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diff --git a/java/com/android/dialer/common/backoff/ExponentialBaseCalculator.java b/java/com/android/dialer/common/backoff/ExponentialBaseCalculator.java
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+++ b/java/com/android/dialer/common/backoff/ExponentialBaseCalculator.java
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+/*
+ * Copyright (C) 2017 The Android Open Source Project
+ *
+ * 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 com.android.dialer.common.backoff;
+
+import com.android.dialer.common.Assert;
+
+/**
+ * Given an initial delay, D, a maximum total backoff time, T, and a maximum number of backoffs, N,
+ * this class calculates a base multiplier, b, and a scaling factor, g, such that the initial
+ * backoff is g*b = D and the sum of the scaled backoffs is T = g*b + g*b^2 + g*b^3 + ... g*b^N
+ *
+ * <p>T/D = (1 - b^N)/(1 - b) but this cannot be written as a simple equation for b in terms of T, N
+ * and D so instead use Newton's method (https://en.wikipedia.org/wiki/Newton%27s_method) to find an
+ * approximate value for b.
+ *
+ * <p>Example usage using the {@code ExponentialBackoff} would be:
+ *
+ * <pre>
+ *   // Retry with exponential backoff for up to 2 minutes, with an initial delay of 100 millis
+ *   // and a maximum of 10 retries
+ *   long initialDelayMillis = 100;
+ *   int maxTries = 10;
+ *   double base = ExponentialBaseCalculator.findBase(
+ *       initialDelayMillis,
+ *       TimeUnit.MINUTES.toMillis(2),
+ *       maxTries);
+ *   ExponentialBackoff backoff = new ExponentialBackoff(initialDelayMillis, base, maxTries);
+ *   while (backoff.isInRange()) {
+ *     ...
+ *     long delay = backoff.getNextBackoff();
+ *     // Wait for the indicated time...
+ *   }
+ * </pre>
+ */
+public final class ExponentialBaseCalculator {
+  private static final int MAX_STEPS = 1000;
+  private static final double DEFAULT_TOLERANCE_MILLIS = 1;
+
+  /**
+   * Calculate an exponential backoff base multiplier such that the first backoff delay will be as
+   * specified and the sum of the delays after doing the indicated maximum number of backoffs will
+   * be as specified.
+   *
+   * @throws IllegalArgumentException if the initial delay is greater than the total backoff time
+   * @throws IllegalArgumentException if the maximum number of backoffs is not greater than 1
+   * @throws IllegalStateException if it fails to find an acceptable base multiplier
+   */
+  public static double findBase(
+      long initialDelayMillis, long totalBackoffTimeMillis, int maximumBackoffs) {
+    Assert.checkArgument(initialDelayMillis < totalBackoffTimeMillis);
+    Assert.checkArgument(maximumBackoffs > 1);
+    long scaledTotalTime = Math.round(((double) totalBackoffTimeMillis) / initialDelayMillis);
+    double scaledTolerance = DEFAULT_TOLERANCE_MILLIS / initialDelayMillis;
+    return getBaseImpl(scaledTotalTime, maximumBackoffs, scaledTolerance);
+  }
+
+  /**
+   * T/D = (1 - b^N)/(1 - b) but this cannot be written as a simple equation for b in terms of T, D
+   * and N so instead we use Newtons method to find an approximate value for b.
+   *
+   * <p>Let f(b) = (1 - b^N)/(1 - b) - T/D then we want to find b* such that f(b*) = 0, or more
+   * precisely |f(b*)| < tolerance
+   *
+   * <p>Using Newton's method we can interatively find b* as follows: b1 = b0 - f(b0)/f'(b0), where
+   * b0 is the current best guess for b* and b1 is the next best guess.
+   *
+   * <p>f'(b) = (f(b) + T/D - N*b^(N - 1))/(1 - b)
+   *
+   * <p>so
+   *
+   * <p>b1 = b0 - f(b0)(1 - b0)/(f(b0) + T/D - N*b0^(N - 1))
+   */
+  private static double getBaseImpl(long t, int n, double tolerance) {
+    double b0 = 2; // Initial guess for b*
+    double b0n = Math.pow(b0, n);
+    double fb0 = f(b0, t, b0n);
+    if (Math.abs(fb0) < tolerance) {
+      // Initial guess was pretty good
+      return b0;
+    }
+
+    for (int i = 0; i < MAX_STEPS; i++) {
+      double fpb0 = fp(b0, t, n, fb0, b0n);
+      double b1 = b0 - fb0 / fpb0;
+      double b1n = Math.pow(b1, n);
+      double fb1 = f(b1, t, b1n);
+
+      if (Math.abs(fb1) < tolerance) {
+        // Found an acceptable value
+        return b1;
+      }
+
+      b0 = b1;
+      b0n = b1n;
+      fb0 = fb1;
+    }
+
+    throw new IllegalStateException("Failed to find base. Too many iterations.");
+  }
+
+  // Evaluate f(b), the function we are trying to find the zero for.
+  // Note: passing b^N as a parameter so it only has to be calculated once
+  private static double f(double b, long t, double bn) {
+    return (1 - bn) / (1 - b) - t;
+  }
+
+  // Evaluate f'(b), the derivative of the function we are trying to find the zero for.
+  // Note: passing f(b) and b^N as parameters for efficiency
+  private static double fp(double b, long t, int n, double fb, double bn) {
+    return (fb + t - n * bn / b) / (1 - b);
+  }
+}