Polly.Contrib.WaitAndRetry is an extension library for Polly containing helper methods for a variety of wait-and-retry strategies.
Resilience , Resiliency , Fault-tolerance , Transient-fault-handling , Retry , Retry-intervals , Jitter , dotnet , fault-handler , jitter-formula , jitter-recommendation , polly , resiliency-patterns , retry-pattern , retry-policies , retry-strategies
Install-Package Polly.Contrib.WaitAndRetry -Version 1.1.1
Polly.Contrib.WaitAndRetry contains several helper methods for defining backoff strategies when using Polly's wait-and-retry fault handling ability.
One common approach when calling occasionally unreliable services is to wrap those calls in a retry policy. For example, if we're calling a remote service, we might choose to retry several times with a slight pause in between to account for infrastructure issues.
We can define a policy to do this in Polly.
While the core Polly package contains core logic and gives examples for a variety of retry strategies, this Contrib packages up a variety of strategies in easy-to-use helper methods.
The following defines a policy that will retry five times and pause 200ms between each call.
var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(retryCount: 5, retryNumber => TimeSpan.FromMilliseconds(200));
We can simplify this by using the
ConstantBackoff helper in Polly.Contrib.WaitAndRetry
var delay = Backoff.ConstantBackoff(TimeSpan.FromMilliseconds(200), retryCount: 5); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
retryCount must be greater than or equal to zero.
Additionally, when using the
ConstantBackoff helper, or any other WaitAndRetry helper, we can signal that the first failure should retry immediately rather than waiting the indicated time. To do this, ensure the
fastFirst parameter is
var delay = Backoff.ConstantBackoff(TimeSpan.FromMilliseconds(200), retryCount: 5, fastFirst: true);
This will still retry five times but the first retry will happen immediately.
It can be desirable to wait increasingly long times between retries. For example, in cases where a remote service is being impacted by a high workload. In this scenario, we want to give the service some time to stabilize before trying again.
The first tool at our disposal is the
var delay = Backoff.LinearBackoff(TimeSpan.FromMilliseconds(100), retryCount: 5); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
This will create a linearly increasing retry delay of 100, 200, 300, 400, 500ms.
The default linear factor is 1.0. However, we can provide our own.
var delay = Backoff.LinearBackoff(TimeSpan.FromMilliseconds(100), retryCount: 5, factor: 2);
This will create an increasing retry delay of 100, 300, 500, 700, 900ms.
Note, the linear factor must be greater than or equal to zero. A factor of zero will return equivalent retry delays to the
We can also specify an exponential back-off where the delay duration is
initialDelay x 2^iteration. Because of the exponential nature, this is best used with a low starting delay or in out-of-band communication, such as a service worker polling for information from a remote endpoint. Due to the potential for rapidly increasing times, care should be taken if an exponential retry is used in the code path for servicing a user request.
var delay = Backoff.ExponentialBackoff(TimeSpan.FromMilliseconds(100), retryCount: 5); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
This will create an exponentially increasing retry delay of 100, 200, 400, 800, 1600ms.
The default exponential growth factor is 2.0. However, can can provide our own.
var delay = Backoff.ExponentialBackoff(TimeSpan.FromMilliseconds(100), retryCount: 5, factor: 4);
The upper for this retry with a growth factor of four is 25,600ms. Care and a calculator should be used when changing the factor.
Note, the growth factor must be greater than or equal to one. A factor of one will return equivalent retry delays to the
If the overall amount of time that an exponential-backoff retry policy could take is a concern, consider placing a TimeoutPolicy outside the wait-and-retry policy using PolicyWrap. A timeout policy used in this way will limit the overall execution time for all tries and waits-between-tries. For instance, you could configure the exponential backoff for your wait-and-retry strategy to be 1, 2, 4, 8 seconds; and also impose an overall timeout, however many tries are invoked, at 45 seconds.
var retryWithBackoff = Policy .Handle<FooException>() .WaitAndRetryAsync(Backoff.ExponentialBackoff(TimeSpan.FromSeconds(1), retryCount: 5)); var timeout = Policy.Timeout(TimeSpan.FromSeconds(45)); var retryWithBackoffAndOverallTimeout = timeout.Wrap(retryWithBackoff);
When the combined time taken to make tries and wait between them exceeds 45 seconds, the TimeoutPolicy will be invoked and cause the current try and further retries to be abandoned.
In a high-throughput scenario, processing many requests at once, a fixed-progression wait-and-retry strategy can have disadvantages.
Sudden issues affecting performance, combined with a fixed-progression wait-and-retry, can lead to subsequent retries being highly correlated. For example, if there are 50 concurrent failures, and all 50 requests enter a wait-and-retry for 10ms, then all 50 requests will hit the service again in 10ms; potentially overwhelming the service again.
One way to address this is to add some randomness to the wait delay. This will cause each request to vary slightly on retry, which decorrelates the retries from each other.
Following exploration by Polly community members, we now recommend a new jitter formula characterised by very smooth and even distribution of retry intervals, a well-controlled median initial retry delay, and broadly exponential backoff.
var delay = Backoff.DecorrelatedJitterBackoffV2(medianFirstRetryDelay: TimeSpan.FromSeconds(1), retryCount: 5); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
Median initial retry delay: The median (50th percentile) of the first retry delay generated by the formula will be close to the configured value.
Broadly exponential characteristic: Averaged over a suitably large sample, the medians of the timing of subsequent retries will be found to fall close to a pattern of 2x, 4x, 8x etc the median of the initial retry delay.
The smoothness and even distribution of the jitter generated by the new recommended formula is illustrated below, and third-from-bottom (dark red line) here, compared against other approaches.
(Note: The distribution is continuous; discontinuity in the graph above is only an effect of the granularity of bucketing samples.)
The Polly team previously recommended the widely-referenced jitter strategy described here.
For completeness and for those wanting continuity with previous implementations, this is still available in Polly.Contrib.WaitAndRetry:
var delay = Backoff.AwsDecorrelatedJitterBackoff(minDelay: TimeSpan.FromMilliseconds(10), maxDelay: TimeSpan.FromMilliseconds(100), retryCount: 5); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
This will set up a policy that will retry five times. Each retry will delay for a random amount of time between the minimum of 10ms and the maximum of 100ms.
The distribution characteristic of this formula is illustrated fifth-from-top (purple line), here.
With both jitter formulae, note that unlike the linear and exponent wait-and-retry helpers, each subsequent retry delay in the jitter providers does not have a deterministic relationship to the previous. Due to the intentional jitter, the fourth retry delay may (for example) be significantly shorter (or longer), than the first or third.
However, both jitter formulae do tend to increasing backoff. That is, averaged over a suitable sample, the timing of later retries will be found to tend to increase compared to earlier retries.
The Polly team reviewed the literature on Random on .Net, including how to ensure optimum randomness and safety in multi-threaded usage.
Internally, both jitter formulae uses a thread-safe, shared
Random to better ensure a random distribution across all calls. You may, optionally, provide your own seed value.
var delay = Backoff.DecorrelatedJitterBackoffV2( medianFirstDelay: TimeSpan.FromSeconds(1), retryCount: 5, seed: 100);
Random will still be used internally in this case, but it will be seeded with your value versus the default used by .NET in a call to
As the new jitter formula has (averaged over a suitably large sample) a
2^n expotential backoff characteristic, configuring a large number of retries will naturally give rise to very long delays. To prevent this, consider imposing a ceiling on retry delays:
var maxDelay = TimeSpan.FromSeconds(45); var delay = Backoff.DecorrelatedJitterBackoffV2(medianFirstRetryDelay: TimeSpan.FromSeconds(1), retryCount: 50) .Select(s => TimeSpan.FromTicks(Math.Min(s.Ticks, maxDelay.Ticks))); var retryPolicy = Policy .Handle<FooException>() .WaitAndRetryAsync(delay);
While the ceiling imposed in the above code example is fixed, this should not generate corrlelated spikes of retries in a real environments, as the jitter provided by preceding tries (iterations of the formula prior to hitting the ceiling) should provide sufficient decorrelation.
Examples in this readme show asynchronous Polly policies, but all backoff helpers in
Polly.Contrib.WaitAndRetry also work with synchronous
All helper methods in Polly.Contrib.WaitAndRetry include an option to retry the first failure immediately. You can trigger this by passing in
fastFirst: true to any of the helper methods.
var delay = Backoff.ConstantBackoff(TimeSpan.FromMilliseconds(200), retryCount: 5, fastFirst: true);
Note, the first retry will happen immediately and it will count against your retry count. That is, this will still retry five times but the first retry will happen immediately.
The logic behind a fast first retry is that a failure may just have been a transient blip rather than reflecting a deeper underlying issue. In that case, trying again with no delay can be the fastest route to success. In this view, it is worth starting to back off (introduce delays) after you have had two failures, indicating a more serious underlying problem.