RBP Access Arrangement Auction Workshop

RBP Access Arrangement Auction Workshop Brisbane 17 May 2012

Current state of play • APTPPL proposed moving from a First Come First Served (FCFS) Queue to an NPV-ranked auction for spare and developable capacity • AER has rejected APTPPL’s proposed approach on a number of grounds • Requires reversion to two FCFS Queues (Spare Capacity and Developable Capacity). • APTPPL must file a Revised Proposal by 25 May 2012 • Purpose of today’s workshop: • To work through various auction designs to define the best option • To identify issues to be addressed in APTPPL’s Revised Proposal • Goal is for APTPPL to submit a revised proposal that is acceptable to both the AER and shippers

NERA Report • APTPPL engaged economics firm NERA to advise on auction processes and structures • This report was not considered by the AER in reaching its draft decision • This report is on the AER’s web site • Ann Whitfield from NERA will discuss this report and its findings • [review Ann Whitfield’s presentation here]

 ---------- Price ----------  Worked example Adam (50 x $10) $500 A capacity auction is held in which seven bids are received. Each bidder defines its own capacity needs and the price it is willing to pay for that capacity.  Quantity Area = Bid Value Bob (30 x $8) $240 Chris (25 x $9) $225 Assume a pipeline with 100 units of available capacity Derek (15 x $10) $150  100 units of capacity  Edith (15 x $9) $135 Fred (10 x $10) $100 Ginger (10 x $9) $90

Total value ranking approach: Adam (50 x $10) $500 Step 1: Arrange bids in descending revenue order. 1 Bob (30 x $8) $240 2 Chris (25 x $9) $225 3 Derek (15 x $10) $150 4  100 units of capacity  Edith (15 x $9) $135 5 Fred (10 x $10) $100 6 Ginger (10 x $9) $90 7

Ranking approach: Adam (50 x $10) $500 Step 2: Assign capacity in that order. 1 Step 3: Stop assigning capacity as soon as the “next largest” cannot be served. Utilisation 80% Revenue $740 Bob (30 x $8) $240 2 Chris (25 x $9) $225 20 units available capacity 3 25 Derek (15 x $10) $150 4  100 units of capacity  Edith (15 x $9) $135 5 Fred (10 x $10) $100 6 Ginger (10 x $9) $90 7

Advanced Ranking 1: Adam (50 x $10) $500 Step 2: Assign capacity in that order. 1 Step 3: Skip any bids that are too big to serve in full. Utilisation 95% Revenue $890 Step 4: Repeat until no more full bids can be served. Bob (30 x $8) $240 2 Chris (25 x $9) $225 5 units available capacity 20 units available capacity 3 25 Derek (15 x $10) $150 4  100 units of capacity  Edith (15 x $9) $135 5 Fred (10 x $10) $100 6 Ginger (10 x $9) $90 7

Advanced Ranking 2: Adam (50 x $10) $500 Step 2: Assign capacity in that order. 1 Step 3: Offer partial capacity to any bids that are too big to serve in full. Utilisation 100% Revenue $920 Workshop: 1) Status of Chris’ bid. 2) Status of unserved request: Interruptible? Priority? Bob (30 x $8) $240 2 5 units unserved capacity Chris (25 x $9) $225 20 units available capacity 3 25 Derek (15 x $10) $150 4  100 units of capacity  Edith (15 x $9) $135 5 Fred (10 x $10) $100 6 Ginger (10 x $9) $90 7

Marginal value ranking approach: Adam (50 x $10) $500 Step 1: Arrange bids in descending price order. 1 Bob (30 x $8) $240 3 Chris (25 x $9) $225 2 Derek (15 x $10) $150 1  100 units of capacity  Edith (15 x $9) $135 2 Fred (10 x $10) $100 1 Ginger (10 x $9) $90 2

Marginal value ranking approach: Adam (50 x $10) $500 Step 2: Assign capacity in that order. 1 Step 3: Same issues as above apply as soon as the “next largest” cannot be served. Not clear how a tie would be decided.The auction would deliver the same value if we accepted Edith and Ginger’s bids instead of Chris’. Utilisation 100% Revenue $975 Bob (30 x $8) $240 3 Chris (25 x $9) $225 2 Derek (15 x $10) $150 1  100 units of capacity  Edith (15 x $9) $135 2 Fred (10 x $10) $100 1 Ginger (10 x $9) $90 2

 ---------- Price ----------  Optimisation approach: Adam (50 x $10) $500  Quantity Step 1: No need to rank bids in any particular order. Bob (30 x $8) $240 Chris (25 x $9) $225 Derek (15 x $10) $150  100 units of capacity  Edith (15 x $9) $135 Fred (10 x $10) $100 Ginger (10 x $9) $90

Optimisation approach: Step 2-∞: Use an optimisation engine to determine the optimal combination of bids delivering the maximum NPV: Utilisation 95% Revenue $890 Chris (25 x $9) $225 5 units available capacity Derek (15 x $10) $150 Bob (30 x $8) $240  100 units of capacity  Adam (50 x $10) $500 Edith (15 x $9) $135 Fred (10 x $10) $100 Ginger (10 x $9) $90

Optimisation approach: Step 2-∞: Use an optimisation engine to determine the optimal combination of bids delivering the maximum NPV: Utilisation 100% Revenue $930 Chris (25 x $9) $225 Ginger (10 x $9) $90 Fred (10 x $10) $100 Bob (30 x $8) $240 Derek (15 x $10) $150  100 units of capacity  Adam (50 x $10) $500 Edith (15 x $9) $135

Optimisation approach: Adam (50 x $10) $500 Step 2-∞: Use an optimisation engine to determine the optimal combination of bids delivering the maximum NPV: Utilisation 100% Revenue $940 Ginger (10 x $9) $90 Note that total revenue is higher, but Adam, the highest value bidder on the day, would miss out under this approach Derek (15 x $10) $150 Bob (30 x $8) $240  100 units of capacity  Chris (25 x $9) $225 Edith (15 x $9) $135 Fred (10 x $10) $100

Optimisation approach: Utilisation 100% Revenue $975 Step 2-∞: Use an optimisation engine to determine the optimal combination of bids delivering the maximum NPV:Q: How do we solve a tie? The auction would deliver the same value if we accepted Edith and Ginger’s bids instead of Chris’. Bob (30 x $8) $240 Lucky for Adam, that was not the optimal solution. But Bob, the second highest value bidder, misses out because there is another combination that generates a higher total revenue. Fred (10 x $10) $100 Derek (15 x $10) $150 Chris (25 x $9) $225  100 units of capacity  Adam (50 x $10) $500 Edith (15 x $9) $135 Ginger (10 x $9) $90

Open forum • Open floor discussion

Delivering Australia’s Energy www.apa.com.au

RBP Access Arrangement Auction Workshop