To balance out the wall, the tension force resisted by
the holddown must be paired with a compression force at the opposite edge.
Proper selection of the compression member involves two checks: one for the
strength of the member and one for where that member meets the remainder
of the construction.
Simple Columns  
When determining the allowable load on a simple column
it is important to know its slenderness ratio  its length relative to its least width dimension (L/D2). The lower the (L/D2) ratio, the stronger the column. 
Braced Columns  
In shear walls we can expect that the sheathing material
will be attached to one or both faces of the end post/column. This will generally act to brace the member against buckling in the weak (D2) direction. Thus the largest L/D ratio may indeed be (L1/D1). Where panel sheathing is not used, calculate the L/D ratio based on the spacing of the intermediate blocking, i.e. L2/D2. 
Beginning with the 1997 Uniform Building Code, the design of wood columns and allowable wood stresses are referred to the ANSI/NFoPA NDS1991 National Design Specification for Wood Construction, published by the American Forest & Paper Association. Here is how it works:
The NDS formulas seek to establish a Column Stability Factor, (Cp), for a column of a given dimension. This is a fancy way of arriving at a value between 0 and 1 by which the maximum axial compression for a given grade of wood is multiplied in order to create an allowable column load.
1. When a column is supported throughout its length to prevent lateral displacement in all directions, Cp = 1.0.
2. Calculate the maximum L/D ratio for the column using
the diagram above. Use (L1/D1) or (L2/D2) whichever is
larger.
Caution: The slenderness ratio may not exceed
50 except that during construction it may not exceed 75.
3. The column stability factor shall be calculated as follows:
A = (1/2c)*(1 + Fce/Fc)B = (1/c)*(Fce/Fc)
Cp = A  SQRT( A*A  B)
in which...
Fc = tabulated compression design value multiplied by all applicable adjustment factors except Cp (see below).Fce = (Kce* E)/SQR(L/D Ratio)
Kce = 0.3 for visually graded lumber and machine evaluated lumber (MEL)
c = 0.8 for sawn lumber
c = 0.9 for glued laminated timber
Fc can be found by referencing the NDS Supplement. Wood values based on grade are listed for various species along with adjustment factors based on the size of member under consideration. The table for DouglasFir Larch is presented below for your reference.
TABLE 4A  BASE DESIGN VALUES FOR VISUALLY GRADED DIMENSION LUMBER  
DOUGLASFIR LARCH  Design values in pounds per square inch (psi)  Grading Rules Agency 

Species and Commercial Grade  Size Classification 
Bending Fb 
Tension parallel to grain Ft 
Shear parallel to grain Fv 
Compression perp. to grain Fcperp. 
Compression parallel to grain Fc 
Modulus of Elasticity E 

Select Structural  2" to 4" thick 2" & wider 
1,450  1,000  95  625  1,700  1,900,000  WCLIB WWPA 
No.1 & Better  1,150  775  95  625  1,500  1,800,000  
No.1  1,000  675  95  625  1,450  1,700,000  
No. 2  875  575  95  625  1,300  1,600,000  
No. 3  500  325  95  625  750  1,400,000  
Stud  675  450  95  625  825  1,400,000  
Construction  2" to 4" thick 2" to 4" wide 
1,000  650  95  625  1,600  1,500,000  
Standard  550  375  95  625  1,350  1,400,000  
Utility  275  175  95  625  875  1,300,000 
Tabulated bending, tension, and compression parallel to
grain design values for dimension lumber 2" to 4" thick shall be multiplied
by the following size factors:
SIZE FACTOR, C_{F}  
Fb  Ft  Fc  
Grades  Width  Thickness  
2" & 3"  4"  
Select Structural, No. 1 & Better, No. 1, No. 2, No. 3 
2", 3" & 4"  1.5  1.5  1.5  1.15 
5"  1.4  1.4  1.4  1.1  
6"  1.3  1.3  1.3  1.1  
8"  1.2  1.3  1.2  1.05  
10"  1.1  1.2  1.1  1.0  
12"  1.0  1.1  1.0  1.0  
14" & wider  0.9  1.0  0.9  0.9  
Stud  2", 3" & 4"  1.1  1.1  1.1  1.05 
5" & 6"  1.0  1.0  1.0  1.0  
Construction & Standard 
2", 3" & 4"  1.0  1.0  1.0  1.0 
Utility  4"  1.0  1.0  1.0  1.0 
2" & 3"  0.4  n/a  0.4  0.6 
And now an example for a 2 x 4, No. 1 D.F., 8'0" tall, with sheathing bracing the member continuously against buckling in the weak direction...
L/D = 96" / 3.5" = 27.43 (< 50, ok)Fc = 1.15 * 1450 = 1667 psi
Fce = (0.30* 1,700,000)/SQR(27.43) = 677.82
A = (1/1.6)*(1 + Fce/Fc) = 0.879
B = (1/0.8)*(Fce/Fc) = 0.508
Cp = A  SQRT( A*A  B) = 0.364
Allowable unit load = (Cp) * (Fc) = 606 psi (< Fcperp. = 625 psi, ok)
Total Load not to exceed (1.5) * (3.5) * (606) = 3,181 pounds.
We've covered the basic forces, overturning, holddowns
and compression posts used in shear walls, but there are some additional
constraints to be considered before we can wrap up the design portion of
this presentation.