<p>Short answer: For big players the cost is mostly negligible.</p>

<p>Longer answer (independent of the current block size debate):</p>

<p>Hard Drive Cost:
A bigger block will take up more disk space. Right now the block size is limited to 1 MB. If the block size is increased to 2 MB and assuming all blocks are always being filled to the max size then you will need about twice the hard drive space to store the same amount of 2 MB blocks as you needed for 1MB blocks.</p>

<p>Network Bandwidth Cost:
If block size is increased, a full node has to relay a bigger size block. Sending and receiving more data will have an impact on cost as well.</p>

<p>Real Time Computational Cost:
A full node has to hash a block and verify that the hash has n amount of leading zeros. Increasing the block size will result in hashing a larger amount of data which increases verification time.</p>

<p>Big O Complexity:
Big O would depend on whatever the Big O is for sha256.</p>

CC BY-SA 3.0
2016-04-11T06:56:09.163Z
You're forgetting: signature validation (which is currently still the majority of the CPU cost) scales linearly with the number of hashes. Signature hash computation scales O(num_transactions * avg_transaction_size^2). Database lookups/updates scale O(inputs) and O(outputs), and each will get slower over time as the UTXO set grows.
Pieter Wuille
CC BY-SA 4.0
2019-12-18T15:34:07.047Z
Only the block header is hashed for the nonce puzzle, so increasing the block size can only increase the transaction merkle root computation cost.
HappyFace
CC BY-SA 3.0
2016-02-21T02:55:10.630Z
2016-02-21T02:55:10.630Z
boris
<p>Does it cost more storage?
For the transaction verification, to which extent bigger blocks make it more computational intensive?
In terms of computational complexity, is the verification process a O(n) computation or anything else?</p>

<nodes><transaction-verification><block-size-increase>
Why do bigger blocks make it more expensive to run a full node?
czheo
# Why do bigger blocks make it more expensive to run a full node?

(Asked by: username_0 on 2016-02-16)

Does it cost more storage?
For the transaction verification, to which extent bigger blocks make it more computational intensive?
In terms of computational complexity, is the verification process a O(n) computation or anything else?

**Tags:** nodes, transaction-verification, block-size-increase


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## 1 answers

### Answer by username_1 on 2016-02-21. Score: 2

Short answer: For big players the cost is mostly negligible.

Longer answer (independent of the current block size debate):

Hard Drive Cost:
A bigger block will take up more disk space. Right now the block size is limited to 1 MB. If the block size is increased to 2 MB and assuming all blocks are always being filled to the max size then you will need about twice the hard drive space to store the same amount of 2 MB blocks as you needed for 1MB blocks.

Network Bandwidth Cost:
If block size is increased, a full node has to relay a bigger size block. Sending and receiving more data will have an impact on cost as well.

Real Time Computational Cost:
A full node has to hash a block and verify that the hash has n amount of leading zeros. Increasing the block size will result in hashing a larger amount of data which increases verification time.

Big O Complexity:
Big O would depend on whatever the Big O is for sha256.

> **Comments:**
> **username_2**: You're forgetting: signature validation (which is currently still the majority of the CPU cost) scales linearly with the number of hashes. Signature hash computation scales O(num_transactions * avg_transaction_size^2). Database lookups/updates scale O(inputs) and O(outputs), and each will get slower over time as the UTXO set grows.  
> **username_3**: Only the block header is hashed for the nonce puzzle, so increasing the block size can only increase the transaction merkle root computation cost.  

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