+/* For the specified period, determine the number of seconds
+ * corresponding to the reset time. There will be a watchdog
+ * exception at approximately 3/5 of this time.
+ *
+ * The formula to calculate this is given by:
+ * 2.5 * (2^(63-period+1)) / timebase_freq
+ *
+ * In order to simplify things, we assume that period is
+ * at least 1. This will still result in a very long timeout.
+ */
+static unsigned long long period_to_sec(unsigned int period)
+{
+ unsigned long long tmp = 1ULL << (64 - period);
+ unsigned long tmp2 = ppc_tb_freq;
+
+ /* tmp may be a very large number and we don't want to overflow,
+ * so divide the timebase freq instead of multiplying tmp
+ */
+ tmp2 = tmp2 / 5 * 2;
+
+ do_div(tmp, tmp2);
+ return tmp;
+}
+
+/*
+ * This procedure will find the highest period which will give a timeout
+ * greater than the one required. e.g. for a bus speed of 66666666 and
+ * and a parameter of 2 secs, then this procedure will return a value of 38.
+ */
+static unsigned int sec_to_period(unsigned int secs)
+{
+ unsigned int period;
+ for (period = 63; period > 0; period--) {
+ if (period_to_sec(period) >= secs)
+ return period;
+ }
+ return 0;
+}
+