Install the Easycpu ENIAC package on a Windows platform. It can be installed on a Windows 10 VM.You can download the file from here:https://sourceforge.net/projects/eniac/Or, you could download the attached file: fattoriable.eniac.Click , and point to the folder that contains the assembly code fattoriable.eniac.You may need to download JAVA, and then run the command listed in the readme file (from the zip file you downloaded).https://www.java.com/en/download/win10.jspClick , and point to the folder that contains the assembly code fattoriable.eniac.The program should be loaded in ASCII. If not, click the menu , , then . It should look as below.Hexadecimal              ASCII===========          ===== ===200003                     JUMP 36                                61                                1a02001                      LOAD @1b88001                      STORE BXc02002                      SUB @22b000c                      JZ 12b88006                     STORE CXf08001                      MUL BXb88001                     STORE BXa08006                     LOAD CX200005                    JMP 510000                      HLTExplanation:(0) address starts at 0x2000, but PC=3 (address 0x2003)(1) stack 1 (@1) has value 6(2) stack 2 (@2) has value 1(3) command load (0xa0) from 0x2001 to register AX (AX=6), PC=4 (address 0x2004)(4) command store (0xb8) the value in 0x2001 to register BX at 0x8001 (BX=6), PC=5(address 0x2005)(5) command substract (0xc0) the value 1 in 0x2002 from the register AX, PC=6 (address 0x2006)(6) command jump (0x2b) to 0xC=12 if the result in AX is zero, PC=7 (address 0x2007)(7) command store (0xb8) the value into register CX at 0x8006 (CX=5), PC=8 (address 0x2008)(8) command mul (0xf0) the value of register AX by the value in register Bx at 0x8001, AX=5*6=30, PC=9 (address 0x2009)(9) command store (0xb8) the value of register AX in register BX at 0x8001 (Bx=30),PC=10 (address 0x200a)(10) command load (0xa0) the value in register CX at 0x8006 to register AX (AX=CX=5), PC=11 (address 0x200b)(11) command jump (0x20) to address 0x5 (0x2005), PC=5 (address 0x2005)(12) command halt (0x1) is exit. The program looped 6 times to exit.When exit, AX=0, BX=720, CX=1, DX=0, PC=12,CA = SI= OV = 0, ZE = EV = 1Lab exercise:Q1. Click menu , then select . Use this function to step through the assembly code.After the statement in address 11 is first executed, what are the value of registers AX, BX, CX,DX, and PC? Explain the function of AX, BX, CX, DX, and PC.A1:Q2. After statement 11 is first executed, what is the next statement to be executed? Can you tellwhat statements 5 through 11 are doing in a simple statement?A2:Q3: How many times are statements 5 and 6 executed before statement 12 is executed?How can you tell, if you don’t step through it? Which register or memory location can allow you to determine that value without stepping through all codes till exit?A3:

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