## 3 Proposed Design
### Numerical Analysis  
In the following, the analysis regarding frame aggregation effect on the performance of IEEE 802.15.4 by incorporating traffic patterns of patient monitoring systems is conducted. The required frame size of MPDU with minimum header overhead is:  
\[\text{MPDU Size}=2\ (\text{data})+9(\text{Minimum MAC layer header})=11\ \text{bytes} \tag{4}\]  
The size of single aggregated MPDU (A-MPDU) frame size is:  
\[\text{A}-\text{MPDU frame size}=4\ \text{bytes}\ (\text{delimiter size})+11\ \text{bytes}\ (\text{MPDU})+4\ \text{bytes}\ (\text{padding})=19\ \text{byte} \tag{5}\]  
Figure 6: Delimiter structure. Figure 7: Block Acknowledgement procedure. The total A-MPDU packet size is:  
\[\text{A}-\text{MPDU packet size}=19\text{ bytes}+6\text{ bytes }(\text{PHY header})=25\text{ bytes} \tag{6}\]  
Similarly, size of A-MPDU with two frames is  
\[\text{A}-\text{MPDU }=19\text{ bytes}+19\text{ bytes}+6\text{ bytes }(\text{PHY header})=44\text{ bytes} \tag{7}\]  
Considering the maximum packet size capacity 127 bytes and the size of single A-MPDU from Equations (5) and (6), we aggregate the maximum six MPDU in an A-MPDU. Therefore, the size of the maximum A-MPDU is:  
\[\text{Total size of A}-\text{MPDU }=(6\times 19)+6\text{ bytes }(\text{PHY header})=120\text{ bytes} \tag{8}\]  
Transmission time of a maximum-sized A-MPDU [31] is:  
\[\text{T}_{\text{frame}}\text{ for A}-\text{MPDU}=\frac{(120)\times 8}{\text{ data rate}}=3.84\text{ ms} \tag{9}\]  
In the following, we compute the value of total delay using [1; 31]:  
\[\text{ED}=\text{macMaxFrameRetries}\times(\text{T}_{\text{WT}}+\text{T}_{ \text{frame}}+\text{T}_{\text{Ack}}+\text{T}_{\text{SIFS}}+\text{ T}_{\tau\text{A}}) \tag{10}\]  
For the scenario mentioned in Figure 1 where multiple nodes try to access the channel, we compute the value of delay by considering the default MAC attribute values mentioned in Table 4.  
This includes total waiting time (TWT), where TWT represents the backoff time period which is assigned to a node before going to channel access attempt. Equation (10) also includes frame transmission time (Tframe), acknowledgement receiving time (TAck), short interframe space time (TSIFS) and turnaround time (T\({}_{\tau\text{A}}\)). For analysis, we consider the bit rate of 250 kbps (2.4 G Hz frequency band is considered with BPSK modulation and a symbol represents 4 bit). T\({}_{\text{BOSlot}}\) denotes the time for a single backoffslot:  
\[\text{A}-\text{MPDU frame size}=4\ \text{bytes}\ (\text{delimiter size})+11\ \text{bytes}\ (\text{MPDU})+4\ \text{bytes}\ (\text{padding})=19\ \text{byte} \tag{5}\]0  
where TBE is the random backoff time, TCCA the clear channel assessment time and TCW the contention windows time. The random backoff time (TBE) is the waiting time for channel access which is computed as:  
\[\text{A}-\text{MPDU frame size}=4\ \text{bytes}\ (\text{delimiter size})+11\ \text{bytes}\ (\text{MPDU})+4\ \text{bytes}\ (\text{padding})=19\ \text{byte} \tag{5}\]1  
Using Equation (12) with ideal channel conditions, Figure 8 presents a comparative analysis of total delay for the transmission of the number of frames with aggregation and without aggregation. It is noted that without aggregation, delay continuously increases as every frame takes a separate channel access attempt and its overheads. We assume that a node gets channel access attempt at \begin{table}
\begin{tabular}{l c} \hline \hline
**Attribute** & **Value** \\ \hline T\({}_{\tau\text{A}}\) & 0.192 ms \\ T\({}_{\text{SIFS}}\) & 0.192 ms \\ T\({}_{\text{Ack}}\) & 0.864 ms \\ T\({}_{\text{CCA}}\) & 0.128 ms \\ T\({}_{\text{CW}}\) & 0.64 ms \\ T\({}_{\text{symbol}}\) & 0.32 ms \\ \hline \hline \end{tabular}
\end{table} 
Table 4: MAC layer parameters and values [1; 31]. the transmission time. On the other hand, for the A-MPDU, a node sends multiple packets in single channel access attempt, only the packet size varies by increasing the frame aggregation size.