“…First iteration: i = 0, then j = (j + S[i] + K[i] ) mod 8 = (j + S[0] + K[0] ) mod 8 = (0 + 0 + 1 ) mod 8 = 1 Swap S[0] and S[1] to produce an array S: 10010111 Second iteration: i = 1, then j = (j + S[i] + K[i] ) mod 8 = (j + S[1] + K[1] ) mod 8 = (1 + 0 + 7 ) mod 8 = 1 Swap S[1] and S[0] to produce an array S: 00010111 Third iteration: i = 2, then j = (j + S[i] + K[i]) mod 8 = (j + S[2] + K[2] ) mod 8 = (0 + 0 + 1 ) mod 8 = 1 Swap S[2] and S[3] to produce an array S: 00110111 Fourth iteration: i = 3, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[3] + K[3] ) mod 8 = (1 + 1 + 7 ) mod 8 = 1 Swap S[3] and S[2] to produce an array S: 01100111 Fourth iteration: i = 4, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[4] + K [4] ) mod 8 = (1 + 0 + 1 ) mod 8 = 0 Swap S[4] and S[3] to produce an array S: 01010111 Fourth iteration: i = 5, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[5] + K[5] ) mod 8 = (0 + 1 + 7 ) mod 8 = 0 Swap S[5] and S[4] to produce an array S: 01100011 Fourth iteration: i = 6, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[6] + K[6] ) mod 8 = (0 + 1 + 1 ) mod 8 = 0 Swap S[6] and S[5] to produce an array S: 01100011 Fourth iteration: i = 7, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[3] + K[3] ) mod 8 = (0 + 1 + 7 ) mod 8 = 0 Swap S[7] and S[6] so as to produce an array S: 01100001 Fourth iteration: i = 8, then j = ( j + S[i] + K[i] ) mod 8 = ( j + S[8] + K[8] ) mod 8 = (0 + 1 + 1 ) mod 8 = 0 Swap S[8] and S[0] to produce an array S: 00110000 After getting the results of the S array from the fourth iteration, the next process is to XOR the pseudo randombyte with plaintext, with the entered plaintext is 10110000. Because the plaintext consists of two characters, two iterations occur.…”