2.4 Batch Control (Type 8), File Control (Type 9) & Entry Hash Calculations

2.4 Batch Control (Type 8), File Control (Type 9) & Entry Hash Calculations

Control records serve as the cryptographic and arithmetic seals of the ACH Network. Without control records, electronic transmission errors, dropped packets, or fraudulent record injections could silently corrupt banking ledgers.

The Company/Batch Control Record (Type 8) summarizes and seals individual batches, while the File Control Record (Type 9) summarizes and seals the entire ACH file. ACH Operators execute strict, automated balance checks against these records before routing transactions for clearing.

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Field-by-Field Specification: Batch Control Record (Type 8)

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Field-by-Field Specification: File Control Record (Type 9)

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Mathematical Mechanics of the Entry Hash

The Entry Hash is a critical mathematical checksum designed to detect corrupted, missing, transposed, or injected routing numbers during data transmission.

Batch Entry Hash Algorithm

1. For every Type 6 Entry Detail Record in the batch, extract the 8-digit Receiving DFI Identification (Positions 04–11). Do NOT include the 9th check digit (Position 12).
2. Add all 8-digit routing numbers together using standard arithmetic addition.
3. If the sum is 10 or fewer digits, format the result with leading zeros to fill exactly 10 character positions.
4. The 10-Digit Overflow Truncation Rule: If the sum exceeds 10 digits (i.e., reaches 10,000,000,000 or greater), truncate the leftmost overflowing digits and retain only the rightmost (least significant) 10 digits (Modulo $10^{10}$).

$\text{Batch Entry Hash} = \left( \sum_{i=1}^{N_{\text{Type 6}}} \text{RDFI Routing}_{i} \right) \bmod 10^{10}$

File Entry Hash Algorithm

To compute the Entry Hash for the Type 9 File Control record, sum the 10-digit Entry Hash values of every Type 8 Batch Control record in the file, applying the same 10-digit rightmost truncation rule:

$\text{File Entry Hash} = \left( \sum_{j=1}^{N_{\text{Batches}}} \text{Batch Hash}_{j} \right) \bmod 10^{10}$

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Step-by-Step Worked Mathematical Examples

Scenario 1: Standard Batch Calculation (No Overflow)

An ODFI batches 4 payroll credit entries (PPD, Service Class Code 220):

• Entry 1: RDFI Routing 071000013 (Pos 04–11 = 07100001), Amount = $1,500.00 (0000150000 in cents)
• Entry 2: RDFI Routing 121000358 (Pos 04–11 = 12100035), Amount = $2,250.50 (0000225050 in cents)
• Entry 3: RDFI Routing 021000021 (Pos 04–11 = 02100002), Amount = $875.25 (0000087525 in cents)
• Entry 4: RDFI Routing 091000019 (Pos 04–11 = 09100001), Amount = $3,100.00 (0000310000 in cents)

Calculations:

1. Entry Count: 4 Type 6 records + 0 Type 7 records = 000004 (Pos 05–10)
2. Debit Dollar Total: 000000000000 ($0.00 in cents, Pos 21–32)
3. Credit Dollar Total: $150,000 + 225,050 + 87,525 + 310,000 = 772,575 \text{ cents} = \mathbf{000000772575} \text{ (Pos 33–44)}$
4. Batch Entry Hash: $\text{Sum} = 07100001 + 12100035 + 02100002 + 09100001 = 30,410,039$ $\text{10-Digit Zero-Padded Hash} = \mathbf{0030410039} \text{ (Pos 11–20)}$

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Scenario 2: High-Volume Commercial Batch (With 10-Digit Overflow)

A large corporate origination batch contains 250,000 debit entries. Standard addition of all 250,000 8-digit routing numbers produces: $\text{Raw Arithmetic Sum} = \mathbf{18,459,283,741,290} \text{ (14 Digits)}$

Truncation Calculation:

1. Identify the rightmost 10 digits: 9283741290.
2. Discard the 4 leftmost overflowing digits (1845).
3. Value placed in Positions 11–20 of Type 8: 9283741290.

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Scenario 3: Multi-Batch File Control Accumulation

Suppose an ACH file contains two batches:

• Batch 1 (Scenario 1): Batch Hash = 0030410039, Debits = $0.00 (000000000000), Credits = $7,725.75 (000000772575), Entry Count = 4
• Batch 2 (Scenario 2): Batch Hash = 9283741290, Debits = $14,500,000.00 (001450000000), Credits = $0.00 (000000000000), Entry Count = 250,000

File Control (Type 9) Field Values:

• Batch Count (Pos 02–07): 000002 (2 batches)
• Entry / Addenda Count (Pos 14–21): 00250004 (250,004 total detail records)
• Total Debit Dollar Amount (Pos 32–43): 001450000000 ($14,500,000.00 in cents)
• Total Credit Dollar Amount (Pos 44–55): 000000772575 ($7,725.75 in cents)
• File Entry Hash (Pos 22–31): $\text{Sum} = 0030410039 + 9283741290 = \mathbf{9314151329} \text{ (Pos 22–31)}$

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Block Count Calculation Formula

The Block Count field (Positions 08–13 of Type 9) verifies that physical blocks were not truncated during transmission.

$\text{Total Logical Records} = N_{\text{Type 1}} + N_{\text{Type 5}} + N_{\text{Type 6}} + N_{\text{Type 7}} + N_{\text{Type 8}} + N_{\text{Type 9}} + N_{\text{Type 99}}$ $\text{Block Count} = \frac{\text{Total Logical Records}}{10} = \lceil \frac{\text{Active Records (Types 1–9)}}{10} \rceil$

Worked Example: Block Count

In our two-batch file above:

• 1 File Header (Type 1)
• 2 Batch Headers (Type 5)
• 250,004 Entry Details (Type 6)
• 0 Addenda Records (Type 7)
• 2 Batch Controls (Type 8)
• 1 File Control (Type 9)

$\text{Active Records} = 1 + 2 + 250,004 + 0 + 2 + 1 = 250,010 \text{ records}$ $\text{Block Count} = \frac{250,010}{10} = \mathbf{25,001} \text{ blocks (Value in Pos 08–13 = `025001`)}$ Since 250,010 is an exact multiple of 10, zero nines padding records are required.