# Throughput of Buffer with Dependent Service Times
## 4 Buffer Throughput
Denoting  
\[d_{i}(n,s)=\int_{0}^{\infty}e^{-su}D_{i}(n,u)\mathrm{d}u, \tag{11}\]  
and exploiting the convolution theorem in (8), we obtain  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]0 \[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]1 \[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]2  
we obtain from (12)  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]3  
with  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]4  
while (16) and (17) yield  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]5 \[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]6  
and  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]7  
respectively. Exploiting Lemma 3.2.1 of [29], we can obtain the general solution of (23) in the form  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]8  
where  
\[D_{i}(n,t)= \sum_{k=0}^{n-1}\sum_{j=1}^{m}\int_{0}^{t}p_{ij}(k,v)\big{[}k{+}D_ {j}(n-k+1,t-v)\big{]}\mathrm{d}G(v)\]9  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]0  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]1  
and \(\mathbf{0}\) denotes the zero matrix. Substituting \(n=1\) in (27), we have  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]2  
Now, (27) accompanied with (28) and (31), we yield  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]3  
where  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]4  
and  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]5  
In this way, the only unknown piece of (32) is \(\mathbf{d}(1,s)\). It, however, can be easily derived using (32) with (25). The results may be arranged as follows:  
**Theorem 1. **_The transform of the expected count of packets departed from the buffer in \((0,t)\) is_  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]6  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]7  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]8  
_where_  
\[+\sum_{j=1}^{m}\int_{0}^{t}r_{ij}(n,v)\big{[}n+D_{j}(1,t-v)\big{]} \mathrm{d}G(v)\]9  
\[+\sum_{k=0}^{n}\sum_{j=1}^{m}kp_{ij}(k,t)(1-G(t)). \tag{8}\]0  
_while \(g(s)\), \(\mathbf{W}(n,s)\), \(\mathbf{u}(n,s)\), \(\mathbf{A}(0,s)\), \(\mathbf{F}(K,s)\), and \(\mathbf{q}(K,s)\) are given in (18), (33), (34), (13), (14), and (24)._  
To obtain numeric results from Theorem 1, we have to calculate matrices \(\mathbf{A}(n,s)\), \(\mathbf{F}(n,s)\), and \(\mathbf{C}(n,s)\) (the latter is needed in \(\mathbf{q}(n,s)\)). \(\mathbf{A}(n,s)\) and \(\mathbf{C}(n,s)\) can be calculated utilizing the well-known method based on the uniformization of a Markov chain, explained in detail in [30] on page 66. \(\mathbf{F}(n,s)\), on the other hand, has been derived in [28] (see Formulae (55)-(57) there). Namely, we have  
\[+\sum_{k=0}^{n}\sum_{j=1}^{m}kp_{ij}(k,t)(1-G(t)). \tag{8}\]1  
where  
\[+\sum_{k=0}^{n}\sum_{j=1}^{m}kp_{ij}(k,t)(1-G(t)). \tag{8}\]2  
\[+\sum_{k=0}^{n}\sum_{j=1}^{m}kp_{ij}(k,t)(1-G(t)). \tag{8}\]3  
\[+\sum_{k=0}^{n}\sum_{j=1}^{m}kp_{ij}(k,t)(1-G(t)). \tag{8}\]4  
Finally, to obtain \(D_{i}(n,t)\) from Theorem 1, we must invert the transforms to the time domain. In the numeric results in Section 6, the formulae of [31] were used to accomplish that.