Coupling of conveyors with different velocities
 Problem:
 two coupled conveyors, lengths l_{1,2} and
constant velocities v_{1,2}
 continuous process for connected conveyors
 → simply change line load λ_{2} by factor
v_{1}/v_{2}
 discrete process with different l_{E1,2} and
∆t_{1,2}
 conveyor 1 sends entities with time intervals ∆t_{1}
 conveyor 2 outputs entities with time intervals
∆t_{2}
 timing doesn't fit
 Basic idea:
 entity ≙ content of a given compartment on a conveyor
 created and filled at the entrance
 emptied and destroyed at the exit
 how to compute mass m_{out} of a new entity
for conveyor 2?
 requirements
 mass conservation on short time scale
 homogeneity, i. e. output distribution similar to
input
 crucial parameter
 k > 1 → add up incoming masses to one
compartment
 k < 1 → distribute incoming mass to several
compartments
 simple case k∊ℕ → add up k incoming masses
 simple case 1/k∊ℕ → distribute incoming mass to 1/k
outgoing masses
 Strategy for k > 1:
 introduce virtual bin with mass m_{acc}
between conveyors
 incoming entities fill bin
 outgoing entity empties bin
 example
 l_{E1} = l_{E2} = 1 m, v_{1}
= 2.5 m/s, v_{2} = 1 m/s, m_{in} ≡ 1 kg
 → ∆t_{1} = 0.4 s, ∆t_{2} =
1 s, k = 2.5

 Strategy A for k < 1:
 again virtual bin with mass m_{acc} between
conveyors
 each time m_{in} arrives: compute partition
mass m_{p}
 mass of outgoing entity
 m_{out} = min(m_{p}, m_{acc})
 good example
 ∆t_{1} = 1 s, ∆t_{2} = 0.35 s → k
= 0.35

 very homogenous distribution
 bad example
 ∆t_{1} = 1 s, ∆t_{2} = 0.8 s → k
= 0.8

 m_{acc} grows → short time mass
conservation violated
 Strategy B for k < 1:
 like A, but changed partition size
 m_{p} = k m_{acc}
 distribute the surplus of m_{acc}
 precise mathematical description in paper
 bad example (k = 0.8)
 comparison
 B: much better short time mass conservation
 A: better homogeneity