minio/cmd/crypto/retry.go
Andreas Auernhammer 9a703befe6
crypto: reduce retry delay when retrying KES requests (#10394)
This commit reduces the retry delay when retrying a request
to a KES server by:
 - reducing the max. jitter delay from 3s to 1.5s
 - skipping the random delay when there are more KES endpoints
   available.

If there are more KES endpoints we can directly retry to the request
by sending it to the next endpoint - as pointed out by @krishnasrinivas
2020-09-02 11:04:10 -07:00

65 lines
2.5 KiB
Go

// MinIO Cloud Storage, (C) 2020 MinIO, Inc.
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
package crypto
import (
"math/rand"
"time"
)
// default retry configuration
const (
retryWaitMin = 100 * time.Millisecond // minimum retry limit.
retryWaitMax = 1500 * time.Millisecond // 1.5 secs worth of max retry.
)
// LinearJitterBackoff provides the time.Duration for a caller to
// perform linear backoff based on the attempt number and with jitter to
// prevent a thundering herd.
//
// min and max here are *not* absolute values. The number to be multiplied by
// the attempt number will be chosen at random from between them, thus they are
// bounding the jitter.
//
// For instance:
// * To get strictly linear backoff of one second increasing each retry, set
// both to one second (1s, 2s, 3s, 4s, ...)
// * To get a small amount of jitter centered around one second increasing each
// retry, set to around one second, such as a min of 800ms and max of 1200ms
// (892ms, 2102ms, 2945ms, 4312ms, ...)
// * To get extreme jitter, set to a very wide spread, such as a min of 100ms
// and a max of 20s (15382ms, 292ms, 51321ms, 35234ms, ...)
func LinearJitterBackoff(min, max time.Duration, attemptNum int) time.Duration {
// attemptNum always starts at zero but we want to start at 1 for multiplication
attemptNum++
if max <= min {
// Unclear what to do here, or they are the same, so return min *
// attemptNum
return min * time.Duration(attemptNum)
}
// Seed rand; doing this every time is fine
rand := rand.New(rand.NewSource(int64(time.Now().Nanosecond())))
// Pick a random number that lies somewhere between the min and max and
// multiply by the attemptNum. attemptNum starts at zero so we always
// increment here. We first get a random percentage, then apply that to the
// difference between min and max, and add to min.
jitter := rand.Float64() * float64(max-min)
jitterMin := int64(jitter) + int64(min)
return time.Duration(jitterMin * int64(attemptNum))
}