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Q1) Write an IAS program to compute the results of the following equation:

Q2) machine code instruction to transfer absolute value of 5

1. 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, 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. and the value of Y into location 5 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 location6 left instruction (20 bits) right instruction (20 bits) 20 28 opcode (8 bits) address (12 bits) opcode (8 bits) address (12 bits) (b) Instruction word 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

Expert 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

 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

Second Method without using direct equation

 Location Instruction/value Comment 7L LOAD M(0) AC=1 7R ADD M(1) AC=1+M(1)=1+3=4 8R ADD M(2) AC=AC+M(2)=4+5=9 9L ADD M(3) AC=AC+M(3)=9+7=16 9R ADD M(4) AC=AC+M(4)=16+9=25 10L STOR M(5) AC->Y 10R JUMP M(10,20:39) Done;Halt

2.Machine code instruction – LOAD DIRECT 5

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