If we assume that the circumference of the unshrunk tubing is twice the flattened dimension that you indicate - 4.2 cm x 2 = 8.4 cm AND it shrinks to 1/2 the original dimension (2:1 as you stated) then the shrunk circumference is 4.2 cm - making the diameter 4.2 cm divided pi or about 1.3 cm
My knowledge of plastics suggests that there is a lot of variability in the performance. As previously suggested, if it's critical it would be best to experiment. One way to do this might be to shrink some samples on to some 1.3 cm pieces of scrap - trying a few smaller to see how that works - and trying larger if 1.3 cm isn't working.