larger than any of the other function index values. The largest offset

in the list, when added to the new index value, must produce a new

variable index value which can be assigned as an output. The first part

is proved.

Range: Define an index range as follows:

 i: L = L. j: L = a(i)+(l-a)L.+kz a{O0,1}

 U = U. U = b(i)+(1-b)U.+k bk{0,1}
 2 1 U
 A = A. A = A.
 2 1

where k and k are offsets. Note that L. cannot be greater than U. as

this could result in a null IDM (for k=kU for example).

Assume L. = U.
 i 2
 then L. = k

 and U, = k

 In order that the IDM be non-null k must be greater than or equal to
 U

 k For one row the IDM is output set assignable since it is non-null.

 Now assume that the IDM is output set assignable for k rows. The number

 of rows can be expressed as:

 U.-L
 r i
 k A.

 and the number of columns as

 b(imax)+(l-b)U.+k -a(imin)-(l-a)L -k~
 uc A --+ 1 3--3


 but imadx = U,

 and imin = L.

 so eqn. 3-3 becomes

 Ui -L +k -ki
 k u + 1
 a.