CRYPTO61674 - Decentralised Finance: Lending and staking: Chargeable Gains: Examples: Example 4: loan of tokens to a platform in exchange for liquidity tokens

Jack wishes to provide liquidity to a Decentralised Finance (DeFi) lending platform. He holds:

  • 100 token A. These are in a section 104 pool with a total acquisition cost of 拢100.
  • 100 token B. These are in a section 104 pool with a total acquisition cost of 拢50.

For more information about section 104 pools see CRYPTO22200.

On 01/06/20XX, Jack transfers 10 token A and 5 token B to the DeFi lending platform. The DeFi lending platform transfers a liquidity token to Jack. At the time of this exchange the token A are valued at 拢1.00 each, the token B are valued at 拢0.60 each. The liquidity token is valued at 拢15.00 each.

Jack decides that a just and reasonable basis for apportioning the value of the liquidity token is in proportion to the market value of the tokens he disposes of. The total market value of the tokens he disposes of is 拢10 (10 token A x 拢1 each) plus 拢3 (5 token B x 拢0.60 each) equals 拢13.

Jack바카라 사이트檚 Chargeable Gains (CG) computation of his disposal of his token A is as follows:

. .
Consideration Liquidity token - 拢15 x 10 / 13 12
Allowable costs Section 104 pool 바카라 사이트� 拢100 x 10 / 100 (10)
Gain . 2

Jack바카라 사이트檚 section 104 pool for token A will be adjusted as follows:

Date Quantity of tokens Allowable costs (拢)
Opening balance 100 100
01/06/20XX (10) (10)
Closing balance 90 90

Jack바카라 사이트檚 CG computation of his disposal of his token B is as follows:

. .
Consideration Liquidity token - 拢15 x 3 / 13 3
Allowable costs Section 104 pool 바카라 사이트� 拢50 x 5 / 100 (3)
Gain . 0

Jack바카라 사이트檚 section 104 pool for token B will be adjusted as follows:

Date Quantity of tokens Allowable costs (拢)
Opening balance 100 50
01/06/20XX (5) (3)
Closing balance 95 47

Jack will be treated as having acquired the liquidity token for the total market value of the token A and token B that Jack transfers to the DeFi lending platform. This is 拢10 (10 token A x 拢1 each) plus 拢3 (5 token B x 拢0.60 each) equals 拢13.