter delta) in terms of the difference between the final values and the initial values, we have D(PV) 5 P2V2 2 P1V1 P DV 5 P(V2 2 V1) V DP 5 V(P2 2 P1) Now, to show Eq. (4-6) gives correct results, let s choose some example numbers. After reading this solution, you should try it yourself with your own numbers. Any numbers will do, but remember that, while the change in pressure or volume can be negative (meaning the pressure or volume decreased), each value of pressure or volume must be positive. For our example, let s say the pressure went from 10 pascals (N/m2) to 7 pascals, while the volume went from 13 m3 to 19 m3. Substituting these numbers into the left side of Eq. (4-6), we get D(PV) 5 P2V2 2 P1V1 5 (7 19) 2 (10 13) 5 133 2 130 5 3 Nm 5 3 J And substituting the same numbers into the right side, we get P DV 5 Pavg(V2 2 V1) 5 [(10 7)/2](19 2 13) 5 8.5(6) 5 51 J V DP 5 Vavg(P2 2 P1) 5 [(13 19)/2](7 2 10) 5 16(23) 5 248 J Then P DV V DP 5 51 J 2 48 J 5 3 J which is the same as D(PV).
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