# Throughput of Buffer with Dependent Service Times
## 5 Lost Packets  
In this section, we deal with three loss-related characteristics: \(B_{i}(n,t)\), \(B_{i}^{\prime}(n,t)\), and \(P\). Namely, \(B_{i}(n,t)\) denotes the expected count of lost packets in \((0,t)\), given that \(N(0)=n\), \(S^{*}(0)=i\). The derivative, \(B_{i}^{\prime}(n,t)\), is temporary loss intensity at \(t\), whereas \(P\) denotes the overall loss probability, i.e.,  
\[P=\lim_{t\to\infty}\frac{B_{1}^{\prime}(0,t)}{\Lambda}. \tag{50}\]  
Making use of total probability law in reference to the first arrival epoch, \(v\), for \(0<n<K\), we obtain  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\] \[\quad+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)B_{j}(1,t-v)\mathrm{d}G (v). \tag{51}\]  
The first component of (51) responds to the case where \(k<n\) packets are departed by \(v<t\), and the buffer is not vacated by \(v\). No losses happen by \(v\) in this scenario, so the new expected number of losses at \(v\) is \(\sum_{j}B_{j}(n-k+1,t-v)\). The second component of (51) responds to the case, where \(n\) packets are departed by \(v<t\), so the buffer becomes vacant at some time \(u<v\). Again, no losses happen by \(v\) in this scenario, so the new expected number of losses at \(v\) is \(\sum_{j}B_{j}(1,t-v)\). In the situation where \(v\geq t\), losses by \(t\) are impossible at all, so the whole component vanishes. For \(n=K\), we have  
\[B_{i}(K,t)= \sum_{j=1}^{m}\int_{0}^{t}p_{ij}(0,v)\big{[}1+B_{j}(K,t-v)\big{]} \mathrm{d}G(v)\] \[+\sum_{k=1}^{K-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(K-k+ 1,t-v)\mathrm{d}G(v)\] \[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(K,v)B_{j}(1,t-v)\mathrm{d}G(v). \tag{52}\]  
This is obtained in a similar manner to (51), but the first component of (51) is now split into two cases, \(k=0\) and \(k>0\), i.e., no completed transmissions and at least one completed transmission, respectively. The only possible case of a packet loss by \(v\) is when \(n=K\) and \(k=0\). It is covered by the first component of (52). The new expected number of losses at \(v\) is then \(\sum_{j}\big{[}1+B_{j}(K,t-v)\big{]}\). The rest of (52) is obtained in a fashion similar to that in (51). For \(n=0\), we obtain  
\[B_{i}(0,t)=\int_{0}^{t}B_{i}(1,t-v)\mathrm{d}G(v), \tag{53}\]  
which is analogous to (10) and can be explained in the same way. Denoting  
\[b_{i}(n,s)=\int_{0}^{\infty}e^{-su}B_{i}(n,u)\mathrm{d}u, \tag{54}\]  
we obtain  
\[b_{i}(n,s)=\sum_{k=0}^{n-1}\sum_{j=1}^{m}a_{ij}(k,s)b_{j}(n-k+1,s)+\sum_{j=1}^{ m}f_{ij}(n,s)b_{j}(1,s),\quad 0<n<K, \tag{55}\]  
\[b_{i}(K,s)= \sum_{j=1}^{m}a_{ij}(0,s)b_{j}(K,s)+\sum_{k=1}^{K-1}\sum_{j=1}^{m} a_{ij}(k,s)b_{j}(K-k+1,s)\] \[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]0  
and  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]1  
from (51), (52), and (53), respectively. Then, using the vector  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]2  
we can rewrite (55)-(57) into  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]3  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]4 \[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]5  
and  
\[B_{i}(n,t) =\sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)B_{j}(n-k+1, t-v)\mathrm{d}G(v)\]6  
respectively. It is readily apparent that systems (59)-(61) have forms similar to (23)-(26). Therefore, they can be resolved with respect to \(\mathbf{b}(n,s)\) using the same method as was used for (23)-(26), with obvious changes. After solving (59)-(61), we arrive at the theorem stated next.