 This time we're going to be converting a couple of jump instructions into machine language. So I'll begin with my opcode again. The opcode for a jump instruction is 2. And then the opcode for a jump and link instruction is 3. So those parts are easy. Now I have the address. So I've got a 32-bit address for both of these instructions. And I'm going to want to encode that into my address field using only 26 bits. I've got two rules that I'll use to do that. I'll drop the upper four bits, which are all zeros in both cases, and the lower two bits, which may not be quite as obvious, but those are also zeros. So I'm going to start by just writing down all of the bits in both of these addresses in binary. So there's my first address. My second address is... So I'm going to drop the first four bits of both of these. And I'm going to drop the last two bits. As expected, the last two bits were zeros. I will convert both of my opcodes. So now I'm going to convert both of these machine language instructions into hexadecimal, so they're easier to read. At 0, 8, 1, 8, I get 2BBF. The second instruction, I get 0, C, 2, 3. And then I have C992. So there's the hexadecimal representations for both of those machine language instructions.