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BOOL CMyWinApp::InitInstance() { CCode10B* pFrame = new CCode10B; m_pMainWnd = pFrame; pFrame->ShowWindow(SW_SHOW); pFrame->UpdateWindow(); return TRUE; } BEGIN_MESSAGE_MAP(CCode10B, CFrameWnd) ON_WM_PAINT() ON_WM_KEYDOWN() END_MESSAGE_MAP() CCode10B::CCode10B() { v=new CUSTOMER [N+1]; c=new COUNTER [C+1]; TopLeft=CPoint(0,0); BottomRight=CPoint(900,620); t=-1; nv=0; tnav=0; for (int k=1;k<=C;k++) { c[k].sta=0; c[k].av[0]=0; } for (int i=1;i<=N;i++) v[i].sta=0; wBar=20; hBar=20; ZeroMemory (&lfTimes,sizeof(lfTimes));
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where A - lI is sometimes called the characteristic matrix of A. How does a determinant become a polynomial The best way to see this is with an example of a symmetric 2 2 matrix: 1 0.7 1 0 1 - l 0.7 A - lI = - l 0 1 = 0.7 1 - l . 0.7 1 Now the determinant det(A - lI) = (1 - l)2 - 0.49, which is seen to be a quadratic (polynomial) in l. Note that the highest power of l is 2, which is also the dimension of A. For n-dimensional matrix A we would have an nth degree polynomial in l. The eigenvalues are also called characteristic roots, proper values, latent roots, and so on. These have fundamental importance in understanding the properties of a matrix. For arbitrary square matrices, eigenvalues are complex roots of the characteristic polynomial. After all, not all polynomials have real roots. For example, the polynomial l2 + 1 = 0 has roots l1 = i and l2 = -i where i denotes -1 . The eigenvalues are denoted by li where i = 1, 2, . . . , n, are in nonincreasing order of absolute values. For complex numbers, the absolute value is de ned as the square root of the product of the number and its complex conjugate. The complex conjugate of i is -i, their product is 1, with square root also 1. Thus |i| = 1 and |-i| = 1, and the polynomial l2 + 1 = (l - l1)(l - l2) holds true. As in the example above, the roots of a polynomial need not be distinct. By the fundamental theorem of algebra, an nth degree polynomial de ned over the eld of complex numbers has n roots. Hence we have to count each eigenvalue with its proper multiplicity when we write |l1(A)| |l2(A)| , . . . , |ln-1(A)| |ln(A)|. If eigenvalues are all real, we distinguish between positive and negative eigenvalues when ordering them by l 1 ( A) l 2 ( A) ,. . . , l n -1 ( A) l n ( A). (8.1.13)
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When people who are rst exposed to ABM hear the phrase, It is better to be approximately correct than precisely inaccurate, they smile because they know exactly what that means in their organization. But they usually do not know what causes ABM to produce substantially better accuracy relative to their existing legacy costing system despite its abundant use of estimates and approximations. The explanation, which is counterintuitive to many, is that the initial errors from estimates in an ABM assignment system cancel out. This is due, in part, because allocating (i.e., reassigning expenses into costs) is a closed system with a zero sum total error in the total costs of the nal cost objects. The starting amount of expenses to be allocated can be assumed to be onehundred percent accurate because they come from the general ledger accounting system, which is specially designed to accumulate and summarize the spending transactions. However, subsequently, as we reassign expenses into calculated costs, imprecise inputs do not automatically result in inaccurate outputs. That is, precision is not synonymous with accuracy. In ABM s cost assignment view, estimating error does not compound, it dampens out. These are properties of statistics found in equilibrium networks (i.e., the amount of expenses and costs remains constant). And ABM is a cost reassignment network much more than it is an accounting system. In ABM, poor model design leads to poor results. A well known and painful lesson about activity-based costing is that when an ABM implementation falls short of its expectations, it is often because the system was overengineered in size and detail. The ABM system usually quickly reached diminishing returns in extra accuracy for incremental levels of effort, but this effect was not recognized by the ABM project team. The system was built so large that the administrative effort to collect the data and maintain the system was ultimately judged to be not worth the perceived bene ts. This results in a death by details ABM project; it is
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PROBLEMS 8.1 Determine the scan angles associated with a grazing grating lobe for an array of square horns of aperture dimension 1.2 0 ( 0 = wavelength in free space) in a square lattice. The element spacing is 1.4 0 in both planes. The scan planes are given by (a) = 0 and (b) = 30 . Assume that the principal polarization is x-directed, that is, = 0 corresponds to the E-plane. 8.2 Repeat problem 8.1 for a triangular lattice with the following lattice parameters: a = b = 1 4 0 = tan 1 2 = 63 4 . 8.3 It is known that the blind spot nearest the bore sight on the = 45 plane occurs at = 15 . Assuming a square lattice and using the periodic nature of the Floquet impedance, find the other possible blind spots on the = 45 scan plane. Element spacing = 1 3 0 . 8.4 Using the equivalent circuit of Figure 8.6, deduce an expression for the input impedance seen by the voltage source V0 . Assume only two waveguide modes on the aperture and two Floquet modes for the radiated fields. Assume R m n as the coupling between the nth waveguide mode to the mth Floquet mode. Set n1 = 1 and n2 = 0 to the final result. Verify the expression in (8.23) assuming f Y02 approaches infinity. 8.5 Using a circle diagram, show that for a rectangular grid an invisible region in the (kx ky )-plane will not exist if the condition in (8.24) holds. Obtain the equivalent condition for an equilateral triangular grid structure. 8.6 Using the saddle point integration method for large , establish the relation in (8.30) starting from (8.29). (Note: The integrand has two saddle points within the range of .) 8.7 An array of rectangular waveguides has the following normalized Floquet impedances:
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(8.12) If the oscillator power had been only 0 dBm, requiring more LO ampli er gain, the increase in mixer noise gure would have been 15.7 dB. This shows the importance of starting with a high-power (as well as low-noise) oscillator. The difference between the value in Eq. (8.12) and that in Eq. (8.11) shows the importance of ltering out some of these ranges.
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Table 17.3
Trading Strategies That Work
stage conceive the product
If you are still in employment and any of your referees work at your company, you may be worried about the prospect of alerting them to the fact that you re looking for a job. If you explain your situation in the following way, you ll probably find that most interviewers are very understanding: I d be happy for you to check my references eventually and I m sure that they will confirm everything that I ve been saying about myself in this interview. But would you mind waiting until you ve decided to make me a firm offer I d rather not draw their attention to the fact that I m looking elsewhere for a job. If you have already left an employer, then your answer can be an unmitigated yes: Please do approach my referees. The contact details for my last boss and the operations director are at the bottom of my CV. I m sure that they will say pretty much the same thing about me as I ve been telling you. Most employers make job offers contingent on receiving satisfactory references. So when you receive such an offer, talk to your referee to make sure that what you have told your new employer corresponds with what your referee is going to tell them.
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