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# tut5_sol - EE4210 Solution to Tutorial 5 1 Use Generalized...

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1 EE4210 Solution to Tutorial 5 1. Use Generalized Delta Rule in Backpropagation Training (a) Backward Phase: (i) δ Y1 = ( d - y Y1 ) y Y1 (1 - y Y1 ) = (0.9 - 0.715)(0.715)(1 - 0.715) = 0.038 (ii) Δ w Y1H1 = α *old Δ w Y1H1 + η δ Y1 y H1 =(0.6)(0.019) + (0.7)(0.038)(0.67) = 0.029 Δ w Y1H2 = α *old Δ w Y1H2 + η δ Y1 y H2 =(0.6)(0.014) + (0.7)(0.038)(0.473) = 0.0206 Δ b Y1 = α *old Δ b Y1 + η δ Y1 ( - 1) = (0.6)( - 0.0285) +(0.7)(0.038)( - 1) = - 0.0435 (iii) δ H1 = k kH k H w v 1 1 ' ) ( δ ϕ = ( δ Y1 w Y1H1 ) y H1 (1– y H1 ) = (0.038) (0.619) (0.67)(1–0.67) = 0.0052 δ H2 = k kH k H w v 2 2 ' ) ( δ ϕ = ( δ Y1 w Y1H2 ) y H2 (1– y H2 ) = (0.038) (–0.686) (0.473)(1–0.473) = –0.00646 (iv) Δ w H1X1 = α *old Δ w H1X1 + η δ H1 X 1 = (0.6)(0) + (0.7)(0.0052)(0) = 0 Δ w H1X2 = α *old Δ w H1X2 + η δ H1 X 2 = (0.6)(0.00379) + (0.7)(0.0052)(1) = 0.00589 Δ b H1 = α *old b H1 + η δ H1 (–1) = (0.6)(–0.00379) +(0.7)(0.0052)( –1) = – 0.00589 Δ w H2X1 = α *old Δ w H2X1 + η δ H2 X 1 = (0.6)(0) + (0.7)(–0.00646)(0) = 0 Δ w H2X2 = α *old Δ w H2X2 + η δ H2 X 2 = (0.6)( –0.00498) + (0.7)(–0.00646)(1) = –0.0075 Δ b H2 = α *old b H2 + η δ H2 (–1) = (0.6)(0.00498) + (0.7)(–0.00646)( –1) = 0.0075 (v) New w Y1H1 = 0.648 New w Y1H2 = –0.666 New b Y1 = –0.872 New w H1X1 = –0.5 New w H1X2 = 0.1097 New b H1 = –0.6097 New w H2X1 = –0.4 New w H2X2 = 0.1875 New b H2 = 0.3125 (b) Feedforward Phase: H 1 : v H1 = ( - 0.5)(0)+(0.1097)(1)+( - 0.6097)( - 1) = 0.719 => y H1 = ϕ (0.719) = 0.672 H 2 : v H2 = ( - 0.4)(0)+(0.1875)(1)+(0.3125)( - 1) = -0 .125 => y

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tut5_sol - EE4210 Solution to Tutorial 5 1 Use Generalized...

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