# Cooperative Admission Control with Network Coding in 5G Underlying D2D-Satellite Communication
## Appendix C Theorem III  
Proof.: In MANC-batch with a lossless second stage, the source or satellite transmits the linear combination of all the missing packets in the devices. In contrast, the local device with the maximum number of the packets available transmits the linear combination via the D2D link. If there are multiple devices with the same number of \(A\), then one is selected randomly. Now in the lossy stage, the transmitter with a maximum number of average receivers is selected to transmit the linear combination of packets. For upper bound, the worse case situation is considered for packet completion. At the start, \(x=argmax_{neR}|A_{n}|\) is selected as a transmitter, which means the device attained a maximum number of the packets. Each device other then transmitter (device x) receives two packets, one from \(\sigma\) probability channel loss from source/ satellite and another from probability loss via D2D links. The device x receives one linear combination of the packet from \(\sigma\) probability channel loss. The size of \(|D_{x}|\) is reduced to \((1-\gamma_{n})\). The \(|D_{n}|\) size is reduced to \((2-\theta_{r}-\tau_{n})\) at each timeslot until the size of the desire \(|D_{n}|\) becomes equal to the size of the \(|D_{x}|\). As per Algorithm 2, a device is selected as the transmitter with the maximum number of acquired \(|A_{n}|\). When innovative packets increase, then the x transmitter changes. After the average k transmission slot, the size of \(|D_{n}|\)\(neR\) is equal to \(|D_{n}|-k((2-\theta_{r}-\tau_{n})\), and the size of the \(|D_{x}|\) is equal to \(|D_{n}|-k(1-\theta_{x})\). The receiver with r as \(argmax_{neR}|A_{n}|=\text{arg min}_{neR}|D_{n}|\) takes maximum \(k=\dfrac{|D_{r}|-|D_{x}|}{(2-\theta_{r}-\tau_{n})}\) timeslots for the transmission and solving, which gives \(|Dx|-k(1-\theta_{x})=|Dr|-k(2-\theta_{r}-\tau_{n})\) for another device as transmitter average. Consider two cases below.
* \(k\geq\frac{|Dx|}{1-\theta_{x}}\): After the time slot of \(\frac{|Dx|}{1-\theta_{x}}\) transmission, the size of the desire \(D\) for x device becomes equal to the size of the \(X\) and r size of desire \(D\) is less than \(X\). In this case, x is the transmitter for the timeslot of the next transmission until all packets are received successfully. So packet completion time is equal to timeslot of transmission of \(T\leq\
icefrac{{|Dx|}}{{1-\theta}}\).
* \(k\leq\frac{|Dx|}{1-\theta_{x}}\): After k timeslots of the transmission, the size of the desire \(D\) of r device becomes equal to the size of the desire \(D\) of the x. The r and x devices selected for the transmission for the next timeslots alternatively until the end of the innovative packets. So the size of \(|D_{r}|\) and \(|D_{x}|\) are \((1-\theta_{r})+(2-\theta_{r}-\tau_{n})\) and \((1-\theta_{x})+(2-\theta_{r}-\tau_{n})\) in alternative timeslots of transmission. It takes \(T_{r}=k+2\frac{|Dr|-k(2-\theta_{r}-\tau_{x,r})}{(3-2\theta_{r}-\tau_{x,r})}\) and \(T_{r}=k+2\frac{|Dx|-k(1-\theta_{x})}{(3-2\theta_{r}-\tau_{x,r})}\). Now placing the value of \(k=|Dr|-|Dx|/(2-\theta_{r}-\tau_{n})-(1-\theta_{x})\). The above conditions conclude proof for upper bound given by Equation (19).