04 Utilization
One Processor, one job type
Utilization Analysis
Utilization (u) = fraction of time a resource is busy.
Calculate u for each resource as shown on
following slides.
In a network (i.e., system) where jobs flow through
multiple resources, the one with the highest u is the
“bottleneck.”
Utilization must be kept below 1 (i.e., 100%), and as we will see later, if there is variability then u may need to even be kept significantly below 100%.
The system capacity is the job flow rate R that yields u = 100% at the bottleneck (unless a lower u is determined necessary for satisfactory operation).
Inventory
Processor
Inventory
(queue of raw materials)
holds
1 job
(finished goods) Departures at rate R
Arrivals at rate R
Interarrival
time
= a =1/R
Inventory
= Iq
Time in queue = Tq
Utilization
=u=R/c=Rp
Actual u cannot exceed 1. If a calculation yields u > 1, it is called ‘implied utilization’.
(The calculation may imply u > 1, but this is not possible.)
Slide ‹#›
Slide ‹#›
Multiple (m) Processors, one job type
Processors
Inventory
Inventory
(queue of raw materials)
(1 job each)
(finished goods) (queue of raw materials)
Inventory
= Iq
Time in queue=Tq …
Interarrival time =1/R
One Processor, multiple (n) job types
Inventory
Arrivals at rate R
Process time = p
Capacity
= c = 1/p
Departures at rate R
Inventory
Processor
(holds 1 job)
Departures at rates
R1, R2,…,Rn
Arrivals at rates
R1, R2,…,Rn
# of processors
=m
(finished goods) For a job of type i,
Process time = pi
Capacity = 1/pi = ci
Utilization for job type i = ui = Ri pi
Process time= p
Capacity
= c = 1/p
Utilization = u = R p / m
Total utilization of each processor = u = u1 + u2 + …. + un
Slide ‹#›
Slide ‹#›
Multiple (m) Processors, multiple (n) job types
Inventory
Processors
Inventory
(queue of raw materials)
(1 job each)
(finished goods) Departures at rates
R1, R2,…,Rn
…
Arrivals at rates
R1, R2,…,Rn
# of processors
=m
For a job of type i,
Process time = pi
Utilization Analysis
Utilization (u) = fraction of time a resource is busy.
Calculate u for each resource as shown on
following slides.
In a network (i.e., system) where jobs flow through
multiple resources, the one with the highest u is the
“bottleneck.”
Utilization must be kept below 1 (i.e., 100%), and as we will see later, if there is variability then u may need to even be kept significantly below 100%.
The system capacity is the job flow rate R that yields u = 100% at the bottleneck (unless a lower u is determined necessary for satisfactory operation).
Inventory
Processor
Inventory
(queue of raw materials)
holds
1 job
(finished goods) Departures at rate R
Arrivals at rate R
Interarrival
time
= a =1/R
Inventory
= Iq
Time in queue = Tq
Utilization
=u=R/c=Rp
Actual u cannot exceed 1. If a calculation yields u > 1, it is called ‘implied utilization’.
(The calculation may imply u > 1, but this is not possible.)
Slide ‹#›
Slide ‹#›
Multiple (m) Processors, one job type
Processors
Inventory
Inventory
(queue of raw materials)
(1 job each)
(finished goods) (queue of raw materials)
Inventory
= Iq
Time in queue=Tq …
Interarrival time =1/R
One Processor, multiple (n) job types
Inventory
Arrivals at rate R
Process time = p
Capacity
= c = 1/p
Departures at rate R
Inventory
Processor
(holds 1 job)
Departures at rates
R1, R2,…,Rn
Arrivals at rates
R1, R2,…,Rn
# of processors
=m
(finished goods) For a job of type i,
Process time = pi
Capacity = 1/pi = ci
Utilization for job type i = ui = Ri pi
Process time= p
Capacity
= c = 1/p
Utilization = u = R p / m
Total utilization of each processor = u = u1 + u2 + …. + un
Slide ‹#›
Slide ‹#›
Multiple (m) Processors, multiple (n) job types
Inventory
Processors
Inventory
(queue of raw materials)
(1 job each)
(finished goods) Departures at rates
R1, R2,…,Rn
…
Arrivals at rates
R1, R2,…,Rn
# of processors
=m
For a job of type i,
Process time = pi