What Are BPA and WPA? Calculating the Time You Need on Your Last Solve for a Target ao5
With four solves of a competition ao5 done, everyone wants to know: “What do I need on the fifth to get a sub-15 average?” BPA (Best Possible Average) and WPA (Worst Possible Average) put numbers on that. This article explains what BPA and WPA mean and how to calculate them, shows with worked examples how to find the fifth-solve time you need for a target ao5, and covers how to handle a DNF and how to decide whether to push or play safe on the last solve at a competition.
BPA and WPA: how far your fifth solve can move your ao5
An ao5 is the mean of the middle three of five solves, with the fastest and slowest removed (for the basics, see how to calculate ao5 and ao12). Once four solves are done, whatever the fifth time is, the ao5 will fall within a certain range. The two ends of that range are BPA and WPA.
| Abbreviation | Full name | Meaning |
|---|---|---|
| BPA | Best Possible Average | Your ao5 if the fifth solve is at least as fast as your fastest of the four. It can’t get better than this |
| WPA | Worst Possible Average | Your ao5 if the fifth solve is at least as slow as your slowest, or a DNF. It can’t get worse than this |
In other words, after four solves your ao5 is locked in “somewhere between BPA and WPA”. If the BPA is slower than your target, no fifth solve, however fast, will get you there; if the WPA is faster than your target, you’ve made it even if the fifth is a DNF. That’s why competition streams and timer apps show BPA and WPA.
How to calculate BPA and WPA
Sort the four times from fastest to slowest as a1, a2, a3, a4. If the fifth time is x, the ao5 takes one of three forms depending on where x falls.
| Where the fifth solve x falls | The two removed | ao5 |
|---|---|---|
| x is faster than a1 | x (fastest) and a4 (slowest) | (a1 + a2 + a3) ÷ 3 = BPA |
| x is between a1 and a4 | a1 (fastest) and a4 (slowest) | (a2 + a3 + x) ÷ 3 |
| x is slower than a4, or a DNF | a1 (fastest) and x (slowest) | (a2 + a3 + a4) ÷ 3 = WPA |
As formulas: BPA = (a1 + a2 + a3) ÷ 3 and WPA = (a2 + a3 + a4) ÷ 3. The middle two, a2 and a3, always count toward the average whatever the fifth solve is, so they’re the foundation of the ao5. It helps to think of the fifth solve as only deciding where it lands between a1 and a4.
For example, say your first four are 12.40, 11.20, 14.10 and 12.90. Sorted fastest first: a1 = 11.20, a2 = 12.40, a3 = 12.90, a4 = 14.10.
- BPA = (11.20 + 12.40 + 12.90) ÷ 3 = 36.50 ÷ 3 = 12.166… → 12.17
- WPA = (12.40 + 12.90 + 14.10) ÷ 3 = 39.40 ÷ 3 = 13.133… → 13.13
- If the fifth is 12.00: (12.40 + 12.90 + 12.00) ÷ 3 = 12.433… → an ao5 of 12.43
With these four, the ao5 will land between 12.17 and 13.13 depending on the fifth. However fast it is, it won’t beat 12.17, and even a DNF won’t make it worse than 13.13.
Finding the fifth-solve time you need for a target ao5
If your target ao5 is T, solving (a2 + a3 + x) ÷ 3 ≤ T gives the fifth-solve time you need: x ≤ 3T − a2 − a3. Averages are rounded to the nearest hundredth, though, so right at the edge you get one hundredth of a second of extra leeway.
| Target ao5 | Verdict | Fifth solve needed |
|---|---|---|
| 12.00 (sub-12) | BPA 12.17 is slower than the target | Not reachable |
| 12.50 | BPA ≤ 12.50 < WPA | 12.21 or faster |
| 13.00 (sub-13) | BPA ≤ 13.00 < WPA | 13.71 or faster |
| 13.50 | WPA 13.13 is faster than the target | Achieved whatever the time (even a DNF) |
Let’s check the 12.50 target. 3 × 12.50 − 12.40 − 12.90 = 12.20, but even a fifth solve of 12.21 gives (12.40 + 12.90 + 12.21) ÷ 3 = 12.503…, which rounds to 12.50 and hits the target. 12.22 gives 12.506… → 12.51 and misses. Knowing the boundary including rounding saves you from celebrating or despairing over a hundredth of a second.
BPA and WPA when there’s a DNF in the first four
A DNF counts as the slowest result and takes the a4 slot. For example, with 12.40, DNF, 11.20 and 12.90: a1 = 11.20, a2 = 12.40, a3 = 12.90, a4 = DNF.
- BPA = (11.20 + 12.40 + 12.90) ÷ 3 = 12.17 (the DNF is removed as the slowest)
- WPA = (12.40 + 12.90 + DNF) ÷ 3 = DNF (if the fifth is also a DNF, two DNFs make the ao5 a DNF)
- If the fifth is a completed solve, however slow, the DNF is removed and the average is (a2 + a3 + x) ÷ 3
If you already have one DNF, the top priority on the fifth solve is “finish rather than be fast”. As long as you complete it, you keep an average, and since the DNF is dropped as the slowest, even a slowish time only affects the average through (a2 + a3 + x) ÷ 3. With two DNFs, the ao5 is a DNF regardless of the fifth solve.
Push or play safe on the fifth solve at a competition?
BPA and WPA are also numbers for deciding your strategy on the fifth solve. Splitting the situation after four solves into these three cases makes the decision easier.
| Situation | How to think about it |
|---|---|
| WPA is within the target | The target is secured. Go all out for a personal-best single |
| BPA ≤ target < WPA | Check the time you need and solve carefully at a matching pace. Rushing into a mistake will cost you |
| BPA is slower than the target | The target is out of reach. The ao5 will land between BPA and WPA, so switch to your next goal (a PB or your placing) |
Some competitions also have a cutoff: unless one of your first two solves beats a set time, you don’t get to do the remaining three. In that case, playing safe through the second solve matters too. Penalties such as going over inspection time apply without mercy on the fifth solve as well, so also review the 15-second inspection rule, +2 and DNF.
Getting BPA, WPA and the time you need automatically
Doing this in your head every time is a chore, so once four results are in, the Rubik’s Cube Timer shows BPA and WPA under “What will your ao5 be after the next solve?”. Enter a target ao5 and it calculates, including the rounding boundary, how fast the next solve needs to be, whether any time will do, or whether even the fastest can’t reach it.
- 1
During practice, on the timer screen
From your last four results, the BPA, WPA and time needed for the next solve appear automatically. You can recreate “fifth-solve pressure” even in practice.
- 2
Mid-competition, in the calculator tab
Paste four times (e.g. 12.40, 11.20, 14.10, 12.90) into the “ao5/ao12 calculator” tab to see the ao5 depending on the fifth solve and the time you need.
- 3
For a “sub-X” goal, enter it minus a hundredth
If you’re going for sub-15, enter 14.99 to get the time needed for the “under 15 seconds” condition.
How to set target times and structure practice to bring down your ao5 is covered in how to use a cube timer and practice to improve your times.
FAQ
- What do BPA and WPA mean?
- BPA stands for Best Possible Average and WPA for Worst Possible Average. After four solves of an ao5, BPA is your ao5 if the fifth is the fastest, and WPA is your ao5 if the fifth is the slowest or a DNF.
- How are BPA and WPA calculated?
- Sort the four times fastest first as a1 to a4. BPA is (a1 + a2 + a3) ÷ 3 and WPA is (a2 + a3 + a4) ÷ 3. For 11.20, 12.40, 12.90 and 14.10, the BPA is 12.17 and the WPA is 13.13.
- What time do I need on the fifth solve to hit my target ao5?
- With target T and the middle two of the four as a2 and a3, the fifth solve needs to be roughly 3T − a2 − a3 or faster. Averages are rounded to the nearest hundredth, so in practice you have a hundredth of leeway. If the BPA is slower than the target you can’t reach it; if the WPA is within the target, any time will do.
- What happens to the WPA if I have one DNF in the first four?
- The WPA becomes a DNF, because a DNF on the fifth would make two DNFs and the ao5 a DNF. If you complete the fifth solve, the DNF is removed as the slowest and you keep an average.
- If I’m aiming for sub-15, can I use 15.00 as my target ao5?
- Sub-15 means “under 15 seconds”, so exactly 15.00 doesn’t count. Using 14.99 as the target is the accurate way to calculate it.
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