Let's do an example:

$475($this value is called the minuend$)$

$-283($this value is called the subtrahend$)$

$----$

$192($this value is called the difference$)$

Working from right to left, we subtract $5-3$ and get $2.$ We then subtract $7-8$ we get $-1$; so we need to do a borrow, which makes it $17-8$ so the answer is $9,$ with a borrow of $1.$ We then subtract $[4-1($the borrow$)]-2$ to get $1.$ The result is therefore $192.$

What is really going on in the computer's memory is a little different. These digits are really stored as their ASCII codes:

$525553$

$-505651$

$----------$

$495750$

This answer is computed almost exactly as described above, but one additional step is required because the numbers used are not equal to the digits they represent: they are larger by $48.$ Consider the subtraction of $51$ from $53($ascii $\text{'}3\text{'}$ from ascii $\text{'}5\text{'})$:

$53-51(48+5)-(48+3)$

In other words, we're subtracting $5-3,$ but we're also adding to that $48-48.$ The result we want is $50,$ which is $(48+2),$ but if we just subtract the two digits, we'll get $2$ because we subtracted out the $48\text{'}$s; this is the ASCII code for the start of text $($STX$),$ which is clearly not what we want. The solution, of course, is to add $48$ to the result $(2+48=50)$; this is the extra step that is required in the algorithm. Create the mips code to do this example