DARPA ULTRALOG Final Report - Industrial and Manufacturing ...

Manuscript for IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED SYSTEMS 11
w i( t )
RA i( t ) ≥ for t ≥ 0 . (7)
ω
n( i )
Service time S i (t) is the time taken to process a task at time t and has a relationship with RA i (t)
as:
t Si(
∫ +
t
t )
RA ( τ )dτ
=
i
f
i
( v
i
). (8)
Suppose a component i receives its tasks at a constant interval of T LB /L i . Then, under
proportional allocation, S i (t) is less than or equal to T LB /L i over time as shown in (9).
f
=
i
( v
i
p∈K
) =
LI
∑
n(
i
LI
i )
∫
p
t+
S (
t
i
t )
RA ( τ )dτ
≥
LI
Si( t ) ≥
T
i
i
LB
S ( t )
i
∫
t+
S (
t
i
t )
T
⇒
L
w i( t ) w
dτ
=
ω ω
LB
i
n(
i )
≥ S ( t )
i
n(
i
i )
S ( t )
i
for t ≥ 0
(9)
So, any component can complete by T LB and generate tasks at a constant interval of T LB /L i
from t=T LB /L i (first task generation time) under proportional allocation when it receives tasks at
a constant interval of T LB /L i from t=0 (first task arrival time). As tasks are infinitesimal and root
tasks increase task availability, each component can receive infinitesimal tasks at a constant
interval in 0≤t≤T LB or more preferably, and complete at less than or equal to T LB . So, the
network completes at T LB under proportional allocation.
From Theorem 1 we can conjecture that a network can achieve a performance close to T LB
under proportional allocation in the limit of large number of tasks. If nodes do not follow the
proportional allocation policy, some components can receive their tasks less preferably than
constant interval resulting in underutilization and consequently increased completion time. Also,
it is optimal for each component to use a pure strategy. Each component’s optimal strategy in the