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RFID-embedded Object in Pervasive Healthcare Applications

OS Group Project
by

farra maidin

on 22 May 2013

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Transcript of RFID-embedded Object in Pervasive Healthcare Applications

Identifying
RFID-embedded
Object in Pervasive
Healthcare
Applications Problem statement 1 3 Goal 2 Although the extend of dropped reads varies across domain and it context-specific 30% of RFID tags read are dropped To be able to deliver necessary healthcare anytime to anyone regardless of location and of the constraints. Sytem type Healthcare system Decision support system (DSS) reduce rates of inappropriate diagnostic test Intelligent support system(IDSS) used to:
1.critique theraphy
2.critique dosage error
3.check for drug interaction main idea of RFID healthcare system:
usage of multiple layer of intelligent IS technology There are more than 3000 RFID case studies in a wide range of application areas. RFID in healthcare is growing rapidly and forecast to become a $2.1 billion global business by 2016. An application system with passive RFID tags that remind userd if they mistakenly leave their personal belongings behind. Deterministic rules are used in the RFID application system to verify whether an object that is supposed to be with a person at a given time does indeed exist with that person. Cost and privacy concerns have generally been recognized as a major factors in success of RFID applications Information from sensors such as RFID tags provide basic context information including the presence of medical staff, devices, instruments and medication in the operating room, the coexistence of entities that are required to be simultaneously present together at a given location among others. Asset management is critical in pervasive healthcare where it is necessary to be able to identify and track mobile objects. Inventory management reduces out-of-stock situations, billing errors, misplaced articles, theft and general shrink age problems. RFID tags and healthcare Proposed Methods Identifying the presence of RFID tags is not always necessarily as straight-forward as it seems. False positives and false negatives can be a problem in RFID-embedded system, especially when signal from a given tag is blocked by an impenetrable object(eg; metal shielding) or when a corrupted signal is read. Example : where two objects are required to be simultaneously present such as a two medications that belong together in a prescription. Solution : algorithm to propose means that reduce false positives and false negatives through a variation Clinical decision support system ALGORITHM 1 Two readers are simultaneously used to read information from each RFID tag that passes through the system. Assuming, tag of interest is T is read twice by each of the two readers within short time duration at each location this system so we have two reads of each tags. For ease of representation, consider the two reads of a tag by a reader as they if are two different tags by T and T' and there are two readers (R, & R2) which are assumed to be oriented at different angles with respect to tagged object. Usage of several readers are to reduce the error rate of tag reads is not new. We compare algorithms against one another so is considered as the base case since it involves minimal processing from the back-end server. #include<iostream.h>int main ()
{ int R1,R2;
cout <<"\nPlease Enter 1 if T is detected by R1 or 0 if T is absent:\n ";
cin>>R1;cout <<"\nPlease Enter 1 if T is detected by R2 or 0 if T is absent:\n";
cin>>R2;
if((R1==1)&&(R2==1))
cout<<"T(tag) is confirmed to be present :)";
else
cout<<"T(tag) is confirmed to be absent :(";
return 0;
}; Coding for Algorithm 1 ALGORITHM 2 #include<iostream.h>
int main ()
{ int R1,R2;
cout<<"\nPlease Enter 1 if T is detected by R1 or 0 if T is absent:\n ";
cin>>R1;cout<<"\nPlease Enter 1 if T is detected by R2 or 0 if T is absent:\n";
cin>>R2;i
f((R1==1)&&(R2==1))
cout<<"T(tag) is confirmed to be present :)";
else if ((R1==0)&&(R2==0))
cout<<"T(tag) is confirmed to be absent :(";
else if((R1==1)||(R2==1))
cout<<"T(tag) is confirmed to be present with probability of 0.5(50%) :)";
else
cout<<"Error!!!!";
return 0;
}; Coding for Algorithm
2 Unlike the base case, we confirm the tag T to be present with probability 0.5 when either R1 or R2 identify it as being present. We include this probability since the fact that one of the readers has identified its presence signifies the possibility that the signal was probably blocked from one of the readers and not the other Affirmative case is refined in this algorithm. Consider two separate reads of the tag within a brief duration of time to more accurately identify the tag's presence. So, we have one tag that is read twice by each reader. Tag T is assumed to be presented when, after the first read, both the readers (R1 and R2) identify its presence just like in the base case. When both the readers (R1 and R2) identify it to be absent after the first read, the tag T is confirmed to be absent in the field of the readers just like in the base case. However, when one of the readers identifies it to be presented after the first read while the other does not identify its presence, we utilize information from the second readings of this tag (T) by the readers in determining the presence/absence of the tag T. When both the readers (R1 and R2) identify T to be present (i.e., the tag is identified to be present during the second reading of T by both the readers), The tag T to be present is confirmed. When neither of the readers (R1 and R2) identifies T to be present during the second read, we confirm the tag T to be absent as well. When only one of the readers (R1 or R2) identifies T to be presented, we confirm that tag T is present with probability 1/8. This is because during the first read the probability of reader R1 confirming tag Ts presence when reader R2 confirms its absence and during the second read one of the readers (R1 or R2) confirms the tag's presence while the other reader confirms tag's absence is (1/4*1/2 =)1/8. ALGORITHM 3
; #include<iostream.h>int
main ()int R1,R2;
cout<<"\nPlease Enter 1 if T is detected by R1 or 0 if T is absent:\n ";
n>>R1;
cout<<"\nPlease Enter 1 if T is detected by R2 or 0 if T is absent:\n";
cin>>R2;
if((R1==1)&&(R2==1))
cout<<"T(tag) is confirmed to be present :)";
else if ((R1==0)&&(R2==0))
cout<<"T(tag) is confirmed to be absent :(";
else if((R1==1)||(R2==1))
cout<<"\nSince only one of R1&R2 has detected T,we will apply a 2nd read for R1,R2";
cout<<"\n\n\n***************************************************************************";
cout<<"\nPlease Enter 1 if T is detected by R1 or 0 if T is absent in the second read:\n ";
cin>>R1;
cout<<"\nPlease Enter 1 if T is detected by R2 or 0 if T is absent in the second read:\n";
cin>>R2;
if((R1==1)&&(R2==1))
cout<<"T(tag) is confirmed to be present in the second read:)";
else if ((R1==0)&&(R2==0))
cout<<"T(tag) is confirmed to be absent in the second read :(";
else if((R1==1)||(R2==1))
cout<<"T(tag) is confirmed to present in the second read with probability of 1/8 :)";
else
cout<<"undefined input";
else
cout<<"Error!!!!";
return 0; Coding for
Algorithm 3 ALGORITHM 4 It varied from the 3rd algorithm in :
1. The number of tags are 2 here.2.It has
2. readers that read each tag four times.
3. When we have T2 in reader 1 present only once then was present also in the 2nd reader twice then , T1 is present with probability of 50%.
4. When T2 is presented in both reads , T1 is present and vice versa . It can be
illustrated
as below #include<iostream.h>
int main ()
{
int R11,R12,R21,R22,R11s,R12s,R21s,R22s;
cout<<"Guidlines:\nT1=tag1\nT2=tag2\nR11=1st read of the first reader for T1\nR21=1st read of the second reader for T1";
cout<<"\nR11s=second read of the first reader for T1\nR21s=second read of the second reader for T1";
cout<<"\nR12=1st read of the first reader for T2\nR22=1st read of the second reader for T2";
cout<<"\nR12s=second read of the first reader for T2\nR22s=second read of the second reader for T2";
cout<<"\n\n************************DETECTING WILL START *****************************";
cout<<"\nPlease Enter 1 if T1 is detected by R11 or 0 if T1 is absent:\n ";
cin>>R11;
cout<<"\nPlease Enter 1 if T1 is detected by R21 or 0 if T1 is absent:\n";
cin>>R21;
cout<<"\nPlease Enter 1 if T1 is detected by R11 in the second read or 0 if T1 is absent:\n ";
cin>>R11s;
cout<<"\nPlease Enter 1 if T1 is detected by R21 or 0 if T1 is absent in the second read:\n";
cin>>R21s;

if((R11==1)&&(R11s==1))
cout<<"T1(tag) is confirmed to be present :)";

else if ((R11==0)&&(R11s==0))
cout<<"T1(tag) is confirmed to be absent :(";

else if((R11==1)||(R11s==1))
{
cout<<"\nSince only one of R11&R11s has detected T1,we will read fron the 2nd reader";
cout<<"\n\n\n***************************************************************************";
cout<<"\nPlease Enter 1 if T1 is detected by R21 or 0 if T1 is absent in the 2nd reader :\n ";
cin>>R21;
cout<<"\nPlease Enter 1 if T is detected by R21 in the 2nd read or 0 if T is absent in the 2nd reader:\n";
cin>>R21s;

if((R21==1)&&(R21s==1))
cout<<"T1(tag) is confirmed to be present in the second reader:)";

else if ((R21==0)&&(R21s==0))
cout<<"T(tag) is confirmed to be absent in the second reader :(";

else if((R21==1)||(R21s==1))
{
cout<<"We will consider the seconed tag (T2)";

cout<<"\nPlease Enter 1 if T2 is detected by R12 or 0 if T2 is absent:\n ";
cin>>R12;
cout<<"\nPlease Enter 1 if T2 is detected by R22 or 0 if T2 is absent:\n";
cin>>R22;
cout<<"\nPlease Enter 1 if T2 is detected by R12 in the second read or 0 if T2 is absent:\n ";
cin>>R12s;
cout<<"\nPlease Enter 1 if T1 is detected by R22 or 0 if T2 is absent in the second read:\n";
cin>>R22s;

if((R12==1)&&(R12s==1))
cout<<"T1(tag) is confirmed to be present :)";

else if ((R12==0)&&(R12s==0))
cout<<"T1(tag) is confirmed to be absent :(";

else if((R12==1)||(R12s==1))
{
if((R22==1)&&(R22s==1))
cout<<"T1(tag) is confirmed to be present :)";
else if ((R22==0)&&(R22s==0))
cout<<"T1(tag) is confirmed to be absent :(";
else if((R22==1)||(R22s==1))
cout<<"T1(tag) is confirmed to be present with a probability of 0.5";
else
cout<<"undefined input";

}
else
cout<<"undefined input";
}}
else
cout<<"Error!!!";
return 0;
}; We only apply T2 readings if and only if T1 or T2 was present because if both were 1 or present means that T1 is present and if both were 0 that means T1 is absent. Coding
for
Algorithm
4 Overview of Journal Main Concept
Full transcript