I'm trying to figure out how to calculate the throughput. Throughput is defined as the rate (bits/sec) bits are transferred between a sender and receiver; also, . I have a source node and a destination node, $k+1$ nodes all together with $k-1$ intermediary nodes. There are $k$ links between each pair of nodes and to transmit one bit it takes $\tau$ seconds and the bandwidth of each link is $W$ bits/sec.
Before a single message can be sent the source sends a connection request message, the destination replies with an accept/deny message to the source, and then the real message transfer begins. Following the source's message a connection disconnect message is sent. These 3 messages are $H$ bits long and occur so that the destination must receive the entire message before the next one is sent out.
First I simplify the problem by only considering the connection between 1 source and 1 destination node. For 1 message that is $H$ bits long the time taken to transmit the entire message should be $H$ bits * $\tau$ seconds. So, to transmit 3 messages across k nodes I have a total time of $3k(H*\tau)$seconds. From here I'm not sure how to calculate the throughput and get lost with the unit cancelation to get to bits/sec.
Any ideas on how to manipulate the problem to get a throughput in bits/sec?
EDIT: Problem explanation.
In steady state after handshaking, a message of maximum segment size ($MSS$) can be sent to the destination by time
$$\text{propagation delay} + \text{transmission delay} = k\tau + k\frac{MSS}{W}$$
The return acknowledgement has a delay of
$$\text{propagation delay} + \text{transmission delay} = k\tau + k\frac{H}{W}$$
The sum of these two gives the whole round trip time. And the actual message length sent without header should be $MSS-H$, which gives throughput
$$\text{throughput} = \frac{MSS-H}{2k\tau+k\frac{MSS+H}{W}}$$
Of course there are quite a lot of assumptions and simplifications here.