Write an IAS program to compute the results of the following equation: Y = 1 + 3

Question

Write an IAS program to compute the results of the following equation: Y = 1 + 3 + 5 + 7 + 9. We have loaded the numbers into memory for you, and the value of Y into location 5. Each integer takes up an entire 40-bit word, but each instruction plus its operand take up only 20 bits, so we have a left and right instruction per 40-bit memory location. Assume when you begin that your constants are stored in order in memory locations 0, 1, 2, 3, and 4. Use memory location 5 for Y, which is initialized to 0. Your job is to fill in the table with the instructions needed to complete the task. Add each integer successively to the accumulator (referred to by the register name "AC"). Assume the accumulator starts with the value 0. Store your instructions beginning in memory location 6. Remember that you have TWO instructions per word when it's used to store instructions (a so-called "instruction word"), each 20 bits, comprising 8 bits of opcode and 12 bits of operand. The system divides up instruction words into Left and Right positions, so 10L represents the left 20 bits. It is read first, then 10R, then 11L and 11R and so on. To terminate the program, use a Jump command, but jump back to your own location. You'll notice the jump command requires a memory location and uses a bit range to specify left or right. Table 1.1 is your friend. Fill in your answers in the table below. Use an many rows as needed.

Explanation / Answer

1. We will calculate as sum of odd number upto 5 terms

using Arthimetic Progression

last_term=1+(n-1)*2

last_term=2*n-1

Sum=(n*(first_term+last_term))/2

sum=n*(1+2*n-1)/2

sum=n*n

Second Method without using direct equation

Location Instruction/Value Comments 6 5 Storing n=5 in address location 7L LOAD M(6) Loading n in AC 7R MULM(6) Multiplying with n in AC for n*n 8L STOR M(5) AC-> Y 8R JUMP M(8,20;39) Done ;Halt

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