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diff --git a/905/CH6/EX6.12/6_12.sce b/905/CH6/EX6.12/6_12.sce
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+clear;
+clc;
+
+// Illustration 6.12
+// Page: 377
+printf('Illustration 6.12 -  Page: 377\n\n');
+
+// solution
+//*****Data*****//
+// Componenets  A-propane  B-pentane  C-methane  D-ethane  E-butane  F-hexane
+// x-mole fraction  a-relative volatility
+xA = 0.25;  aA = 4.08;
+xB = 0.11;  aB = 1.00;
+xC = 0.26;  aC = 39.47;
+xD = 0.09;  aD = 10.00;
+xE = 0.17;  aE = 2.11;
+xF = 0.12;  aF = 0.50;
+// Since propane and pentane are light and heavy key respectively
+// Methane and ethane are LNK, hexane is a HNK, while butane is a “sandwich component,” meaning that it has a volatility intermediate between the keys.
+
+FRlkd = 0.98;
+FRhkd = 0.01;
+// For methane
+D_CR = (aC-1)/(aA-1)*FRlkd + (aA-aC)/(aA-1)*FRhkd;
+// For ethane
+D_DR = (aD-1)/(aA-1)*FRlkd + (aA-aD)/(aA-1)*FRhkd;
+// For butane
+D_ER = (aE-1)/(aA-1)*FRlkd + (aA-aE)/(aA-1)*FRhkd;
+// For hexane
+D_FR = (aF-1)/(aA-1)*FRlkd + (aA-aF)/(aA-1)*FRhkd;
+// Since the feed is 66% vaporized
+q = 1-0.66;
+
+// Now equation 6.82 is solved for two values of Q
+deff('[y] = f14(Q1)','y = 0.66 - aA*xA/(aA-Q1)-aB*xB/(aB-Q1)-aC*xC/(aC-Q1)-aD*xD/(aD-Q1)-aE*xE/(aE-Q1)-aF*xF/(aF-Q1)');
+Q1 = fsolve(1.2,f14);
+
+deff('[y] = f15(Q2)','y = 0.66 - aA*xA/(aA-Q2)-aB*xB/(aB-Q2)-aC*xC/(aC-Q2)-aD*xD/(aD-Q2)-aE*xE/(aE-Q2)-aF*xF/(aF-Q2)');
+Q2 = fsolve(2.5,f15);
+
+// Basis: 100 mole of feed
+F = 100; // [mole]
+// Let d1 = Dxad, d2 = Dxbd, d3 = Dxcd, and so on
+d1 = F*xA*FRlkd; // [moles of propane]
+d2 = F*xB*FRhkd; // [moles of pentane]
+d3 = F*xC; // [moles of methane]
+d4 = F*xD; // [moles of ethane]
+d6 = F*xF*0; // [moles of hexane]
+// And d5 is unknown
+// Applying equation 6,78 for each value of Q
+
+// Solution of simultaneous equation 
+function[f]=H(e)
+    f(1) = e(1) - aA*d1/(aA-Q1)-aB*d2/(aB-Q1)-aC*d3/(aC-Q1)-aD*d4/(aD-Q1)-aE*e(2)/(aE-Q1)-aF*d6/(aF-Q1);
+    f(2) = e(1) - aA*d1/(aA-Q2)-aB*d2/(aB-Q2)-aC*d3/(aC-Q2)-aD*d4/(aD-Q2)-aE*e(2)/(aE-Q2)-aF*d6/(aF-Q2);
+    funcprot(0);
+endfunction
+
+// Initial guess
+e = [90 5];
+y = fsolve(e,H);
+Vmin = y(1); // [mole]
+d5 = y(2); // [d5 = Dxed, mole]
+
+// From equ 6.84
+D = d1+d2+d3+d4+d5+d6; // [mole]
+// From mass balance 
+Lmin = Vmin-D; // [mole]
+// For minimum reflux ratio
+Rmin = Lmin/D;
+printf("The minimum reflux ratio is %f\n\n",Rmin);
+\ No newline at end of file