The way to determine IV for AES encryption, possible security problems with this approach

Question

I am trying to make a folder encryption with maximum automation (except password input). Inside a bash script I do the following:

echo $RANDOM | md5sum | cut -d' ' -f1 > iv

and then using it:

openssl aes-128-cbc -md sha256 -iv $(cat iv) -in folder.7z -out folder.7z.enc

Assuming that echo $RANDOM returns good random value, then I use md5sum on that value, and md5sum has collision weaknesses, so I assume that there is some security problem with that, but this is the way I found of how to produce a correct iv string for openssl. Can someone point what exactly can be a problem with this approach, and may be give an advice how to produce a pseudo-random iv in a better way?

Answer 1

... md5sum has collision weaknesses, so assume that there is some security problem with that,

Collision attacks are irrelevant here. All what matters is that the IV used is in no way predictable (i.e. no bias - all outputs have the same probability) and the same IV is not used with the same input. For more see How do poor-quality initialization vectors affect the security of CBC mode?.

Answer 2

Assuming that echo $RANDOM returns good random value…

It does not. $RANDOM in bash is implemented using a LCRNG with 32 bits of state and a 15-bit output.