Example 9 - Failure

Summary

This is the same as the previous example but we disable transparent inputs. The transaction fails because it cannot pay for the fees.

Let’s consider the following orders:

The number is the quantity for a given address type.

• Order 1 is a t-addr for 10

Suppose we have the following notes:

• T: 0
• S: 12
• O: 10

Note Selection

Let’s assume we choose (Sapling, Orchard, Transparent) in the Pool Usage Order.

We begin with (0, 12, 10) in each pool respectively.

• Order 1: we use 10 from S-pool. Now we have (0, 2, 10)

Fee ZIP 327

In the T-pool, we have

• 0 input,
• 1 output for Order #1

In the S-pool, we have

• 1 input,
• 0 output

In the O-pool, we have

• 0 input,
• 0 output

1. T-pool contributes 1 logical actions = max(0, 1).
2. S-pool contributes 1 logical actions = max(1, 0).
3. O-pool contributes 0 logical actions

The number of logical actions = 2 and the fee is 2 * 5 = 10

However, we haven’t paid for the fee, and we haven’t considered the change outputs yet.

Paying for the fee and making change outputs

We pay for the fee using 2 from S-pool and 8 from O-pool, following our pool usage preferences.

We also have change outputs:

• S: we used 12, pay 10 to Order #1, pay 2 in fees, we get 0 change
• O: we used 10, pay 8 in fees, we get 2 in change

These outputs add up to the transaction and modify the fee.

1. T-pool contributes 1 logical actions = max(0, 1).
2. S-pool contributes 1 logical actions = max(1, 0).
3. O-pool contributes 1 logical actions

The number of logical actions = 3 and the fee is 3 * 5 = 15

Let’s adjust for the new fee.

But we cannot pay the additional 5 in fees since we only have 2 left.

Final transaction

Transaction failed because of unsufficient funds (cannot pay fees)

If we add another orchard note worth 10, we get the final transaction:

• Inputs: all notes are used
  • S: 12
  • O: 10, 10
• Outputs:
  • Order 1: T/10
  • Change: O/2
• Fee: 20

The analysis is left as an exercise.

Hint:

1. T-pool contributes 1 logical actions = max(0, 1).
2. S-pool contributes 1 logical actions = max(1, 0).
3. O-pool contributes 2 logical actions = max(2, 1)