Constant Learning Rate for Gradient Decent

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Given, we have a learning rate, $alpha_n$ for the $n^{th}$ step
of the gradient descent process. What would be the impact of using a constant value for $alpha_n$ in gradient descent?






gradient-descent learning-rate







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  • $begingroup$
    Do you mean a constant value of $alpha$ for each step?
    $endgroup$
    – Wes
    4 hours ago
















1












$begingroup$


Given, we have a learning rate, $alpha_n$ for the $n^{th}$ step
of the gradient descent process. What would be the impact of using a constant value for $alpha_n$ in gradient descent?






gradient-descent learning-rate







share|improve this question







New contributor




Umbrage is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Do you mean a constant value of $alpha$ for each step?
    $endgroup$
    – Wes
    4 hours ago














1












1








1





$begingroup$


Given, we have a learning rate, $alpha_n$ for the $n^{th}$ step
of the gradient descent process. What would be the impact of using a constant value for $alpha_n$ in gradient descent?






gradient-descent learning-rate







share|improve this question







New contributor




Umbrage is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Given, we have a learning rate, $alpha_n$ for the $n^{th}$ step
of the gradient descent process. What would be the impact of using a constant value for $alpha_n$ in gradient descent?






gradient-descent learning-rate




gradient-descent learning-rate






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  • $begingroup$
    Do you mean a constant value of $alpha$ for each step?
    $endgroup$
    – Wes
    4 hours ago


















  • $begingroup$
    Do you mean a constant value of $alpha$ for each step?
    $endgroup$
    – Wes
    4 hours ago
















$begingroup$
Do you mean a constant value of $alpha$ for each step?
$endgroup$
– Wes
4 hours ago




$begingroup$
Do you mean a constant value of $alpha$ for each step?
$endgroup$
– Wes
4 hours ago










2 Answers
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$begingroup$

Intuitively, if $alpha$ is too large you may "shoot over" your target and end up bouncing around the search space without converging. If $alpha$ is too small your convergence will be slow and you could end up stuck on a plateau or a local minimum.



That's why most learning rate schemes start with somewhat larger learning rates for quick gains and then reduce the learning rate gradually.






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    0












    $begingroup$

    Gradient descent has the following rule:



    $theta_{j} := theta_{j} - alpha frac{delta}{delta theta_{j}} J(theta)$



    Here $theta_{j}$ is a parameter of your model, and $J$ is the cost/loss function. At each step the product $alpha frac{delta}{delta theta_{j}} J(theta)$ gets smaller as we get closer to the gradient $frac{delta}{delta theta_{j}} J(theta)$ converging to 0. $alpha$ can be constant, and in many cases, it is, but varying $alpha$ might help converge faster.






    share|improve this answer








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      2 Answers
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      2 Answers
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      active

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      0












      $begingroup$

      Intuitively, if $alpha$ is too large you may "shoot over" your target and end up bouncing around the search space without converging. If $alpha$ is too small your convergence will be slow and you could end up stuck on a plateau or a local minimum.



      That's why most learning rate schemes start with somewhat larger learning rates for quick gains and then reduce the learning rate gradually.






      share|improve this answer









      $endgroup$


















        0












        $begingroup$

        Intuitively, if $alpha$ is too large you may "shoot over" your target and end up bouncing around the search space without converging. If $alpha$ is too small your convergence will be slow and you could end up stuck on a plateau or a local minimum.



        That's why most learning rate schemes start with somewhat larger learning rates for quick gains and then reduce the learning rate gradually.






        share|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Intuitively, if $alpha$ is too large you may "shoot over" your target and end up bouncing around the search space without converging. If $alpha$ is too small your convergence will be slow and you could end up stuck on a plateau or a local minimum.



          That's why most learning rate schemes start with somewhat larger learning rates for quick gains and then reduce the learning rate gradually.






          share|improve this answer









          $endgroup$



          Intuitively, if $alpha$ is too large you may "shoot over" your target and end up bouncing around the search space without converging. If $alpha$ is too small your convergence will be slow and you could end up stuck on a plateau or a local minimum.



          That's why most learning rate schemes start with somewhat larger learning rates for quick gains and then reduce the learning rate gradually.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          oW_oW_

          3,046729




          3,046729























              0












              $begingroup$

              Gradient descent has the following rule:



              $theta_{j} := theta_{j} - alpha frac{delta}{delta theta_{j}} J(theta)$



              Here $theta_{j}$ is a parameter of your model, and $J$ is the cost/loss function. At each step the product $alpha frac{delta}{delta theta_{j}} J(theta)$ gets smaller as we get closer to the gradient $frac{delta}{delta theta_{j}} J(theta)$ converging to 0. $alpha$ can be constant, and in many cases, it is, but varying $alpha$ might help converge faster.






              share|improve this answer








              New contributor




              Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$


















                0












                $begingroup$

                Gradient descent has the following rule:



                $theta_{j} := theta_{j} - alpha frac{delta}{delta theta_{j}} J(theta)$



                Here $theta_{j}$ is a parameter of your model, and $J$ is the cost/loss function. At each step the product $alpha frac{delta}{delta theta_{j}} J(theta)$ gets smaller as we get closer to the gradient $frac{delta}{delta theta_{j}} J(theta)$ converging to 0. $alpha$ can be constant, and in many cases, it is, but varying $alpha$ might help converge faster.






                share|improve this answer








                New contributor




                Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Gradient descent has the following rule:



                  $theta_{j} := theta_{j} - alpha frac{delta}{delta theta_{j}} J(theta)$



                  Here $theta_{j}$ is a parameter of your model, and $J$ is the cost/loss function. At each step the product $alpha frac{delta}{delta theta_{j}} J(theta)$ gets smaller as we get closer to the gradient $frac{delta}{delta theta_{j}} J(theta)$ converging to 0. $alpha$ can be constant, and in many cases, it is, but varying $alpha$ might help converge faster.






                  share|improve this answer








                  New contributor




                  Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  Gradient descent has the following rule:



                  $theta_{j} := theta_{j} - alpha frac{delta}{delta theta_{j}} J(theta)$



                  Here $theta_{j}$ is a parameter of your model, and $J$ is the cost/loss function. At each step the product $alpha frac{delta}{delta theta_{j}} J(theta)$ gets smaller as we get closer to the gradient $frac{delta}{delta theta_{j}} J(theta)$ converging to 0. $alpha$ can be constant, and in many cases, it is, but varying $alpha$ might help converge faster.







                  share|improve this answer








                  New contributor




                  Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|improve this answer



                  share|improve this answer






                  New contributor




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                  answered 4 hours ago









                  WesWes

                  965




                  965




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