# Optimal interconnect wire sizing

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    % Section 5.1, L. Vandenberghe, S. Boyd, and A. El Gamal
    % "Optimizing dominant time constant in RC circuits"
    % Original by Lieven Vandenberghe
    % Adapted for CVX by Joelle Skaf - 11/25/05
    % Modified by Michael Grant - 3/8/06
    %
    % we consider the problem of sizing an interconnect wire that connects
    % a voltage source and conductance G to a capacitive load C. We divide the
    % wire into n segments of length li, and width xi, i = 1,...,n, which is
    % constrained as 0 <= xi <= Wmax. The total area of the interconnect wire
    % is therefore sum(li*xi). We use a pi-model of each wire segment, with
    % capacitors beta_i*xi and conductance alpha_i*xi.
    % To minimize the total area subject to the width bound and a bound Tmax on
    % dominant time constant, we solve the SDP
    %               minimize        sum_{i=1}^20 xi*li
    %                   s.t.        Tmax*G(x) - C(x) >= 0
    %                               0 <= xi <= Wmax

    %
    % Circuit parameters
    %

    n=21;          % number of nodes; n-1 is number of segments in the wire
    m=n-1;         % number of segments
    beta = 0.5;    % segment has two capacitances beta*xi
    alpha = 1;     % conductance is alpha*xi per segment
    Go = 1;        % driver conductance
    Co = 10;       % load capacitance
    wmax = 1.0;    % upper bound on x

    %
    % Construct the capacitance and conductance matrices
    %   C(x) = C0 + x1 * C1 + x2 * C2 + ... + xn * Cn
    %   G(x) = G0 + x1 * G1 + x2 * G2 + ... + xn * Gn
    % We assemble the coefficient matrices together as follows:
    %   CC = [ C0(:) C1(:) C2(:) ... Cn(:) ]
    %   GG = [ G0(:) G1(:) G2(:) ... Gn(:) ]
    %

    CC = zeros(n,n,m+1);
    GG = zeros(n,n,m+1);
    % constant terms
    CC(n,n,1) = Co;
    GG(1,1,1) = Go;
    % segment values
    for i = 1 : n - 1,
        CC(i,  i,  i+1) = beta;
        CC(i+1,i+1,i+1) = beta;
        GG(i,  i,  i+1) = +alpha;
        GG(i+1,i,  i+1) = -alpha;
        GG(i,  i+1,i+1) = -alpha;
        GG(i+1,i+1,i+1) = +alpha;
    end
    % Reshape for easy Matlab use
    CC = reshape(CC,n*n,m+1);
    GG = reshape(GG,n*n,m+1);

    %
    % Compute points the tradeoff curve, and the four sample points
    %

    npts    = 50;
    delays  = linspace(400,2000,npts);
    xdelays = [ 370, 400, 600, 1800 ];
    xnpts   = length(xdelays);
    areas   = zeros(1,npts);
    xareas  = zeros(1,xnpts);
    sizes   = zeros(m,xnpts);
    for i = 1 : npts + xnpts,

        if i > npts,
            xi = i - npts;
            delay = xdelays(xi);
            disp( sprintf( 'Particular solution %d of %d (Tmax = %g)', xi, xnpts, delay ) );
        else,
            delay = delays(i);
            disp( sprintf( 'Point %d of %d on the tradeoff curve (Tmax = %g)', i, npts, delay ) );
        end

        %
        % Construct and solve the convex model
        %

        cvx_begin sdp quiet
            variable x(m)
            variable G(n,n) symmetric
            variable C(n,n) symmetric
            minimize( sum(x) )
            G == reshape( GG * [ 1 ; x ], n, n );
            C == reshape( CC * [ 1 ; x ], n, n );
            delay * G - C >= 0;
            0 <= x <= wmax;
        cvx_end

        if i <= npts,
            areas(i) = cvx_optval;
        else,
            xareas(xi) = cvx_optval;
            sizes(:,xi) = x;

            %
            % Plot the step response
            %

            figure(xi+2);
            A = -inv(C)*G;
            B = -A*ones(n,1);
            T = linspace(0,2000,1000);
            Y = simple_step(A,B,T(2),length(T));
            hold off; plot(T,Y,'-');  hold on;
            xlabel('time');
            ylabel('v');

            % compute threshold delay, elmore delay, dominant time constant
            tthres=T(min(find(Y(n,:)>0.5)));
            GinvC=full(G\C);
            tdom=max(eig(GinvC));
            telm=max(sum(GinvC'));
            plot(tdom*[1;1], [0;1], '--', telm*[1;1], [0;1],'--', ...
                 tthres*[1;1], [0;1], '--');
            text(tdom,0,'d');
            text(telm,0,'e');
            text(tthres,0,'t');
            title(sprintf('Step responses at the 21 nodes for solution (%d), Tmax=%g', xi, delay ));

        end

    end

    %
    % Plot the tradeoff curve
    %

    figure(1)
    ind = isfinite(areas);
    plot(areas(ind), delays(ind));
    xlabel('Area');
    ylabel('Tdom');
    title('Area-delay tradeoff curve');
    hold on
    for k = 1 : xnpts,
        text( xareas(k), xdelays(k), sprintf( '(%d)', k ) );
    end

    %
    % Draw wires for the four solutions
    %

    figure(2)
    m2 = 2 * m;
    x1 = reshape( [ 1 : m ; 1 : m ], 1, m2 );
    x2 = x1( 1, end : -1 : 1 );
    y  = [ - 0.5 * sizes(x1,:) ; + 0.5 * sizes(x2,:) ; - 0.5 * sizes(1,:)  ];
    x1 = reshape( [ 0 : m - 1 ; 1 : m ], m2, 1 );
    x2 = x1( end : -1 : 1, 1 );
    x  = [ x1 ; x2 ; 0 ];
    h = fill( x, y, ones(4*m+1,1)*[0.9,0.8,0.7,0.6] );
    hold on
    h2 = plot( x, y, '-' );
    axis([ -0.1, m + 0.1, min(y(:))-0.25, max(y(:))+0.1 ]);
    colormap(gray);
    caxis([-1,1]);
    title('Solutions at points on the tradeoff curve');
    legends = {};
    for k = 1 : xnpts,
        set( h(k), 'EdgeColor', get( h2(k), 'Color' ) );
        legends{k} = sprintf( 'Tmax=%g', xdelays(k) );
    end
    legend(legends{:},4);

    Point 1 of 50 on the tradeoff curve (Tmax = 400)
    Point 2 of 50 on the tradeoff curve (Tmax = 432.653)
    Point 3 of 50 on the tradeoff curve (Tmax = 465.306)
    Point 4 of 50 on the tradeoff curve (Tmax = 497.959)
    Point 5 of 50 on the tradeoff curve (Tmax = 530.612)
    Point 6 of 50 on the tradeoff curve (Tmax = 563.265)
    Point 7 of 50 on the tradeoff curve (Tmax = 595.918)
    Point 8 of 50 on the tradeoff curve (Tmax = 628.571)
    Point 9 of 50 on the tradeoff curve (Tmax = 661.224)
    Point 10 of 50 on the tradeoff curve (Tmax = 693.878)
    Point 11 of 50 on the tradeoff curve (Tmax = 726.531)
    Point 12 of 50 on the tradeoff curve (Tmax = 759.184)
    Point 13 of 50 on the tradeoff curve (Tmax = 791.837)
    Point 14 of 50 on the tradeoff curve (Tmax = 824.49)
    Point 15 of 50 on the tradeoff curve (Tmax = 857.143)
    Point 16 of 50 on the tradeoff curve (Tmax = 889.796)
    Point 17 of 50 on the tradeoff curve (Tmax = 922.449)
    Point 18 of 50 on the tradeoff curve (Tmax = 955.102)
    Point 19 of 50 on the tradeoff curve (Tmax = 987.755)
    Point 20 of 50 on the tradeoff curve (Tmax = 1020.41)
    Point 21 of 50 on the tradeoff curve (Tmax = 1053.06)
    Point 22 of 50 on the tradeoff curve (Tmax = 1085.71)
    Point 23 of 50 on the tradeoff curve (Tmax = 1118.37)
    Point 24 of 50 on the tradeoff curve (Tmax = 1151.02)
    Point 25 of 50 on the tradeoff curve (Tmax = 1183.67)
    Point 26 of 50 on the tradeoff curve (Tmax = 1216.33)
    Point 27 of 50 on the tradeoff curve (Tmax = 1248.98)
    Point 28 of 50 on the tradeoff curve (Tmax = 1281.63)
    Point 29 of 50 on the tradeoff curve (Tmax = 1314.29)
    Point 30 of 50 on the tradeoff curve (Tmax = 1346.94)
    Point 31 of 50 on the tradeoff curve (Tmax = 1379.59)
    Point 32 of 50 on the tradeoff curve (Tmax = 1412.24)
    Point 33 of 50 on the tradeoff curve (Tmax = 1444.9)
    Point 34 of 50 on the tradeoff curve (Tmax = 1477.55)
    Point 35 of 50 on the tradeoff curve (Tmax = 1510.2)
    Point 36 of 50 on the tradeoff curve (Tmax = 1542.86)
    Point 37 of 50 on the tradeoff curve (Tmax = 1575.51)
    Point 38 of 50 on the tradeoff curve (Tmax = 1608.16)
    Point 39 of 50 on the tradeoff curve (Tmax = 1640.82)
    Point 40 of 50 on the tradeoff curve (Tmax = 1673.47)
    Point 41 of 50 on the tradeoff curve (Tmax = 1706.12)
    Point 42 of 50 on the tradeoff curve (Tmax = 1738.78)
    Point 43 of 50 on the tradeoff curve (Tmax = 1771.43)
    Point 44 of 50 on the tradeoff curve (Tmax = 1804.08)
    Point 45 of 50 on the tradeoff curve (Tmax = 1836.73)
    Point 46 of 50 on the tradeoff curve (Tmax = 1869.39)
    Point 47 of 50 on the tradeoff curve (Tmax = 1902.04)
    Point 48 of 50 on the tradeoff curve (Tmax = 1934.69)
    Point 49 of 50 on the tradeoff curve (Tmax = 1967.35)
    Point 50 of 50 on the tradeoff curve (Tmax = 2000)
    Particular solution 1 of 4 (Tmax = 370)
    Particular solution 2 of 4 (Tmax = 400)
    Particular solution 3 of 4 (Tmax = 600)
    Particular solution 4 of 4 (Tmax = 1800)