Bir **Web API** için bu süreç normalde uygulamayı **uzak bir makineye** yerleştirmeyi, iyi performans, kararlılık vb. özellikler sağlayan bir **sunucu programı** ile **kullanıcılarınızın** uygulamaya etkili ve kesintisiz bir şekilde **erişebilmesini** kapsar.

	// Tolerance is not set, defaults to half of the locker clients.
	tolerance := len(restClnts) / 2

	// Quorum is effectively = total clients subtracted with tolerance limit
	quorum := len(restClnts) - tolerance
	if !isReadLock {
		// In situations for write locks, as a special case
		// to avoid split brains we make sure to acquire
		// quorum + 1 when tolerance is exactly half of the
		// total locker clients.

estivéramos
esteja
estejamos
estejam
estivesse
estivéssemos
estivessem
estiver
estivermos
estiverem
hei
há
havemos
hão
houve
houvemos
houveram
houvera
houvéramos
haja
hajamos
hajam
houvesse
houvéssemos
houvessem
houver
houvermos
houverem
houverei
houverá
houveremos
houverão
houveria
houveríamos
houveriam
sou
somos
são

Existen varias formas de hacerlo dependiendo de tu caso de uso específico y las herramientas que uses.

Puedes **desplegar un servidor** tú mismo usando un conjunto de herramientas, puedes usar **servicios en la nube** que haga parte del trabajo por ti, o usar otras posibles opciones.

MinIO protects data against hardware failures and silent data corruption using erasure code and checksums. With the highest level of redundancy, you may lose up to half (N/2) of the total drives and still be able to recover the data.

## What is Erasure Code?

  private final long k1;

  /**
   * @param c the number of compression rounds (must be positive)
   * @param d the number of finalization rounds (must be positive)
   * @param k0 the first half of the key
   * @param k1 the second half of the key
   */
  SipHashFunction(int c, int d, long k0, long k1) {
    checkArgument(
        c > 0, "The number of SipRound iterations (c=%s) during Compression must be positive.", c);

   *   <li>awaits executor service termination for half of the specified timeout.
   *   <li>if the timeout expires, it calls {@link ExecutorService#shutdownNow()}, cancelling
   *       pending tasks and interrupting running tasks.
   *   <li>awaits executor service termination for the other half of the specified timeout.
   * </ol>
   *

      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

      return entry.value;
    }
    entry = cacheEntry2;
    if (entry != null && entry.key == key) {
      // Promote second cache entry to first so the access pattern
      // [K1, K2, K1, K3, K1, K4...] still hits the cache half the time.
      addToCache(entry);
      return entry.value;
    }
    return null;
  }

  @Override
  void clearCache() {
    super.clearCache();
    cacheEntry1 = null;
    cacheEntry2 = null;
  }

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