@Benchmark
  int hashMultiset(int reps) {
    int sum = 0;
    for (int i = 0; i < reps; i++) {
      for (Object value : hashMultiset) {
        sum += value.hashCode();
      }
    }
    return sum;
  }

  @Benchmark
  int linkedHashMultiset(int reps) {
    int sum = 0;
    for (int i = 0; i < reps; i++) {
      for (Object value : linkedHashMultiset) {
        sum += value.hashCode();
      }
    }

    powerSet = Sets.powerSet(set);
  }

  @Benchmark
  int iteration(int reps) {
    int sum = 0;
    for (int i = 0; i < reps; i++) {
      for (Set<Integer> subset : powerSet) {
        for (Integer value : subset) {
          sum += value;
        }
      }
    }
    return sum;
  }

      return new CacheStats(
          negativeToMaxValue(hitCount.sum()),
          negativeToMaxValue(missCount.sum()),
          negativeToMaxValue(loadSuccessCount.sum()),
          negativeToMaxValue(loadExceptionCount.sum()),
          negativeToMaxValue(totalLoadTime.sum()),
          negativeToMaxValue(evictionCount.sum()));
    }

      }
    }
  }

  /**
   * Returns the sum of all values in this map.
   *
   * <p>This method is not atomic: the sum may or may not include other concurrent operations.
   */
  public long sum() {
    long sum = 0L;
    for (AtomicLong value : map.values()) {
      sum = sum + value.get();
    }
    return sum;
  }

  @LazyInit @CheckForNull private transient Map<K, Long> asMap;

    // We define the arithmetic means X_n = 1/n \sum_{i=1}^n x_i, and Y_n = 1/n \sum_{i=1}^n y_i.
    // We also define the sum of the products of the differences from the means
    //           C_n = \sum_{i=1}^n x_i y_i - n X_n Y_n
    // for all n >= 1. Then for all n > 1:
    //       C_{n-1} = \sum_{i=1}^{n-1} x_i y_i - (n-1) X_{n-1} Y_{n-1}

    CacheStats sum = two.plus(one);
    assertEquals(124, sum.requestCount());
    assertEquals(64, sum.hitCount());
    assertEquals(64.0 / 124, sum.hitRate());
    assertEquals(60, sum.missCount());
    assertEquals(60.0 / 124, sum.missRate());
    assertEquals(56, sum.loadSuccessCount());
    assertEquals(52, sum.loadExceptionCount());
    assertEquals(52.0 / 108, sum.loadExceptionRate());

        sum += array[index].hashCode();
      }
    }
    return sum;
  }

  @Benchmark
  int arrayFor(int reps) {
    int sum = 0;
    for (int i = 0; i < reps; i++) {
      for (Object value : array) {
        sum += value.hashCode();
      }
    }
    return sum;
  }

  @Benchmark
  int arrayListIndexed(int reps) {
    int sum = 0;
    for (int i = 0; i < reps; i++) {

  public void testSum() {
    assertThat(emptyAccumulator.sum()).isWithin(0.0).of(0.0);
    assertThat(emptyAccumulatorByAddAllEmptyIterable.sum()).isWithin(0.0).of(0.0);
    assertThat(emptyAccumulatorByAddAllEmptyStats.sum()).isWithin(0.0).of(0.0);
    assertThat(oneValueAccumulator.sum()).isWithin(ALLOWED_ERROR).of(ONE_VALUE);
    assertThat(oneValueAccumulatorByAddAllEmptyStats.sum()).isWithin(ALLOWED_ERROR).of(ONE_VALUE);

  }

  public void testSum() {
    assertThat(EMPTY_STATS_VARARGS.sum()).isEqualTo(0.0);
    assertThat(EMPTY_STATS_ITERABLE.sum()).isEqualTo(0.0);
    assertThat(ONE_VALUE_STATS.sum()).isWithin(ALLOWED_ERROR).of(ONE_VALUE);
    assertThat(TWO_VALUES_STATS.sum()).isWithin(ALLOWED_ERROR).of(TWO_VALUES_MEAN * 2);
    assertThat(MANY_VALUES_STATS_VARARGS.sum())
        .isWithin(ALLOWED_ERROR)

    // this case only, we can compute the sum as a long without risking overflow or loss of
    // precision. So we do that, as it's slightly quicker than the Knuth algorithm.
    long sum = 0;
    for (int index = 0; index < values.length; ++index) {
      sum += values[index];
    }
    return (double) sum / values.length;
  }

  /**