  Two mobile agents, starting from different nodes of a network at possibly
different times, have to meet at the same node. This problem is known as
$\mathit{rendezvous}$. Agents move in synchronous rounds. Each agent has a
distinct integer label from the set $\{1,\dots,L\}$. Two main efficiency
measures of rendezvous are its $\mathit{time}$ (the number of rounds until the
meeting) and its $\mathit{cost}$ (the total number of edge traversals). We
investigate tradeoffs between these two measures. A natural benchmark for both
time and cost of rendezvous in a network is the number of edge traversals
needed for visiting all nodes of the network, called the exploration time.
Hence we express the time and cost of rendezvous as functions of an upper bound
$E$ on the time of exploration (where $E$ and a corresponding exploration
procedure are known to both agents) and of the size $L$ of the label space. We
present two natural rendezvous algorithms. Algorithm $\mathtt{Cheap}$ has cost
$O(E)$ (and, in fact, a version of this algorithm for the model where the
agents start simultaneously has cost exactly $E$) and time $O(EL)$. Algorithm
$\mathtt{Fast}$ has both time and cost $O(E\log L)$. Our main contributions are
lower bounds showing that, perhaps surprisingly, these two algorithms capture
the tradeoffs between time and cost of rendezvous almost tightly. We show that
any deterministic rendezvous algorithm of cost asymptotically $E$ (i.e., of
cost $E+o(E)$) must have time $\Omega(EL)$. On the other hand, we show that any
deterministic rendezvous algorithm with time complexity $O(E\log L)$ must have
cost $\Omega (E\log L)$.
