“…As y 1 , y 2 , z 1 , and z 2 are different versions of a common least squares problem that result from rearrangements only, methods z 1 and z 2 clearly fail in the case of r 1 ¼ 1/r 2 , in the same way as do y 1 and y 2 . [8] In Figure 1 the applicability of all four methods is demonstrated by use of synthetic data (r 1 ¼ 0.5, r 2 ¼ 0.5, k 111 ¼ 150 L Á mol À1 Á s À1 , k 222 ¼ 450 L Á mol À1 Á s À1 , s 1 ¼ 0.4, and s 2 ¼ 0.6). In Figure 1a the input data (k p vs. f 1 ) are shown.…”