BIometrics- plotting DET and EER curve using Matlab

BIometrics- plotting DET and EER curve using Matlab

• 1.
  Name: Shivakumar Koppad Date:12/03/2015 Submitted to: Dr. Richard Conway Description
• 2.
  Two samplefiles G.datand I.datwith genuine and imposter sample scores respectively were given and the following should be obtained from those.  Plot the scoredistributions for both  DET curve  Determining EER  Operation point to minimize the cost MATLAB 2013a has been used to show the requirements Scores Distributioncurve The following code has been used to generate the graph to plot the score for distribution curve. %Name: Shivakumar Koppad clc; clear all; close all; % Importing data load('G.dat'); load('I.dat'); % Implementing G file gen=[0 0 0 0 0 0 0 0 0 0]; for g=1:1:1200 if 0<=G(g)<0.1 gen(1)=gen(1)+1; elseif 0.1<=G(g)<0.2 gen(2)=gen(2)+1; elseif 0.2<=G(g)<0.3 gen(3)=gen(3)+1; elseif 0.3<=G(g)<0.4 gen(4)=gen(4)+1; elseif 0.4<=G(g)<0.5 gen(5)=gen(5)+1; elseif 0.5<=G(g)<0.6 gen(6)=gen(6)+1; elseif 0.6<=G(g)<0.7 gen(7)=gen(7)+1; elseif 0.7<=G(g)<0.8 gen(8)=gen(8)+1;
• 3.
  elseif 0.8<=G(g)<0.9 gen(9)=gen(9)+1; else gen(10)=gen(10)+1; end end % ImplementingI file imp=[0 0 0 0 0 0 0 0 0 0]; for i=1:1:1200 if 0<=I(i)<0.1 imp(1)=imp(1)+1; elseif 0.1<=I(i)<0.2 imp(2)=imp(2)+1; elseif 0.2<=I(i)<0.3 imp(3)=imp(3)+1; elseif 0.3<=I(i)<0.4 imp(4)=imp(4)+1; elseif 0.4<=I(i)<0.5 imp(5)=imp(5)+1; elseif 0.5<=I(i)<0.6 imp(6)=imp(6)+1; elseif 0.6<=I(i)<0.7 imp(7)=imp(7)+1; elseif 0.7<=I(i)<0.8 imp(8)=imp(8)+1; elseif 0.8<=I(i)<0.9 imp(9)=imp(9)+1; else imp(10)=imp(10)+1; end end % Defining Score Score=[0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9]; hold on; plot(Score,imp,'color','r'); plot(Score,gen,'color','b'); hold off; xlabel('Scores') ylabel('No. of users') title('Distribution Diagram') legend('genuine','imposter') % End Result: In the below output of the graph, X-axis represents the scores and Y-axis represents the count for both genuine and imposter users.
• 4.
  DET curve Detection ErrorTradeoff curves areROC (receiver operating characteristic) type curves showing therange of operating points of systems performing detection tasks as a threshold is varied to alter the miss and false alarmrates and plotted using a normal deviate scale for each axis. DET curves have the property that if the underlying scoredistributions for the two types of trials are normal, the curve becomes a straightline. They havebeen widely used to present the performance characteristics of speaker recognition systems The below code is used to generate the DET curve. A new function name called DET curveis created to generate the graph and to find the EER equal error rate on the curve. %Name: Shivakumar Koppad %ID: 14043513 %Title: Assignment 1 % %Submitted to: Dr.Richard Conway
• 5.
  function [TPR,FPR] =DETCurve(G,I) load('G.dat'); load('I.dat'); % Creates a matrix TPR = zeros(1, 1000); FPR = zeros(1, 1000); P = 1200; N = 1200; index = 0; % Assume the threshold as 0.005 for threshold = 0:0.005:1 TP = 0; FP = 0; %Provides the genuine users count for i = 1:1200 if (G(i) >= threshold) TP = TP + 1; end end % Provides the Imposters count for i1 = 1:1200 if(I(i1) >= threshold) FP = FP + 1; end end index = index + 1; % Calculating true positive rate TPR(index) = TP/P; % Calculating false positive rate FPR(index) = FP/N; end % Calculating false negative rate(FNR) using TPR+FNR=1 FNR = (1-TPR); x = 0:0.1:1; y = x; [x(i),y(i)] = polyxpoly(x,y,FPR,FNR); fprintf('EER(X): %d n', x(i)); fprintf('EER(Y): %d n', y(i)); plot(FPR,FNR,'LineWidth',2, 'color','g'); hold on; plot(x,y,x,1-y, 'color','r'); plot (x(i),y(i),'X','MarkerSize',10, 'LineWidth', 2,'Color','b'); hold off; title('DET CURVE'); xlabel('False Positive Rate (FPR) '); ylabel('False Neagtive Rate (FNR) '); end
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  Result The below isthe DET curve generated
• 7.
  Equal error rate Abiometric security systempredetermines the threshold values for its false acceptance rate and its false rejection rate, and when the rates are equal, the common value is referred to as the equal error rate. The value indicates that the proportion of falseacceptances is equal to the proportion of false rejections. The lower the equal error rate value, the higher the accuracy of the biometric system. Fromabove figurewe can determine EER fromthe DET curve EER(X, Y) = (FPR, FNR) = (0.08611, 0.08611) Cost minimization OP_min_costis created to find a suitable point on the DET curve which minimizes the cost FromBayesian statistics, the Cost (C) is calculated as: C=CFAPIFAR(T) + CFRPGFRR(T) Where, CFA is the costof a single false accept CFR is the cost of a single false reject PI is probability of imposter users PG is probability of genuine users Given Cost of False accept rate is 17 Cost of false reject rate is 27 Probability of Imposter users is 0.5 Probability of Genuine users is 0.5 (Weare assuming equal probabilities) C = 17*0.5*FAR(T) +27*0.5*FRR(T)
• 8.
  C = 8.5*FAR(T)+13.5*FRR(T) 13.5*FRR(T) =C- 8.5*FAR(T) FRR(T) = [C- 8.5*FAR(T)]/13.5 FRR(T) = -0.6296*FAR(T) +C/13.5 This is in the form of y= mx+c The following code has been used to generate the graph which can be used to find FRR(T) and FAR(T) the point operated to minimize the cost
• 9.
  function [FAR, FRR]= OPcost(G,I) load('G.dat'); load('I.dat'); %Creates a matrix FRR = zeros(1, 1000); FAR = zeros(1, 1000); i = 1200; j = 1200; index = 0; %Assume the threshold as 0.001 for threshold = 0:0.001:1 FN = 0; FP = 0; for i = 1:1200 if (G(i) <= threshold) FN = FN + 1; end end for i1 = 1:1000 if(I(i1) >= threshold) FP = FP + 1; end end index = index + 1; FRR(index) = FN/i; FAR(index) = FP/j; end hold on; plot(FAR,FRR,'LineWidth',1,'color','c'); %To start with assume the minimum cost as 100 min = 100; min_new=0; for i1=1:1000 if 17*0.5*FAR(i1) + 27*0.5*FRR(i1) < min min = 17*0.5*FAR(i1) + 27*0.5*FRR(i1); min_new = i1; end end plot(FAR(min_new),FRR(min_new),'x','MarkerSize',14); Xmin = 0:0.05:1; Ymin = (-0.33)*Xmin + 0.06989*min; plot (Xmin,Ymin,'color', 'm'); hold off; title('Operating point to minimise cost'); xlabel('FAR '); ylabel('FRR ');
• 10.
  Result According to thegraph FAR(T) = 0.02333 FRR(T) = 0.1092 Now, 13.5*FRR(T) =C- 8.5*FAR(T) 13.5*0.1092=C-8.5*0.02333 1.4742=c-0.198305 Therefore, C = 1.6725 So the cost C is 1.6725