Pt2520 Unit 1 Problem Analysis Paper

III SYNTHESIS AND SIMULATIONS RESULTS The simulation and synthesis work is finally done by the xilinix and modelsim respectively. Figure 5:synthesis results of Fault FFT. The figures intimate the fault injected FFT,which is checked by the manual error injected via all diferent possibilities by using RTL scripting. Eventhough the soft error is added in the FFT the error detector code 100% detect the errors and corrector correct the errors. Figure 6:synthezised diagram of DMC with Sum of square algm The above synthezised diagram is sucessfully completed by using Xilinix Syntheziser.

For matrices A_ij with i = j, the diagonal elements are made zero by replacing diagonal elements by zero. First, two matrices with the same k value are horizontally concatenated and then the matrices obtained after the horizontal concatenations are vertically concatenated to produce N1 - N2 adjacency matrix A. The numbers N1 and N2 represent the numbers of nodes of type 1 and type 2 respectively. The value of N1 is obtained by rounding the product of the network size N and the node proportion q1 of type 1 to the nearest integer. Then the remaining number of nodes N-N1 is the number N2 of nodes of

Training Sets For training it is necessary to define suitable training sets. We used photos from ground truth image database[11]. In this group of photos a secret message will be inserted using the program steghide. The message is unique in every image due to a random generator of strings that will be used. Huffman’s coding data from JPEGSnoop will be transferred to training set-all four columns are to given line by line which will create a vector.

Then it passes through canny edge detection technique. It provides a binary image with wrinkle edges as shown in Fig. 4(a). The white pixels of the wrinkle area give information about wrinkle present in the facial image. In binary image, binary value 1 is used for white pixel, and binary value 0 is for black pixel.

On the Asymptomatic significance test, the significance level is 0.05. Table showing the hypothesis summary Hypothesis Test Summary Null Hypothesis Test Sig. 1 Median of differences between the GPA and the final equals 0 Related-samples Wilcoxon Signed rank test .000 Reject the null hypothesis Section 4: Interpretation Table Showing the Correlation output: Correlations GPA final gender total gpa Pearson Correlation 1 .499** -.193*

Using instruction paint $ Cat f1 b. Displays files in full screen $ More f1 $ Pg f1 13: Renaming files a. Using instruction mv $ Mv f1 prog.txt $ Ls b. Transferring files to another directory. If the last argument is the name of the directory, the files will be transferred to the directory.

The knowledge base consists of a collection of fuzzy if-then rules of the following form: $R^{l}$: if $x_1$ is $F_1^{l}$ and $x_2$ is $F_2^{l}$ and $ldots$ and $x_n$ is $F_n^{l}$, then is $G^{l},~l=1,2, cdots ,n$, where $x=[x_1,cdots,x_i]^{T}$ and $y$ are the FLS input and output, respectively. Fuzzy sets $F_i^{l}$ and $G^{l}$, associated with the fuzzy functions $mu_{{F_i}^{l}}(x_i)$ and $mu_{{G}^{l}}(y)$, respectively. $N$ is the rules inference number. \Through singleton function, center average defuzzification and product inference cite{shaocheng2000fuzzy}, the FLS can be expressed as: For any continuous function $f(x)$ defined on a compact set $Omegain R^n$, there exists a fuzzy system $y(x) = heta ^T xi (x)$ and any positive constant $delta$, such that: where there exists an ideal parameter vector $ heta ^*$ to estimate the value of the vector $ heta$. The optimal parameter vector $ heta ^*$ can be defined as: According to Lemma 1, in order to estimate the

Week 6 – Resource Management Plan To the WBS that was developed earlier, I had made resource assignments by selecting the right resources in the Gantt chart entry form. This I did by selecting from a drop down in under the resources column, and by selecting more than one resource using a comma to separate them. Post this I printed out the resource sheet, by first viewing the resource sheet. This allowed me to identify resources that are over allocated as they were red in color. Next I printed the Gantt chart and entry table by first selecting the Gantt chart and then selecting the entry table from the more tables’ option.

From the design specifications, we know that Q = 0 if DG = 01 and Q = 1 if DG = 11 because D must be equal to Q when G = 1. We assign these conditions to states a and b. When G goes to 0, the output depends on the last value of D. Thus, if the transition of DG is from 01 to 10, the Q must remain 0 because D is 0 at the time of the transition from 1 to 0 in G. If the transition of DG is from 11 to 10 to 00, then Q must remain 1. First, we fill in one square belonging to the stable state in that row. Next a dash is entered in the squares, where both inputs change simultaneously.

5. RESULTS AND DISCUSSIONS 5.1 Unit Root Test Results Unit root test was the first test done on each of the independent variables in both models to check for stationarity. As putting the result of each test here would make it congested, the T-static and probability values from the results are selected from each result and compiled in the tables hereunder. The unit root test results for each variable are attached in the appendix. Table 5.

The mapper extracts the support the call identifier (pass to reducer as a key) and the support call description (pass to the reducer as the value). Each map tasks receives a subset of the initial centroids and is responsible for assign each input data point, to the nearest centroid (cluster). Every time the mapper generate the a key / value pair , where the key is the cluster identifier and the value corresponds to the coordinates of the point. The algorithm uses a combiner to reduce the amount of the data to be transferred from the mapper to the reducer. The Hadoop system follow the different task before approaching the map/reduce

This distinct location identifies the first position where vector [a] and [b] differ. 1. C1 then performs a permutation on vector [Y] and sends it to C2. C2 decrypts the vector and informs C1 where the distinct bit is located. By performing reverse permutation C1 knows precisely where the bit flip occurs and the two key bits that must be compared 2.

The preceding figures shows the Fibonacci and Galois implementations of maximal length shift register m-sequences. As can be seen in these figures, m-sequences contain m shift registers. The shift register set is filled with an m-bit initial seed that can be any value except 0 (if the m bits in the m shift registers are all zero, then it is a degenerate case and the output of the generator is 0). The following examples demonstrate bit generation. 1.