Consider a 32-bit microprocessor , with a 16-bit external data bus, driven by an 8-MHz input clock. Assume that this microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate across the bus that this microprocessor can sustain in bytes/s? To increase its performance, would it be better to make its external data bus 32 bits or to double the external clock frequency supplied to the microprocessor? State any other assumptions you make and explain. Hint: Determine the number of bytes that can be transferred per bus cycle.

We as humans write math expression in infix notation, e. g. 5 + 2 (the operators are written in-between the operands). in a computer’s language, however, it is preferred to have the operators on the right side of the operands, i. e. 5 2 +. for more complex expressions that include parenthesis and multiple operators, a compiler has to convert the expression into postfix first and then evaluate the resulting postfix. write a program that takes an “infix” expression as input, uses stacks to convert it into postfix expression, and finally evaluates it. it must support the following operations: + - / * ^ % (example infix expression: (7 - 3) / (2 + 2)postfix expression: 7 3 - 2 2 + /result: 1guidelines: 1. you will need to use stacks in three placesa. one for the parenthesis check [char stack]b. one during infix to postfix [char stack]c. one during evaluation [int stack]for a and b above, you can use same array and same push, pop method as both ofthem are char. but for evaluation you have int stack and you might consider to createanother push pop method to handle it. maybe push_int, pop_int, etc. or find otherstrategy to utilize existing push pop method2. you can create a function for obtaining operator priority. that function should take anoperator as input and return its priority as an integer. this function will you a lot andreduce repeated code3. during evaluation you will need to convert char into integer. example for single digit: char c = '5'; int x = c - '0';

If an improvement creates no significant change in a product’s performance, then it is a(n) ? a0 design improvement. (there are no answer choices)

Writing a modular program in visual c++. i am new to this and not sure what i am missing. i am getting the following error: baddate. cpp: in function ‘int main()’: baddate. cpp: 50: 3: error: ‘else’ without a previous ‘if’elsehere are the instructions and code: writing a modular program in c++in this lab, you add the input and output statements to a partially completed c++ program. when completed, the user should be able to enter a year, a month, and a day. the program then determines if the date is valid. valid years are those that are greater than 0, valid months include the values 1 through 12, and valid days include the values 1 through 31.notice that variables have been declared for you. write the simulated housekeeping() function that contains the prompts and input statements to retrieve a year, a month, and a day from the user. include the output statements in the simulated endofjob() function. the format of the output is as follows: month/day/year is a valid date. ormonth/day/year is an invalid date. execute the program entering the following date: month = 5, day = 32, year = 2014. record the output of this program. execute the program entering the following date: month = 9, day = 21, year = 2002. record the output of this /* program name: baddate. cppfunction: this program determines if a date entered by the user is valid. input: interactiveoutput: valid date is printed or user is alerted that an invalid date was entered*/#include bool validatedate(int, int, int); using namespace std; int main(){// declare variablesint year; int month; int day; const int min_year = 0, min_month = 1, max_month = 12, min_day = 1, max_day = 31; bool validdate = true; // this is the work of the housekeeping() method// get the year, then the month, then the daycout< < "enter the year"< > year; cout< < "enter the month"< > month; cout< < "enter the day"< > day; // this is the work of the detailloop() method// check to be sure date is validif(year < = min_year) // invalid yearvaliddate = false; else if (month < min_month || month > max_month) // invalid monthvaliddate = false; else if (day < min_day || day > max_day) // invalid dayvaliddate = false; // this is the work of the endofjob() method// test to see if date is valid and output date and whether it is valid or notif(validdate == true); {// output statementcout<