## 4 Proposed Scheme
### Analysis of Transmission Probability  
The proposed medium access scheme is analyzed below. Suppose all the \(n\) users contend for channel access at certain frame. The scores are denoted by \(y_{1},y_{2},\ldots,y_{n}\). Without loss of generality we assume \(y_{1}\geq y_{2}\geq\ldots\geq y_{n}\). For Sensor \(i\), we would like to find out the probability that Sensor \(i\) will win the frame. We found that it is complicated to solve for the probability for nonzero \(\delta\). Instead we will consider an approximation which is a continuous time analogue to the system. We will again assume the ideal case such that \(m\to\infty\) and \(\delta=t_{c}/m\to 0\). For further simplicity we will ignore \(\beta\) and assume \(\gamma=1\) and \(c=0\), and define \(t_{c}=1\). Under these assumptions the time instant for transmission, which is analogous to choosing a minislot in the discrete time case, is chosen uniformly over the interval \([1-y_{i},1]\), \(\forall i\in I\). Thus the size of ECW for Sensor \(i\) is \(y_{i}\). **Proposition 1**. : _The probability that Sensor \(i\) wins the frame is given by_  
\[\sum_{j=0}^{n-i}\left[\left(\prod_{k=1}^{i+j}z_{k}^{-1}\right)\left\{\frac{z_ {i+j}^{i+j}-z_{i+j+1}^{i+j}}{i+j}\right\}\right] \tag{3}\]  
_where \(z_{j}:=y_{j}/y_{i}\) for \(j=1,\ldots,n\) and \(z_{n+1}:=0\)._  
Proof. : We will assume \(i>1\): it is easy to verify that Equation (3) holds for the case where \(i=1\) using the similar arguments as below. Let us define the time variable \(t\) such that \(t=0\) at the instant of the beginning of the ECW of Sensor \(i\). Consider the probability of Sensor \(i\) winning the frame during the time interval \([0,y_{i}-y_{i+1}]\), _i.e._, Sensor \(i\) wins the frame prior to the beginning of the ECW of Sensor \(i+1\). Since the probability that Sensor \(i\) transmits during \([t,t+dt]\) is given by \(dt/y_{i}\), and in order for this transmission to be successful on Sensors \(1,\ldots,i-1\) must not have transmitted until time \(t\). Notethe ECW of Sensor \(k\) for \(1\leq k\leq i-1\) begins at time \(y_{i}-y_{k}\) which is a negative number. Thus Sensor \(k\) must not transmit during \([y_{i}-y_{k},t]\), and the probability for this event is  
\[1-\frac{y_{k}-y_{i}+t}{y_{k}}=\frac{y_{i}-t}{y_{k}}\]  
Since the above events are mutually independent, the probability of Sensor \(i\)'s transmission is successful during \([0,y_{i}-y_{i+1}]\) is given by  
\[\int_{0}^{y_{i}-y_{i+1}}\prod_{k=1}^{i-1}\left(\frac{y_{i}-t}{y_{k}}\right) \frac{dt}{y_{i}} \tag{4}\]  
By making the change of integration variables \(s=t/y_{i}\), Equation (4) is equal to  
\[\left(\prod_{k=1}^{i-1}\frac{y_{i}}{y_{k}}\right)\int_{0}^{1-z_{i+1}}(1-s)^{i- 1}ds=\left(\prod_{k=1}^{i-1}z_{k}^{-1}\right)\left\{\frac{1-z_{i+1}^{i}}{i} \right\}=\left(\prod_{k=1}^{i}z_{k}^{-1}\right)\left\{\frac{z_{i}^{i}-z_{i+1} ^{i}}{i}\right\} \tag{5}\]  
where \(z_{k}\)'s follow the above definition. Note the Equation (5) results simply from \(z_{i}=1\). Let us consider the probability that Sensor \(i\)'s transmission is successful during \([y_{i}-y_{i+1},y_{i}-y_{i+2}]\). This time interval corresponds to the duration between the beginning of ECW of Sensor \(i+1\) and \(i+2\). Similar to the previous case, when Sensor \(i\) transmits during \([t,t+dt]\), Sensor \(k\) for \(k\in\{1,\ldots,i-1,i+1\}\) must not transmit until time \(t\).