I need to know number of encryption/decryption operations? If i want to break quadruple DES (say EDED), what is complexity? Like for EDE, it is O(2^112).
I'm not sure you understand the purpose of EDE. EDE only specifies the order of operations of the DES primitive in 3DES is Encrypt-Decrypt-Encrypt. It does not specify the amount of key material. That's determined by the key option.

In 3DES, key option 1 uses three independent keys for a total of 168 bits, meaning a brute force attack requires 2^168 attempts. Note that in this case using EDE results in the same amount of work and protection as EEE would.

Key option 2 uses two keys for a total of 112 bits, or 2^112 attempts. They are used in a 1-2-1 pattern, helping protect against a meet-in-the-middle attack.

Key option 3 means the same 56-bit key is used for all three DES operations - the EDE causes the second pass to decrypt the first pass, yielding the plaintext to be reencrypted by the third pass. It's backwards compatible with single DES. This is the reason EDE was created, to serve old 56-bit applications back when they were trying to upgrade from DES.

So that's why EDED doesn't actually make much sense. There is no backwards compatibility it would preserve, so EEEE would be a less complex choice. However, going forward with the idea that 4DES is a decision someone would make over AES-256 (which would probably be a bad idea), I would assume they would do so with four independent keys, for a total strength of 224 bits, or 2^224 attempts.

Note that all these answers are excluding any benefits in performance gained by cryptanalysis of the algorithms.