Appendix: Proof of Theorem 2.1

Proof:

We only need to consider two cases.

1. Case 1: t belongs to a time interval in which busy periods of Server 1 and Server 2 do not overlap each other. (see Figure 7).

    
   

   

   Figure 7: Case 1.

    
   

   If , because Server 2 is not busy, the time of the -th bit leaving Server 2 can be evaluated as

   

   On the other hand, the time of the -th bit arriving at Server 1 can be evaluated as

   

   By using , we have

   

   If , because Server 1 is not at busy state, the delay suffered by the traffic of Connection 2 or Connection 3 is bounded by

   

   If , the queuing delay suffered by Connection 2 or Connection 3 at this subsystem is zero.

2. Case 2: t belongs to a time interval in which busy periods of Server 1 and Server 2 overlap each other. (see Figure 8).

    
   

   

   Figure 8: Case 2.

    
   

   First, we consider the situation . If , the time of bit leaving Server 2 can be evaluated as

   

   On the other hand, the time of the bit arriving at Server 1 can be evaluated as

   

   So, we have

   

   where is the length of maximum busy period of Server 1. If , this situation has been discussed in Case 1.

   Finally, we consider the situation . Similarly, if , we have

   

   and

   

   So,

   

   If , this situation has been discussed in Case 1.

Q.E.D