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Deterministic model of throughput in a (single tread) production line


Operational Research. (Ressource Management)What is the Deterministic Traffic Generation Model?Is there a Little law for a network of two connected queues?optimizing contractor schedules - operations research linear programmingModel linearly: Determine amount of units for productionAlgorithm for partitioning works to workersHow do I minimize a queue with two servers, or can I?Little's Law, Queueing Theory, and the Universal Scalability LawQueuing system - advice needed on what models to use, and suitable free simulation softwareHow to model this constraint linearly?













0












$begingroup$


I am looking for a simple and canonical model of throughput (number of batches per unit of time) of a sequential and deterministic production line. For this simple model, I am searching for theorems relating throughput to the throughput of the slowest step and lead time (time to transform inputs into the final product).



The production line (P) is sequential and single threaded, comprises S steps. Each step, s, comprises a throughput per worker and a number of workers. The throughput per worker in each step is deterministic.



Is there a canonical model for this?



I found references for Jackson networks in this Operations Research textbook, such as this, but these seem much more complex and flexible than I am looking for (every node can connect to any other node, the productivity of each step is stochastic). I suppose this is related to work on queues, but I was neve exposed to such theories.



This is for a research project. Any references will be appreciated.










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
    $endgroup$
    – Paul Sinclair
    Mar 30 at 0:58










  • $begingroup$
    @PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
    $endgroup$
    – LucasMation
    Mar 31 at 7:43






  • 1




    $begingroup$
    I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
    $endgroup$
    – Paul Sinclair
    Mar 31 at 17:02















0












$begingroup$


I am looking for a simple and canonical model of throughput (number of batches per unit of time) of a sequential and deterministic production line. For this simple model, I am searching for theorems relating throughput to the throughput of the slowest step and lead time (time to transform inputs into the final product).



The production line (P) is sequential and single threaded, comprises S steps. Each step, s, comprises a throughput per worker and a number of workers. The throughput per worker in each step is deterministic.



Is there a canonical model for this?



I found references for Jackson networks in this Operations Research textbook, such as this, but these seem much more complex and flexible than I am looking for (every node can connect to any other node, the productivity of each step is stochastic). I suppose this is related to work on queues, but I was neve exposed to such theories.



This is for a research project. Any references will be appreciated.










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
    $endgroup$
    – Paul Sinclair
    Mar 30 at 0:58










  • $begingroup$
    @PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
    $endgroup$
    – LucasMation
    Mar 31 at 7:43






  • 1




    $begingroup$
    I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
    $endgroup$
    – Paul Sinclair
    Mar 31 at 17:02













0












0








0





$begingroup$


I am looking for a simple and canonical model of throughput (number of batches per unit of time) of a sequential and deterministic production line. For this simple model, I am searching for theorems relating throughput to the throughput of the slowest step and lead time (time to transform inputs into the final product).



The production line (P) is sequential and single threaded, comprises S steps. Each step, s, comprises a throughput per worker and a number of workers. The throughput per worker in each step is deterministic.



Is there a canonical model for this?



I found references for Jackson networks in this Operations Research textbook, such as this, but these seem much more complex and flexible than I am looking for (every node can connect to any other node, the productivity of each step is stochastic). I suppose this is related to work on queues, but I was neve exposed to such theories.



This is for a research project. Any references will be appreciated.










share|cite|improve this question









$endgroup$




I am looking for a simple and canonical model of throughput (number of batches per unit of time) of a sequential and deterministic production line. For this simple model, I am searching for theorems relating throughput to the throughput of the slowest step and lead time (time to transform inputs into the final product).



The production line (P) is sequential and single threaded, comprises S steps. Each step, s, comprises a throughput per worker and a number of workers. The throughput per worker in each step is deterministic.



Is there a canonical model for this?



I found references for Jackson networks in this Operations Research textbook, such as this, but these seem much more complex and flexible than I am looking for (every node can connect to any other node, the productivity of each step is stochastic). I suppose this is related to work on queues, but I was neve exposed to such theories.



This is for a research project. Any references will be appreciated.







operations-research queueing-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 29 at 15:40









LucasMationLucasMation

1012




1012







  • 2




    $begingroup$
    It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
    $endgroup$
    – Paul Sinclair
    Mar 30 at 0:58










  • $begingroup$
    @PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
    $endgroup$
    – LucasMation
    Mar 31 at 7:43






  • 1




    $begingroup$
    I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
    $endgroup$
    – Paul Sinclair
    Mar 31 at 17:02












  • 2




    $begingroup$
    It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
    $endgroup$
    – Paul Sinclair
    Mar 30 at 0:58










  • $begingroup$
    @PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
    $endgroup$
    – LucasMation
    Mar 31 at 7:43






  • 1




    $begingroup$
    I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
    $endgroup$
    – Paul Sinclair
    Mar 31 at 17:02







2




2




$begingroup$
It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
$endgroup$
– Paul Sinclair
Mar 30 at 0:58




$begingroup$
It seems rather simple, the throughput of the entire line will be the throughput of its slowest step. Every other step must slow production to match the slowest - those before to prevent the built-up of inventory, and those after because they don't receive input any faster.
$endgroup$
– Paul Sinclair
Mar 30 at 0:58












$begingroup$
@PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
$endgroup$
– LucasMation
Mar 31 at 7:43




$begingroup$
@PaulSinclair, I agree. But is there a proof for this? Or how to go about proving this?
$endgroup$
– LucasMation
Mar 31 at 7:43




1




1




$begingroup$
I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
$endgroup$
– Paul Sinclair
Mar 31 at 17:02




$begingroup$
I gave the reasoning for the proof in the comment above. Anything more is just window dressing. If the steps before run at full capacity, their output will just pile up, as it cannot be cleared any faster by the slowest step, and the steps after simply experience downtime as they wait for the slowest step to produce another output for them to work on.
$endgroup$
– Paul Sinclair
Mar 31 at 17:02










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