2.(2 points) Show how the value ASCII “MIRIAM” is stored as hexadecimal in memory in Little Endian format starting at location 100 hexadecimal. Assume that each memory location stores two ASCII characters.Memory LocationValue Stored100101102 3.(2 points) Draw the block diagram for the hardware that implements the following:y + xz: AR BR + CRwhere AR, BR and CR are n-bit registers and x, y, and z are control variables. Include the logic gates for the control function and a block labeled ‘adder’ for the addition function. (Remember that that the symbol + designates an OR operation in the control function, but an arithmetic plus in a microoperation.) 4. (8 points) Perform the computations below given the two 8-bit binary words:1100 10101001 0110ComputeAnswer(A v B)’(A NOR B)(A ^ B) (A AND B)5. (12 points) The content of AC in the basic computer is hexadecimal 79A0 and the initial value of E is 1. Determine the contents of AC, E, PC, AR, and IR in hexadecimal in each of the register –reference instructions from CLA to HLT. The initial value of PC is hexadecimal 156.EACPCARIRInitial179A0156–CLACLECMACMECIRCILINCSPASNASZASZEHLT6. (8 points) An instruction at address 106 in the basic computer has I = 0, and an address part equal to 417 (all numbers are hexadecimal). The memory word at address 417 contains 16AB and the content of AC is 5E34. Go over the instruction cycle and determine the contents of registers PC, AR, DR, AC, and IR for each of the seven memory-reference instructions.PCARDRACIRInitial106–5E34-ANDADDLDASTABUNBSAISZ7.(10 ¬¬¬¬¬¬points) The following program is a list of instructions for the Basic Computer in hexadecimal code. The computer executes the instructions starting from address 100. What are the contents of AC and the memory at address 104 when the computer halts? Show the instructions in the sequence that they are executed.See Mano’s problem 6-2 and the answer in the Module 3 conference in Note 5. LocationInstruction100210710172001025104103700110400001057020106C104107AA55LocationInstructioncommentAC1002107The accumulator (AC) has the value ____; memory location 104 has the value ____8. (12 points) For X = 1110 0011 1001 0100, show the result of the following independent operations (i.e. each instruction occurs with X starting at the value above):________________________Write code that performs the computation:X = A + B + C * D +* E / Fusing CPUs that have the following instruction formats: Do not modify the values of A, B, C, D, E, or F. If necessary, use a temporary location T to store the intermediate results.a. three-operand instructions b. two-operand instructionsc. one-operand instructionsd. stack instructions10. (16 points)A two-word instruction is stored at location 200 with its address field at location 201 (all numbers in hexadecimal). The first word of the instruction specifies the operation code and mode; the second word specifies the address part. The values of the program counter (PC register), a general register (R1), the index register (XR), the base register (BR), and certain addresses in memory are as shown below.Evaluate the effective address and the value that is loaded into the AC if the addressing mode of the instruction is:Addressing ModeEffective addressValue of the operand loaded into the ACa)Immediate b)Direct addressingc)Indirect addressingd)Register e)Register indirectf)Relative addressg)Base register addressingh)Indexed addressingMemoryContent PC =98200Load to AC; Mode201Address = 600R1 = 900202Next instruction::XR = 300300600::BR = 200400300::AC = ?500700::600800::700500::800950::90085011. (10 points) For the ALU of figure 8-2 with tables 8-1 and 8-2.a)Specify the 14-bit control word that must be applied to implement the following microoperation:(3 bits)SELB (3 bits)SELD (3 bits)OPR (5bits)Control Wordb)Determine the microoperation that will be executed in the processor when the following 14-bit control word is applied:00101110101000SELA (3 bits)SELB (3 bits)SELD (3 bits)OPR (5bits)Microoperation12. (6 points) Given the following set of events, show which routines the CPU is executing for times 0 to 100 ns. Each handler routine (with its interrupt request) takes 20 ns to complete. The priority of the interrupts ranges from IRQ6 as the highest priority interrupt to IRQ0 as the lowest priority interrupt.TimeAction0 nsStart of main program10 nsIRQ120 nsIRQ335 nsIRQ450 nsIRQ6TimeAction0 ns:Start of Main Program10 ns:IRQ1
