C3 Week 2 Case Study

K 13. C 14. C 15. E 16. D 17.

Use the network diagram below and the additional information provided to answer the corresponding questions. [15 points]

keep track of its state.If the first node fails, the second node takes up the operations from where the

The CSC's first step consists on creating all the possible direct edges by employing the maximum available transmission power $P_{MAX}$. The algorithm for this step takes as input: the node set $V = {v_1, v_2, ..., v_n}$ with location information; the maximum transmission power $P_{MAX}$. As output, a direct graph $\overline{G}$ is produced. Figure \ref{top_nos} shows an input graph which consists of 500$\times$500 area with $n = 70$ nodes and no edges. Figure \ref{top_original} show the resulting graph when nodes create edges based on its maximum transmission power. In this examples, the maximum transmission power is $P_{MAX} = 4900$ and the maximum transmission range is $R_{MAX}$ = 70 meters. Note that the resulting graph is not necessarily connected.

● When the expected zone does excludes host S, a way from host S to host D must incorporate has outside the expected zone. In this manner, extra region must be incorporated

with the quantization step of ${\rm N} = \left[ {\frac{H}{{\Delta Y_k}}} \right]$. The nodes are divided into ${L}$ steps and we decide whether node $j$ is selected

Next, based on the classical structure/arrangement model Carr strongest section is his Confirmation. This section includes evident and arguments that support

Proposed algorithm consider three types of nodes every type have different initial energy level. normal nodes have E_0 energy. m advanced nodes have a times energy more than normal nodes with E_0 (1+a) energy level. m_0 super nodes have b times energy more than normal nodes with E_0 (1+b) energy level, where a and b are energy factors. As N is the number of total nodes in network, then for number of normal nodes, advanced nodes and super nodes N(1-m) , Nm〖(1-m〗_0) and Nmm_0 in the network, respectively.

16. Consider the i-node shown in Fig. 4-13. If it contains 10 direct addresses and these

The state sequence of MKB is change as N0 ⊢ N1 ⊢ ··· ⊢ Nc, where the initial N0 is

It traces the distance based group of users with same distance from the source .Now we use the probability function to form the probability based group.

Similarly to problem number one, problem number four returns to the use of a Venn diagram to solve the problem. However, this time there is an additional complement within the brackets. Let U = {a, b, c, d, e, f, g}, A= {a, b, c, f}, B= {c, d}, C= {a, d, e, f, g}, find (‘B∩C)∩A’. Step one, fill out a three circle Venn diagram including all sets, U, A, B, C. Next, find the complement of B, B’= {a, b, e, f, g}. Then, find the intersection of B’ and C, (‘B∩C) = {a, d, e, f, g}. Next, one must solve for A’, A’ = {e,g} Lastly, solve for pug in set values and solve, {a, d, e, f, g} ∩ {e, g} = {e. g}. Therefore, one can now resolve, (‘B∩C)∩A’ = {e, g}.

2) Which of the following models determines the path through the network that connects all the points?

c) The same as part b above, but now there are four index records at the lowest level of the tree index.

B. True/False. Encircle T if the statement is correct and F if it is incorrect.