© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2: Waste Sludge Incineration: Steady-State Nonlinear DR and Detection of Gross Errors Through Analysis *Flowsheet taken from Felder & Rousseau “Elementary Principles of Chemical Processes”, page 502.

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Dryer Incinerator Boiler Sludge F1 Sat. Vap. F2 Conc. Sludge F3Waste Gas F14 Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) Waste Gas F8 Boiler Feed Water F6 #6 Fuel Oil F7 Air F10 Cool Water F16Hot Water F15 Preheated Air F9 Natural Gas F11 Hot Water F17 Cool Water F18 Air F13 Preheated Air F12

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Incinerator ProblemEngineering Problem Accounting ProblemManagement Problem

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Stream #Flow Rate (kg/day) Temperature (ºC) Mass Fraction Solid N/A 3? ? 4? 5? N/A 73500N/A 8? 9?127N/A Table 2.1a: List of measured and unmeasured variables

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Stream #Flow Rate (kg/day) Temperature (ºC) Mass Fraction Solid N/A N/A 12?124N/A N/A 14?N/A N/A N/A N/A N/A Table 2.1b: List of measured and unmeasured variables

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Gross Errors? Need to change the problem from NONLINEAR to BILINEAR!! See the end of this module.

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 #6 Fuel Oil: 87% C, 10% H, 0.84% S, and 2.16% inert (weight percent) HHV = 3.75 x 10 4 kJ/kg Natural Gas: 90% CH 4, 10% C 2 H 6 (mole percent) Boiler: Efficiency = 62% 25% Excess Air Sludge: Cp of solids = 2.5 kJ/kg·ºC Liquid ~ Water

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Dryer: Efficiency = 55% Pressure at 1 bar on sludge side, and 4 bar on steam side All steam and condensate flows are saturated Incinerator: Sludge must enter at higher than 75% consistency HHV of concentrated sludge = kJ/kg dry solids 195 SCM natural gas/tonne wet sludge 2.5 SCM air/10000 kJ of sludge HHV for complete combustion 100% Excess Air for sludge and natural gas Standard Deviation: Flows 500 kg/day, Temperature 2ºC, Composition 0.03

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 1)Define the y and z matrices for the measured variables and unmeasured variables. 2)Set up the V matrix, scaling down if necessary. 3)Fill in the matrix with the measured values and the matrix with guesses for the unmeasured variables, scaling down appropriately. 4)Determine the mass, component, and heat balances for the process, f(, ) = 0, scaling down parameters where necessary. 5)Determine the Jacobian matrices, A y and A z, and solve them using the values from the and matrices. 6)Calculate b 0 = A y + A z – f (, ) 7)Carry out QR Factorization of the A z matrix, separating the Q matrix into Q 1 (m x n) and Q 2 (m x m-n) matrices, and the R matrix into an R 1 (n x n) matrix.

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 8) Calculate = – V(Q 2 T A y ) T [(Q 2 T A y )V(Q 2 T A y ) T ] -1 (Q 2 T A y – Q 2 T b 0 ) 9)Calculate = R 1 -1 Q 1 T b 0 – R 1 -1 Q 1 T A y - R 1 -1 R 2 (the last term is not necessary if R 2 is a zero matrix) 10)Replace,, and b 0 with,, and b 1. 11)Repeat steps 5 to 10 until the difference between both and, and and are very small.

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 1:,

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 2: V(1,1), V(2,2), V(3,3), …, V(11,11) = (after scaling) V(12,12) = (not necessary to scale) V(13,13), V(14,14), …, V(21,21) = (after scaling) Assumed that Covariance = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 3:,

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: F 3 + F 11 + F 12 - F 14 = 0 F 4 - F 5 = 0 F 15 – F 16 = 0 F 5 - F 6 = 0 Dryer Incinerator Boiler Sludge F1 Sat. Vap. F2 Conc. Sludge F3Waste Gas F14 Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) Waste Gas F8 Boiler Feed Water F6 #6 Fuel Oil F7 Air F10 Cool Water F16Hot Water F15 Preheated Air F9 Natural Gas F11 Hot Water F17 Cool Water F18 Air F13 Preheated Air F12 F 12 - F 13 = 0 F 9 - F 10 = 0 F 7 + F 9 - F 8 = 0 F 17 – F 18 = 0 Mass Balances:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Component Balances: Dryer Sludge F1 Sat. Vap. F2 Conc. Sludge F3 Solids: F 1 x 1 - F 3 x 3 = 0 Water: F 1 y 1 - F 3 y 3 - F 2 = 0 Normalization Constraints: x 3 + y = 0 x 1 + y = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Balances on Heat Exchangers: Air F10 Cool Water F16 Hot Water F15 Preheated Air F9 Hot Water F17 Cool Water F18 Air F13 Preheated Air F F 9 T 9 – 1.046F 10 T F 16 T 16 – 4.18F 15 T 15 = F 12 T 12 – 1.046F 13 T F 18 T 18 – 4.18F 17 T 17 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Heat Flow Through Boiler: (0.62)(HHV OIL )(F 7 ) = (h 5 )(F 5 ) – (h 6 )(F 6 ) 23250F 7 – F F 6 = F 7 – F F 6 = 0 Boiler Sat. Vap. F5 (p=4 bar) Waste Gas F8 Boiler Feed Water F6 #6 Fuel Oil F7 Preheated Air F9

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Air / Oil Ratio in Boiler: Boiler Sat. Vap. F5 (p=4 bar) Waste Gas F8 Boiler Feed Water F6 #6 Fuel Oil F7 Preheated Air F9 For 1 kg of oil: 0.87 kg C kmoles C 0.10 kg H kmoles H kg S kmoles S O 2 required (25% excess): For C(0.0724)(1.25)(1) For H(0.0992)(1.25)(0.25) For S( )(1.25)(1) Total kmoles O F 7 – F 9 = kmoles Air kg Air

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Heat Flow Through Dryer: Dryer Sludge F1 Sat. Vap. F2 Conc. Sludge F3 Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) (x 1 )(F 1 )(Cp s )(T B -T 1 ) + (y 1 )(F 1 )(Cp H2O )(T B -T 1 ) + (F 2 )(Lv H2O ) – (0.55)(h 5 -h 4 )(F 5 ) = 0 250x 1 F 1 – 2.5x 1 F 1 T y 1 F 1 – 4.18y 1 F 1 T F 2 – F 5 = x 1 F 1 – x 1 F 1 T y 1 F 1 – y 1 F 1 T F 2 – F 5 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: NG / Sludge Ratio: Incinerator Conc. Sludge F3 Natural Gas F SCM/tonne wet sludge kmol NG/tonne F kmol CH kg tonnes kmol C 2 H kg tonnes tonnes NG/tonne F F 3 – F 11 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Air / Sludge + NG Ratio: Therefore for 1 kg of NG: kg CH kmoles CH kg C 2 H kmoles C 2 H 6 O 2 required (100% excess): For CH 4 (0.0516)(2)(2) For C 2 H 6 (0.0057)(2)(7/2) Total kmoles O kmoles Air kg Air/kg NG 0.9 kmole CH kg (82.75%) 0.1 kmole C 2 H kg (17.25%) Since NG is 10 mol% C 2 H 6 and 90 mol% CH 4 : Incinerator Conc. Sludge F3 Natural Gas F11 Preheated Air F12 To burn NG:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Air / Sludge + NG Ratio: Incinerator Conc. Sludge F3 Natural Gas F11 Preheated Air F12 To burn sludge: (2.5 SCM air/10000 kJ)(19000 kJ/kg solid) (x 3 kg solid/kg sludge) = 4.75x 3 SCM air/kg sludge x 3 kmol air/kg sludge x 3 kg air/kg sludge 100% Excess: 12.3x 3 kg air/kg sludge 12.3F 3 x F 11 – F 12 = 0 Combining the air needed for the both the sludge and the NG:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Summary of Constraints: 1)2.325y 4 – z y 3 = 0 2)0.25y 12 y 1 – y 12 y 1 y z 8 y 1 – z 8 y 1 y y 2 – z 3 = 0 3)y 1 y 12 – z 1 z 10 = 0 4)y 1 z 8 – z 1 z 9 – y 2 = 0 5)z 10 + z 9 – 1 = 0 6)y 12 + z 8 – 1 = 0 7)z 2 – z 3 = 0 8)z 1 + y 6 + z 6 – z 7 = 0 9)z 6 – y 7 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 4: Summary of Constraints: 10) z 5 – y 5 = 0 11) z 3 – y 3 = 0 12) y 4 + z 5 – z 4 = 0 13) 16.83y 4 – z 5 = 0 14) z 1 – y 6 = 0 15)12.3z 1 z y 6 – z 6 = 0 16)1.046z 5 y 14 – 1.046y 5 y y 9 y 19 – 4.18y 8 y 18 = 0 17)1.046z 6 y 16 – 1.046y 7 y y 11 y 21 – 4.18y 10 y 20 = 0 18) y 8 – y 9 = 0 19)y 10 – y 11 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 5:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 5:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 5:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 6:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 7:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 7:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 7:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 8:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Step 9:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Steps 10 & 11:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Results: Variables Raw Measurement Reconciled Measurement Adjustment (%) F F F3?10170N/A F4?31470N/A F5?31470N/A F F F8?64040N/A F9?60450N/A F Table 2.2a: Measured and reconciled data

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Results: Variables Raw Measurement Reconciled Measurement Adjustment (%) F F12?159880N/A F F14?171600N/A F F F F x x3?0.8576N/A Table 2.2b: Measured and reconciled data

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Results: Variables Raw Measurement Reconciled Measurement Adjustment (%) y1?0.6197N/A y3?0.1424N/A T T T T T T T T T Table 2.2c: Measured and reconciled data

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Interpretation: Dryer Incinerator F1 = F2 = F3 = – = 6700 F11 = 6700(0.1519) = 1018 Raw Measurements:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Interpretation: Dryer Incinerator F1 = F2 = F3 = – = F11 = 10170(0.1519) = 1545 ! Reconciled Measurements:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Interpretation: Control natural gas feed based on incinerator temperature Incorrect measurement of vapor flow from dryer (F2) Insufficient natural gas fed to incinerator (F11) Incinerator temperature too low Install more measurement devices throughout the process

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 MATLAB code used to solve Case Study #1: f=[0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0]; V=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ]; for I=1:11, V(I,I)=0.005^2; end;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 for I=12, V(I,I)=0.03^2; end; for I=13:21, V(I,I)=0.02^2; end; y=[0.238; 0.171; 0.303; 0.035; 0.600; 0.010; 1.600; 0.146; 0.145; 0.356; 0.357; 0.36; 0.23; 1.27; 0.24; 1.24; 0.26; 1.35; 0.35; 1.35; 0.39]; z=[0.11; 0.3; 0.3; 0.64; 0.6; 1.6; 1.73; 0.64; 0.22; 0.78]; flag=1; while flag>0, SSEy=0; SSEz=0;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Ay=[ ;0.25*y(12) *y(12)*y(13) *z(8) *z(8)*y(13) *y(1) *y(1)*y(13) *y(12)*y(1) *z(8)*y(1) ;y(12) y(1) ;z(8) ; ; ; ; ; ; ; ; ; ; ; ; *y(15) *y(18) 4.18*y(19) *z(5) *y(5) *y(8) 4.18*y(9) 0 0; *y(17) *y(20) 4.18*y(21) *z(6) *y(7) *y(10) 4.18*y(11); ; ]; Az=[ ; *y(1) *y(1)*y(13) 0 0; - z(10) z(1); -z(9) y(1) -z(1) 0; ; ; ; ; ; ; ; ; ; ; 12.3*z(10) *z(1); *y(14) ; *y(16) ; ; ];

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 f(1)=2.3250*y(4) *z(3) *y(3); f(2)=0.25*y(12)*y(1) *y(12)*y(1)*y(13)+0.418*z(8)*y(1) *z(8)*y(1)*y(13) *y(2) *z(3); f(3)=y(1)*y(12)-z(1)*z(10); f(4)=y(1)*z(8)-z(1)*z(9)-y(2); f(5)=z(10)+z(9)-1; f(6)=y(12)+z(8)-1; f(7)=z(2)-z(3); f(8)=z(1)+y(6)+z(6)-z(7); f(9)=z(6)-y(7); f(10)=z(5)-y(5); f(11)=z(3)-y(3); f(12)=y(4)+z(5)-z(4); f(13)=16.83*y(4)-z(5); f(14)=0.1519*z(1)-y(6); f(15)=12.3*z(1)*z(10)+34.01*y(6)-z(6); f(16)=1.046*z(5)*y(14)-1.046*y(5)*y(15)+4.18*y(9)*y(19)-4.18*y(8)*y(18);

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 f(17)=1.046*z(6)*y(16)-1.046*y(7)*y(17)+4.18*y(11)*y(21)-4.18*y(10)*y(20); f(18)=y(8)-y(9); f(19)=y(10)-y(11); b=Ay*y+Az*z-f; [Q,R]=qr(Az); for I=1:19, for J=1:10, Q1(I,J)=Q(I,J); end; for I=1:19, for J=11:19, Q2(I,J-10)=Q(I,J); end;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 for I=1:10, for J=1:10, R1(I,J)=R(I,J); end; yhat=y-V*(Q2'*Ay)'*inv((Q2'*Ay)*V*(Q2'*Ay)')*(Q2'*Ay*y-Q2'*b); zhat=inv(R1)*Q1'*b-inv(R1)*Q1'*Ay*yhat; for I=1:21, SSEy=SSEy+(y(I)-yhat(I))^2; end; for I=1:10, SSEz=SSEz+(z(I)-zhat(I))^2; end;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 if and(SSEy<1.0e-6,SSEz<1.0e-6), flag=-1; end; y=yhat; z=zhat; end

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Bilinear Method with Gross Error Detection y 12 = x 1 F 1 = y 13 = T 1 F 1 = y 14 = T 10 F 10 = y 15 = T 13 F 13 = y 16 = T 15 F 15 = z 8 = y 1 F 1 z 9 = y 3 F 3 z 10 = x 3 F 3 z 11 = T 9 F 9 z 12 = T 12 F 12 z 13 = T 1 F 1 y 1 y 17 = T 16 F 16 = y 18 = T 17 F 17 = y 19 = T 18 F 18 = y 20 = T 1 F 1 x 1 = Altered Bilinear Variables:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 1)2.325y 4 – z y 3 = 0 2)0.25y 12 – y z 8 – z y 2 – z 3 = 0 3)y 12 – z 10 = 0 4)z 8 – z 9 – y 2 = 0 5)z 10 + z 9 – z 1 = 0 6)y 12 + z 8 – y 1 = 0 7)z 2 – z 3 = 0 8)z 1 + y 6 + z 6 – z 7 = 0 9)z 6 – y 7 = 0 Altered Bilinear Constraints:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Altered Bilinear Constraints: 10) z 5 – y 5 = 0 11) z 3 – y 3 = 0 12) y 4 + z 5 – z 4 = 0 13) 16.83y 4 – z 5 = 0 14) z 1 – y 6 = 0 15)12.3z y 6 – z 6 = 0 16)1.046z 11 – 1.046y y 17 – 4.18y 16 = 0 17)1.046z 12 – 1.046y y 19 – 4.18y 18 = 0 18) y 8 – y 9 = 0 19)y 10 – y 11 = 0

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Variables Raw Measurement Reconciled Measurement Adjustment (%) F F F3?7840N/A F4?31030N/A F5?31030N/A F F F8?63160N/A F9?59620N/A F Table 2.3a: Measured and reconciled data using a bilinear approach

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Variables Raw Measurement Reconciled Measurement Adjustment (%) F F12?159880N/A F F14?168910N/A F F F F x x3?1.2387N/A Case Study #2 Table 2.3b: Measured and reconciled data using a bilinear approach

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Variables Raw Measurement Reconciled Measurement Adjustment (%) y1?0.6016N/A y3? N/A T T T T T T T T T Case Study #2 Table 2.3c: Measured and reconciled data using a bilinear approach

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 NO GROSS ERROR?!? Need to impose inequality constraints on mass fractions IMPOSSIBLE WITH BILINEAR DR!

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Calculating the bias in measurement y 2 : Adjusting the measurement: y 2 = = Faulty measurement is known from previous analysis (bilinear DR can’t properly detect error).

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Variables Raw Measurement Reconciled Measurement F F F3?10190 F4?31030 F5?31030 F F F8?63160 F9?59620 F Table 2.4a: Bilinear reconciliation without bias

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Variables Raw Measurement Reconciled Measurement F F12? F F14? F F F F x x3? Table 2.4b: Bilinear reconciliation without bias

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Variables Raw Measurement Reconciled Measurement y1? y3? T T T T T T T T T Table 2.4c: Bilinear reconciliation without bias

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 NONLINEAR DR!!!

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 y=[0.238; 0.171; 0.303; 0.035; 0.600; 0.010; 1.600; 0.146; 0.145; 0.356; 0.357; ; ; 0.144; 0.416; ; ; ; ; ]; V=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ]; MATLAB code used for bilinear DR:

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 Ay=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ]; Az=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ];

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 [Q,R]=qr(Az); for I=1:19, for J=1:13, Q1(I,J)=Q(I,J); end;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 for I=1:19, for J=14:19, Q2(I,J-13)=Q(I,J); end; for I=1:13, for J=1:13, R1(I,J)=R(I,J); end;

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 G=Q2'*Ay; yhat=y-V*G'*inv(G*V*G')*G*y; zhat=-inv(Az'*Az)*Az'*(Ay*yhat); yhat2=yhat; yhat2(12)=yhat2(12)/yhat2(1); yhat2(13)=yhat2(13)/yhat2(1); yhat2(14)=yhat2(14)/yhat2(5); yhat2(15)=yhat2(15)/yhat2(7); yhat2(16)=yhat2(16)/yhat2(8); yhat2(17)=yhat2(17)/yhat2(9); yhat2(18)=yhat2(18)/yhat2(10); yhat2(19)=yhat2(19)/yhat2(11);

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 zhat2=zhat; zhat2(8)=zhat2(8)/yhat2(1); zhat2(9)=zhat2(9)/zhat2(1); zhat2(10)=zhat2(10)/zhat2(1); zhat2(11)=zhat2(11)/zhat2(5); zhat2(12)=zhat2(12)/zhat2(6); Vr=G*V*G'; r=G*y; tau=r'*inv(Vr)*r; Id=[ ; ; ; ; ; ];

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2 for I=1:20, for J=1:6, Ac(J,1)=G(J,I); end; Vri=inv(Vr)*(Id-Ac*inv(Ac'*inv(Vr)*Ac)*Ac'*inv(Vr)); O(I)=r'*Vri*r; end; B=[0;1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0]; P=G*B; m=inv(P'*inv(G*V*G')*P)*P'*inv(G*V*G')*G*y;