5. Below is a Little Man program that solves exercise 6.9, p. 164 , from textbook. The program is very similar to the LMC program which you will find in the lecture notes on Chapter 6 posted on Blackboard and discussed in Panopto. The difference is that the program below is somewhat simpler as it uses only 2 branches (BRZ 09 and BR 01), whereas the program in the lecture notes uses 3 branches (BRP 05, BR 10, and BR 01). First, try to understand each instruction thoroughly and then trace the execution of each instruction. Address Instruction (Mnemonics) Address Contents 151617DATDATDAT?210(decrementedby1eachtimetheloopisexecuted)10 Assume now that the above program will only read 3 numbers. That is, the following numbers in this order will be placed, one at a time, in the In-basket: 2,28 , and 25 , where 2 is the count of numbers that follow, and 28 and 25 are the numbers that are to be added. The first column in the table on page 3 shows the order in which the instructions from the program will be executed. Trace the execution of these instructions and determine the contents of the PC before and after each instruction is executed. Also, write down in the table the contents of the In-basket; Accumulator; Memory locations 15, 16, and 17; and Out-basket after each instruction is executed. Memory location 15 controls the loop. It initially contains an unknown value (?), then 2, next 1 , and finally 0. Memory location 16 always contains 1 . It is used to decrease the loop count by 1 . Memory location 17 is initialized with 0 , and finally it stores 53 , the sum of 28 and 25 . The entry 0001 in the PC column means that the PC is 00 when the instruction IN started and is changed to 01 when the instruction IN is finished. 5. Below is a Little Man program that solves exercise 6.9, p. 164 , from textbook. The program is very similar to the LMC program which you will find in the lecture notes on Chapter 6 posted on Blackboard and discussed in Panopto. The difference is that the program below is somewhat simpler as it uses only 2 branches (BRZ 09 and BR 01), whereas the program in the lecture notes uses 3 branches (BRP 05, BR 10, and BR 01). First, try to understand each instruction thoroughly and then trace the execution of each instruction. Address Instruction (Mnemonics) Address Contents 151617DATDATDAT?210(decrementedby1eachtimetheloopisexecuted)10 Assume now that the above program will only read 3 numbers. That is, the following numbers in this order will be placed, one at a time, in the In-basket: 2,28 , and 25 , where 2 is the count of numbers that follow, and 28 and 25 are the numbers that are to be added. The first column in the table on page 3 shows the order in which the instructions from the program will be executed. Trace the execution of these instructions and determine the contents of the PC before and after each instruction is executed. Also, write down in the table the contents of the In-basket; Accumulator; Memory locations 15, 16, and 17; and Out-basket after each instruction is executed. Memory location 15 controls the loop. It initially contains an unknown value (?), then 2, next 1 , and finally 0. Memory location 16 always contains 1 . It is used to decrease the loop count by 1 . Memory location 17 is initialized with 0 , and finally it stores 53 , the sum of 28 and 25 . The entry 0001 in the PC column means that the PC is 00 when the instruction IN started and is changed to 01 when the instruction IN is finished