2 files changed, 97 insertions, 77 deletions
diff --git a/2471/CH1/EX1.9/Ex1_9.sce b/2471/CH1/EX1.9/Ex1_9.sce
index e7d6fd5d7..8510ddb59 100755
--- a/2471/CH1/EX1.9/Ex1_9.sce
+++ b/2471/CH1/EX1.9/Ex1_9.sce
@@ -1,24 +1,33 @@
-clear ;
-clc;
-// Example 1.9
-printf('Example 1.9\n\n');
-printf('Page No. 17\n\n');
-//given
-
-p= [50, 55, 65, 50, 95, 90, 85, 80, 60, 90, 70, 110, 60, 105];// weakly production in tonnes
-s= [0.4, 0.35, 0.45, .31, 0.51,0.55, 0.45, 0.5, 0.4, 0.51, 0.4, 0.6, 0.45, 0.55];// weakly steam consumption in 10^6 kg
-coefs = regress(p,s);
-new_p = 0:120
-new_s = coefs(1) + coefs(2)*new_p;
-plot(p,s,'r*');
-set(gca(),"auto_clear","off")
-
-plot(new_p,new_s);// please see the corresponding graph in graphic window
-xtitle('weakly steam consumption-production','weakly output (tonnes)','steam consumption/week (10^6 kg)')
-l = legend([_('Given data'); _('Fitting function')],2);
-
-in= coefs(1)*10^6;// intercept of graph in kg/weak
-printf('At zero output the steam consumption is %3.0f in kg/weak \n',in)
-
-
-
+clear ;
+clc;
+function [coefs]=regress(x,y)
+coefs=[]
+  if (type(x) <> 1)|(type(y)<>1) then error(msprintf(gettext("%s: Wrong type for input arguments: Numerical expected.\n"),"regress")), end
+  lx=length(x)
+  if lx<>length(y) then error(msprintf(gettext("%s: Wrong size for both input arguments: same size expected.\n"),"regress")), end
+  if lx==0 then error(msprintf(gettext("%s: Wrong size for input argument #%d: Must be > %d.\n"),"regress", 1, 0)), end
+  x=matrix(x,lx,1)
+  y=matrix(y,lx,1)
+  xbar=sum(x)/lx
+  ybar=sum(y)/lx
+  coefs(2)=sum((x-xbar).*(y-ybar))/sum((x-xbar).^2)
+  coefs(1)=ybar-coefs(2)*xbar
+endfunction
+// Example 1.9
+printf('Example 1.9\n\n');
+printf('Page No. 17\n\n');
+//given
+
+p= [50, 55, 65, 50, 95, 90, 85, 80, 60, 90, 70, 110, 60, 105];// weakly production in tonnes
+s= [0.4, 0.35, 0.45, .31, 0.51,0.55, 0.45, 0.5, 0.4, 0.51, 0.4, 0.6, 0.45, 0.55];// weakly steam consumption in 10^6 kg
+coefs = regress(p,s);
+new_p = 0:120
+new_s = coefs(1) + coefs(2)*new_p;
+plot(p,s,'r*');
+mtlb_hold on
+plot(new_p,new_s);// please see the corresponding graph in graphic window
+xtitle('weakly steam consumption-production','weakly output (tonnes)','steam consumption/week (10^6 kg)')
+l = legend([_('Given data'); _('Fitting function')],2);
+
+in= coefs(1)*10^6;// intercept of graph in kg/weak
+printf('At zero output the steam consumption is %3.0f in kg/weak \n',in)
+\ No newline at end of file
diff --git a/2471/CH3/EX3.14/Ex3_14.sce b/2471/CH3/EX3.14/Ex3_14.sce
index 84b34e6bd..c897eb5cb 100755
--- a/2471/CH3/EX3.14/Ex3_14.sce
+++ b/2471/CH3/EX3.14/Ex3_14.sce
@@ -1,53 +1,64 @@
-clear ;
-clc;
-// Example 3.14
-printf('Example 3.14\n\n');
-printf('Page No. 74\n\n');
-
-// given
-n = 5;//years
-C = 80000;// Cost of the project in Pound
-Cash_in = [10000 20000 30000 40000 50000]// Cash inflow in Pound
-r_d1 = 15;// Discount factor of 15%
-r_d2 = 18 ;// Discount factor of 18%
-r_d3 = 20;// Discount factor of 20%
-
-//At discount of 15%
-df_1 = [0.870 0.756 0.658 0.572 0.497]// Discount factor for every year
-PV_1 = [8700 15120 19740 22880 24850]// Present value
-Net_1 = sum (PV_1);// net present value
-
-
-//At discount of 18%
-df_2 = [0.847 0.718 0.609 0.516 0.437]// Discount factor for every year
-PV_2 = [8470 14360 18270 20640 21850]// Present value
-Net_2 = sum (PV_2);// net present value
-
-
-//At discount of 20%
-df_3 = [0.833 0.694 0.579 0.482 0.402]// Discount factor for every year
-PV_3 = [8330 13880 17370 19280 20100]// Present value
-Net_3 = sum (PV_3);// net present value
-
-// f = N.P.V. cash inflow - N.P.V. cash outflow
-//(1) By Numerical Method
-ff = 2*((sum (PV_2) - C)/(sum (PV_2) - sum(PV_3)));// in percentage
-f = 18 + ff;
-printf('the internal rate of return in percentage is %3.2f \n\n',f)// Deviation in answer due to direct substitution
-
-//(2) By Graphical Interpolation
-f_1 = (sum (PV_1) - C)/10^3;//At discount factor of 15%
-f_2 = (sum (PV_2) - C)/10^3;//At discount factor of 18%
-f_3 = (sum (PV_3) - C)/10^3;//At discount factor of 20%
-
-x = [f_1 f_2 f_3];
-y = [r_d1 r_d2 r_d3];
-plot(x,y,'r*');
-
-plot2d (x,y);// please see the corresponding graph in graphic window
-xtitle('Discount factor against f','f ( *10^3 Pound)','Discount factor(%)')
-regress(x,y)
-coefs = regress(x,y);
-printf('the internal rate of return in percentage is %3.1f \n',coefs(1))// Deviation in answer due to direct substitution
-
-
+clear ;
+clc;
+function [coefs]=regress(x,y)
+coefs=[]
+  if (type(x) <> 1)|(type(y)<>1) then error(msprintf(gettext("%s: Wrong type for input arguments: Numerical expected.\n"),"regress")), end
+  lx=length(x)
+  if lx<>length(y) then error(msprintf(gettext("%s: Wrong size for both input arguments: same size expected.\n"),"regress")), end
+  if lx==0 then error(msprintf(gettext("%s: Wrong size for input argument #%d: Must be > %d.\n"),"regress", 1, 0)), end
+  x=matrix(x,lx,1)
+  y=matrix(y,lx,1)
+  xbar=sum(x)/lx
+  ybar=sum(y)/lx
+  coefs(2)=sum((x-xbar).*(y-ybar))/sum((x-xbar).^2)
+  coefs(1)=ybar-coefs(2)*xbar
+endfunction
+// Example 3.14
+printf('Example 3.14\n\n');
+printf('Page No. 74\n\n');
+
+// given
+n = 5;//years
+C = 80000;// Cost of the project in Pound
+Cash_in = [10000 20000 30000 40000 50000]// Cash inflow in Pound
+r_d1 = 15;// Discount factor of 15%
+r_d2 = 18 ;// Discount factor of 18%
+r_d3 = 20;// Discount factor of 20%
+
+//At discount of 15%
+df_1 = [0.870 0.756 0.658 0.572 0.497]// Discount factor for every year
+PV_1 = [8700 15120 19740 22880 24850]// Present value
+Net_1 = sum (PV_1);// net present value
+
+
+//At discount of 18%
+df_2 = [0.847 0.718 0.609 0.516 0.437]// Discount factor for every year
+PV_2 = [8470 14360 18270 20640 21850]// Present value
+Net_2 = sum (PV_2);// net present value
+
+
+//At discount of 20%
+df_3 = [0.833 0.694 0.579 0.482 0.402]// Discount factor for every year
+PV_3 = [8330 13880 17370 19280 20100]// Present value
+Net_3 = sum (PV_3);// net present value
+
+// f = N.P.V. cash inflow - N.P.V. cash outflow
+//(1) By Numerical Method
+ff = 2*((sum (PV_2) - C)/(sum (PV_2) - sum(PV_3)));// in percentage
+f = 18 + ff;
+printf('the internal rate of return in percentage is %3.2f \n\n',f)// Deviation in answer due to direct substitution
+
+//(2) By Graphical Interpolation
+f_1 = (sum (PV_1) - C)/10^3;//At discount factor of 15%
+f_2 = (sum (PV_2) - C)/10^3;//At discount factor of 18%
+f_3 = (sum (PV_3) - C)/10^3;//At discount factor of 20%
+
+x = [f_1 f_2 f_3];
+y = [r_d1 r_d2 r_d3];
+plot(x,y,'r*');
+
+plot2d (x,y);// please see the corresponding graph in graphic window
+xtitle('Discount factor against f','f ( *10^3 Pound)','Discount factor(%)')
+regress(x,y)
+coefs = regress(x,y);
+printf('the internal rate of return in percentage is %3.1f \n',coefs(1))// Deviation in answer due to direct substitution
+\ No newline at end of file