Consider the following program using the instruction set of the processor given in the Table 5-6 in the textbook. CLA INP STAX ADD Y OUT HLT a) Obtain running time of the above program as a function of number clocks cycles consumed. b) Assuming that processor clock frequency f = 2 GHZ, obtain the time it takes to run the above program. TABLE 5-6 Control Functions and Microoperations for the Basic Computer ADD Feich R'T AR-PC RT: IR --MAR). PC --PC + 1 Decode R'Ts D. ....D.-Decode IR (12-14). AR-IR(0-11). I--IR(15) Indirect DUIT: ARMAR] Interrupt: TETITIEN)(FGI + FGO): R-1 RT: AR-0, IR-P RT: MAR]--TR, PC -- RTS PC - PC + 1, IEN -0. R--0, SC-0 Memory-reference: AND DT.: DR MAR) DT: AC-AC A DR. SC--0 DT. DR-MAR) DT: AC-AC + DR, E SCO LDA DT: DR --MAR) DT: AC-DR, SCO STA DT: MAR --ACSC-0 BUN DTA PC+AR, SC-0 BSA D.T.: MAR-PC. AR-AR + 1 D,T: PC+AR, SCO ISZ D.T.: DR -- M[AR] DT: DR DR + 1 DTs MAR] --DR, (DR = 0) then (PC - PC + 1). SC-0 Register-reference DI'T, -r (common to all register-reference instructions) IR() = B. (i = 0, 1, 2,...,11) T: SC-0 CLA B. AC -- CLE BE0 BACAC B.EE B A C - AC, AC(15) --E, E-AC(0) R AC-shl AC, AC(O)-E, E-AC(15) INC B AC. AC + 1 B. If (AC(15) - ) then (PC +PC+1) By: If (AC(15)-1) then (PC PC + 1) SZA B, IF (AC = 0) then PC - PC + 1) SZE B: (E = 0) then (PC - PC + 1) HLT FRS = Input-output: DJT, (common to all input-output instructions) IR(I) = B. (i = 6, 7, 8, 9, 10, 11) P: SCO pB, AC(0-7) --INPR, FGI-O OUT pBas: OUTR --AC(0-7). FGO--0 pB IF (FG! - 1) then (PC --PC +1) SKO pB. IF (FGO - 1) then (PC --PC +1) PB: JENI 1OF pENO MA CIR CIL SPA SNA INP SKI Consider the following program using the instruction set of the processor given in the Table 5-6 in the textbook. CLA INP STAX ADD Y OUT HLT a) Obtain running time of the above program as a function of number clocks cycles consumed. b) Assuming that processor clock frequency f = 2 GHZ, obtain the time it takes to run the above program. TABLE 5-6 Control Functions and Microoperations for the Basic Computer ADD Feich R'T AR-PC RT: IR --MAR). PC --PC + 1 Decode R'Ts D. ....D.-Decode IR (12-14). AR-IR(0-11). I--IR(15) Indirect DUIT: ARMAR] Interrupt: TETITIEN)(FGI + FGO): R-1 RT: AR-0, IR-P RT: MAR]--TR, PC -- RTS PC - PC + 1, IEN -0. R--0, SC-0 Memory-reference: AND DT.: DR MAR) DT: AC-AC A DR. SC--0 DT. DR-MAR) DT: AC-AC + DR, E SCO LDA DT: DR --MAR) DT: AC-DR, SCO STA DT: MAR --ACSC-0 BUN DTA PC+AR, SC-0 BSA D.T.: MAR-PC. AR-AR + 1 D,T: PC+AR, SCO ISZ D.T.: DR -- M[AR] DT: DR DR + 1 DTs MAR] --DR, (DR = 0) then (PC - PC + 1). SC-0 Register-reference DI'T, -r (common to all register-reference instructions) IR() = B. (i = 0, 1, 2,...,11) T: SC-0 CLA B. AC -- CLE BE0 BACAC B.EE B A C - AC, AC(15) --E, E-AC(0) R AC-shl AC, AC(O)-E, E-AC(15) INC B AC. AC + 1 B. If (AC(15) - ) then (PC +PC+1) By: If (AC(15)-1) then (PC PC + 1) SZA B, IF (AC = 0) then PC - PC + 1) SZE B: (E = 0) then (PC - PC + 1) HLT FRS = Input-output: DJT, (common to all input-output instructions) IR(I) = B. (i = 6, 7, 8, 9, 10, 11) P: SCO pB, AC(0-7) --INPR, FGI-O OUT pBas: OUTR --AC(0-7). FGO--0 pB IF (FG! - 1) then (PC --PC +1) SKO pB. IF (FGO - 1) then (PC --PC +1) PB: JENI 1OF pENO MA CIR CIL SPA SNA INP SKI