11 characters at the end? Should take seconds, not hours or days.
It depends on whether the key is compressed or uncompressed. For a compressed key you have to check about 22 million while for an uncompressed key the number is 5.8 billion.
// Numbers are approximates, values usually are ±1
// Uncompressed ; Compressed
// 1-5 -> 1 ; 1
// 6 -> 9 ; 1
// 7 -> 514 ; 3
// 8 -> 29,817 ; 117
// 9 -> 1,729,387 ; 6,756
// 10 -> 100,304,420 ; 391,815
// 11 -> 5,817,656,406 ; 22,725,222
// 12 -> ; 1,318,062,780
For the first case it takes about 2 minutes to check all the keys using FinderOuter
Ok, that might seem like a lot for your program, the FinderOuter, but I could check 5.8 billion in seconds. It's like people want to make finding 11 missing characters from the end harder, or only use a program that they know or have used before. Now all I can wonder is why the merit for you, for just telling how long it would take for you to find it with a slower program? Who knows...