📄 payoff_amount.m
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function varargout = payoff_amount(p)%% %% [y] = payoff_amount(x)% [y,yp] = payoff_amount(x) returns the function value in y and % the derivative in yp%% Matthias Heinkenschloss% Department of Computational and Applied Mathematics% Rice University% April 17, 2001%a = 100000;n = 20;r = 0.06;% return the function valuevarargout{1} = a*(1+r).^n - p*((1+r).^n-1)./r;if( nargout > 1 ) % return the derivative as the second argument varargout{2} = -((1+r).^n-1)./r;end
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