With the transfer event, D,T and L events must be assigned costs to define most parsimonious reconciliation. 
Different costs give different most parsimonious solutions. In the top case, D is assigned a cost of 1, T of 3 and L of 1, and if we look at the number of events in the proposed scenario we find 1 duplication and 1 loss, so a cost of 2. If we compute that cost again with the bottom cost assignment we have a scenario cost of 3. With that new cost assignment the bottom scenario is more parsimonious as its cost is 2, while it had a cost of 3 with the top cost assignment.
Own work
Hugo Menet