test: scale amounts in test_doublespend_tree down by factor 10

This is done in order to prepare the make_utxo helper to use MiniWallet,
which only supports creating transactions with single inputs, i.e. we
need to create amounts small enough to be funded by coinbase transactions
(50 BTC).

Author: theStack

test/functional/feature_rbf.py

@@ -201,10 +201,10 @@ def test_doublespend_chain(self):
     def test_doublespend_tree(self):
         """Doublespend of a big tree of transactions"""
 
-        initial_nValue = 50 * COIN
+        initial_nValue = 5 * COIN
         tx0_outpoint = self.make_utxo(self.nodes[0], initial_nValue)
 
-        def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001 * COIN, _total_txs=None):
+        def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.00001 * COIN, _total_txs=None):
             if _total_txs is None:
                 _total_txs = [0]
             if _total_txs[0] >= max_txs:
@@ -235,7 +235,7 @@ def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001 * COIN, _to
                                   _total_txs=_total_txs):
                     yield x
 
-        fee = int(0.0001 * COIN)
+        fee = int(0.00001 * COIN)
         n = MAX_REPLACEMENT_LIMIT
         tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
         assert_equal(len(tree_txs), n)
@@ -248,10 +248,10 @@ def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001 * COIN, _to
         # This will raise an exception due to insufficient fee
         assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, 0)
 
-        # 1 BTC fee is enough
+        # 0.1 BTC fee is enough
         dbl_tx = CTransaction()
         dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
-        dbl_tx.vout = [CTxOut(initial_nValue - fee * n - 1 * COIN, DUMMY_P2WPKH_SCRIPT)]
+        dbl_tx.vout = [CTxOut(initial_nValue - fee * n - int(0.1 * COIN), DUMMY_P2WPKH_SCRIPT)]
         dbl_tx_hex = dbl_tx.serialize().hex()
         self.nodes[0].sendrawtransaction(dbl_tx_hex, 0)
 
@@ -264,7 +264,7 @@ def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001 * COIN, _to
         # Try again, but with more total transactions than the "max txs
         # double-spent at once" anti-DoS limit.
         for n in (MAX_REPLACEMENT_LIMIT + 1, MAX_REPLACEMENT_LIMIT * 2):
-            fee = int(0.0001 * COIN)
+            fee = int(0.00001 * COIN)
             tx0_outpoint = self.make_utxo(self.nodes[0], initial_nValue)
             tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
             assert_equal(len(tree_txs), n)