Does encrypt time less than decrypt time in RSA cryptograohy? - cryptography

Question

I going to do implementation RSA cryptography. I want know . How many times encryption faster than decryption in RSA cryptography. I try compute elapsed time in java by use System.currentTimeMillis(); but give me time encrypt = 0.05 ms while time decrypt 0.55 ms mean from that 1:11. I think this result is not rational my code is the follow
//here my key has 256 bits
for (;;) {
long begin = System.currentTimeMillis();
for (int i = 0; i < num; i++) {
decrypt();
}
long end = System.currentTimeMillis();
long time = end - begin;
if (time >= 10000) {
System.out.printf("Average Encryption takes: %.2f ms\n",
(double) time / num);
break;
}
num *= 2;
}
p = BigInteger.probablePrime(128, random);
q = BigInteger.probablePrime(128, random);
N = (p.subtract(one)).multiply(q.subtract(one));
e = BigInteger.probablePrime(32, random);
d = e.modInverse(N);
private void encrypt()
{
C= M.modPow(e,N);
}
private void decrypt()
{
RM = C.modPow(d, N);
}
please any explanation for these results

Answer 1

Please, don't implement RSA yourself, it is very easy to do it wrong and it takes months to write version which will be resistant to 3-4 older cryptographic attacks.
All the crypto code you’ve ever written is probably broken -- Tony Arcieri
RSA encryption is more difficult. The 'best practice' in implementing RSA is: don't implement RSA. Other people have done it better than you can. -- Matthew Green (Johns Hopkins University)
Why Cryptography Is Harder Than It Looks -- Bruce Schneier, 1997:
Most systems are not designed and implemented in concert with cryptographers, but by engineers who thought of cryptography as just another component. It's not.
In industrial implementations of RSA, encrypting using someone's public key is faster then decrypting using private key, because public key has short public exponent e, usually 65537 (0x10001). This is true, when fast exponentiation is used (method named Exponentiation_by_squaring). Time of this operation depends linearly on bit length and linearly on 1 bits count in exponent's value, both length and count are small for 65537 (17 bit length and 2 bits are in state 1).
In your pseudocode of RSA-like operation, e is 32 bit and it is usually shorter than d, therefore operation using e exponent is faster than same with d.