Error (170040): Can't place all RAM cells in design

[Altera_Forum]

Dear All.... 

  I have an issue here.... quartus gives this error while compiling... my compilation report is saying 74% of memory used. But i am getting this error still !!!! Is there anything i can do to avoid this and use the whole memory ???? or is this unavoidable !!? Kindly help asap, its a bit urgent. Thanks in advance.  The design contains NIOS II with two processors using 203776 bits... and around 14 ROMs  error:  Error (170040): Can't place all RAM cells in design Info (170034): Selected device has 66 memory locations of type M9K. The current design requires 70 memory locations of type M9K to successfully fit.  Info (170033): Memory usage required for the design in the current device: 106% M9K memory block locations required   compilation report:  Flow Status Flow Failed - Tue Jul 23 10:27:53 2013 Quartus II 32-bit Version 12.0 Build 232 07/05/2012 SP 1 SJ Web Edition Revision Name qcon_v0 Top-level Entity Name qcon_v0 Family Cyclone III Device EP3C25Q240C8 Timing Models Final Total logic elements 17,352 / 24,624 ( 70 % ) Total combinational functions 14,339 / 24,624 ( 58 % ) Dedicated logic registers 8,059 / 24,624 ( 33 % ) Total registers 8059 Total pins 130 / 149 ( 87 % ) Total virtual pins 0 total memory bits 450,288 / 608,256 ( 74 % ) Embedded Multiplier 9-bit elements 132 / 132 ( 100 % ) Total PLLs 1 / 4 ( 25 % )

[Altera_Forum]

its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)

[Altera_Forum]

--- Quote Start ---  its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)  --- Quote End ---    Ok.. so i must check the individual ROM's and try to make use of whole M9K blocks and try to avoid overflowing into a new M9K blocks. am i right ? i shall try it, thanks Tricky...

[Altera_Forum]

--- Quote Start ---  its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)  --- Quote End ---    Hello,in my design I use M4K with the depth of 256 and the width of 13, but after fitter processing I saw in the "fitter ram summary" the so-called implentation depth of 256 and the implentation width of 2, where is the other 11 bits, so strange? Waiting the answer.Thank you!