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Aligned free-block capacity is the usable payload that a particular free extent can supply after meeting memory alignment and metadata requirements. It differs from the raw extent length and from the largest free memory block. For an illustrative block $[a,b)$, header requirement $h$ and payload alignment $A>0$, the first permissible payload start isIf the allocator reserves no trailer or additional rounding, usable payload capacity is $\max (b-p,0)$. This formula belongs to the stated layout; it is not a universal allocator ABI.

For a free interval $[3,67)$ in byte offsets, an8-byte header and16-byte alignment give $p=16$. The raw64-byte extent then has51 bytes after the aligned start. Allocation alignment padding is $p-(a+h)=5$ bytes, in addition to the8-byte header. A48-byte payload fits this model, whereas a52-byte payload does not.

If allocated payload size must also be rounded to a multiple $q$, a requested size $R$ needs $q\lceil R/q\rceil$ usable bytes. For $q=8$ in the same example, the largest such payload is48 bytes: a49-byte request rounds to56 and fails. Header information and allocation granularity can therefore change the answer even when the alignment condition itself is satisfied.^[ref]

For integer addresses, the padding before the payload lies between0 and $A-1$ bytes. Splitting a chosen block can introduce another constraint: a leftover interval may be too small to hold a free record. A capacity calculation must consequently identify the actual header, alignment, rounding and remainder conventions rather than treating every raw free byte as payload.

Allocation alignment padding is the unused space inserted before an allocated payload so its address meets memory alignment. If a free extent starts at integer byte address $a$, an allocation header consumes $h$ bytes, and payload addresses must be multiples of positive integer $A$, the earliest permitted address and its padding areEquivalently, $P$ is the least nonnegative residue of $-(a+h)$ modulo $A$. The alignment padding bound is $0\le P\le A-1$. Already aligned payload starts have zero padding; adding a whole extra alignment unit in that case would waste space unnecessarily.

For an original example, let a free extent be $[3,67)$, with $h=8$ and $A=16$. Then $a+h=11$, $p=16$ and $P=5$ bytes. A24-byte payload ends at40, so the used prefix occupies37 bytes:8 for the header,5 for padding and24 for data. The remaining interval $[40,67)$ has27 bytes before other block-rounding or minimum free-block size requirements are applied. This accounting distinguishes padding from header storage and from payload-size rounding.

The same header and alignment can produce different padding at another address: $a=8$ gives $a+h=16$ and $P=0$. If the residues of $a+h$ were uniformly distributed over all16 possibilities, the mean padding would be7.5 bytes, obtained by averaging0 through15. An actual allocator can constrain its block starts, so that distribution is an explicit model assumption rather than a measured universal overhead.

Aligned free-block capacity must deduct this padding when testing whether a request fits. Allocation alignment padding can thus make two equally long free extents offer different usable capacities. OSTEP's separation of the allocated payload from its bookkeeping header is the basic layout issue behind this calculation; alignment adds the stated address constraint.^[ref]

The largest free memory block is the longest contiguous free region available within a specified allocation pool and address space. If free-block lengths are $s_1,\dots ,s_n$,For an empty free-block set, take $L_{\max }=0$. The largest block and total free capacity answer different questions: a contiguous request can fail even when $F_{\mathrm{free}}$ exceeds its size. This is the central layout issue in external memory fragmentation.^[ref]

For free lengths24,40 and16 bytes, $F_{\mathrm{free}}=80$ bytes and $L_{\max }=40$ bytes. A48-byte contiguous request fails in that fixed, nonmoving pool before metadata and memory alignment are considered. Coalescing an additional freed separator can increase $L_{\max }$ without requiring the allocator to obtain another region from the operating system.

Raw extent length can overstate payload capacity. For a block $[a,b)$, an illustrative header requirement $h$ and payload alignment $A$, choose aligned payload startThe aligned free-block capacity in this simple layout is $\max (b-p,0)$. A32-byte block starting at0 with an8-byte header and16-byte payload alignment has $p=16$ and only16 bytes of payload capacity. A24-byte payload therefore fails despite a raw32-byte extent. Actual allocator layouts can impose further constraints.

The quantity also depends on the allocation layer. A large virtually contiguous block need not occupy contiguous physical page frames. Comparing $L_{\max }$ values without specifying virtual versus physical space, allocator pool and usable payload convention can therefore mix distinct fragmentation problems.