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estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
#!/usr/bin/env python
#
# Test fee estimation code
#
from test_framework import BitcoinTestFramework
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
from bitcoinrpc.authproxy import AuthServiceProxy, JSONRPCException
from util import *
class EstimateFeeTest(BitcoinTestFramework):
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
def setup_network(self, test_dir):
nodes = []
nodes.append(start_node(0, test_dir,
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
["-debug=mempool", "-debug=estimatefee"]))
# Node1 mines small-but-not-tiny blocks, and allows free transactions.
# NOTE: the CreateNewBlock code starts counting block size at 1,000 bytes,
# so blockmaxsize of 2,000 is really just 1,000 bytes (room enough for
# 6 or 7 transactions)
nodes.append(start_node(1, test_dir,
["-blockprioritysize=1500", "-blockmaxsize=2000",
"-debug=mempool", "-debug=estimatefee"]))
connect_nodes(nodes[1], 0)
# Node2 is a stingy miner, that
# produces very small blocks (room for only 3 or so transactions)
node2args = [ "-blockprioritysize=0", "-blockmaxsize=1500",
"-debug=mempool", "-debug=estimatefee"]
nodes.append(start_node(2, test_dir, node2args))
connect_nodes(nodes[2], 0)
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
sync_blocks(nodes)
return nodes
def run_test(self, nodes):
# Prime the memory pool with pairs of transactions
# (high-priority, random fee and zero-priority, random fee)
min_fee = Decimal("0.001")
fees_per_kb = [];
for i in range(12):
(txid, txhex, fee) = random_zeropri_transaction(nodes, Decimal("1.1"),
min_fee, min_fee, 20)
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
tx_kbytes = (len(txhex)/2)/1000.0
fees_per_kb.append(float(fee)/tx_kbytes)
# Mine blocks with node2 until the memory pool clears:
count_start = nodes[2].getblockcount()
while len(nodes[2].getrawmempool()) > 0:
nodes[2].setgenerate(True, 1)
sync_blocks(nodes)
all_estimates = [ nodes[0].estimatefee(i) for i in range(1,20) ]
print("Fee estimates, super-stingy miner: "+str([str(e) for e in all_estimates]))
# Estimates should be within the bounds of what transactions fees actually were:
delta = 1.0e-6 # account for rounding error
for e in filter(lambda x: x >= 0, all_estimates):
if float(e)+delta < min(fees_per_kb) or float(e)-delta > max(fees_per_kb):
raise AssertionError("Estimated fee (%f) out of range (%f,%f)"%(float(e), min_fee_kb, max_fee_kb))
# Generate transactions while mining 30 more blocks, this time with node1:
for i in range(30):
for j in range(random.randrange(6-4,6+4)):
(txid, txhex, fee) = random_transaction(nodes, Decimal("1.1"),
Decimal("0.0"), min_fee, 20)
tx_kbytes = (len(txhex)/2)/1000.0
fees_per_kb.append(float(fee)/tx_kbytes)
nodes[1].setgenerate(True, 1)
sync_blocks(nodes)
all_estimates = [ nodes[0].estimatefee(i) for i in range(1,20) ]
print("Fee estimates, more generous miner: "+str([ str(e) for e in all_estimates]))
for e in filter(lambda x: x >= 0, all_estimates):
if float(e)+delta < min(fees_per_kb) or float(e)-delta > max(fees_per_kb):
raise AssertionError("Estimated fee (%f) out of range (%f,%f)"%(float(e), min_fee_kb, max_fee_kb))
# Finish by mining a normal-sized block:
while len(nodes[0].getrawmempool()) > 0:
nodes[0].setgenerate(True, 1)
sync_blocks(nodes)
final_estimates = [ nodes[0].estimatefee(i) for i in range(1,20) ]
print("Final fee estimates: "+str([ str(e) for e in final_estimates]))
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
10 years ago
if __name__ == '__main__':
EstimateFeeTest().main()