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Add the following 2's complement binary numbers. Also express the answer in decimal. a. 01+ 1011b. 11+ 01010101c. 0101+ 110d. 01

kenny6666 [7]

Answer:

(a) 01 + 1011

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 1011

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1 ------------------> + 1 (in decimal)

1 0 1 1 ------------------> - 5 (in decimal)

1 1 0 0 -------------------> - 4 (in decimal)

Explanation for (a)

01 or 0001

Since the most significant bit is 0 (leftmost bit), it shows that it is a positive number. Therefore, it can be directly converted to its decimal representation as follows:

0001 = 0 x $2^{3}$ + 0 x $2^{2}$ + 0 x $2^{1}$ + 1 x $2^{0}$ = 1

0001 is + 1 in decimal.

1011

Its most significant bit(MSB) is 1 showing that it is negative.

Flipping all its bits and adding 1 to the result, will convert it to it's positive counterpart.

i.e 1011 => 0100 + 1 = 0101.

Converting the result (0101) to decimal gives

0101 = 0 x $2^{3}$ + 1 x $2^{2}$ + 0 x $2^{1}$ + 1 x $2^{0}$ = 5

1011 is -5 in decimal

01 + 1011 or 0001+ 1011 = 1100

The result (1100), of the addition of the two numbers, is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1100 => 0011 + 1 = 0100.

Converting the result (0100) to decimal gives

0100 = 0 x $2^{3}$ + 1 x $2^{2}$ + 0 x $2^{1}$ + 0 x $2^{0}$ = 4

Therefore, the result, 1100 is -4 in decimal.

===================================================

(b) 11 + 01010101

Convert the two numbers into their 8-bit representation since one of them is in 8-bit.

PS: Taking 11, its MSB is 1 showing that it is a negative number. Also, it's 8-bit representation will mean pre-padding it with ones(1s) rather than zeros as follows. In other words, pre-padding a 2's complement number is done using its sign bit.

=> 11 = 11111111

=> 01010101

Addition of the two numbers (in their 2's complement form) gives the following:

1 1 1 1 1 1 1 1 ------------------> - 1 (in decimal)

0 1 0 1 0 1 0 1 ------------------> + 85 (in decimal)

1 0 1 0 1 0 1 0 0 -------------------> + 84 (in decimal)

Discard the carry-out bit (1) making the result 01010100

Explanation for (b)

11 or 11111111

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

11111111 => 00000000 + 1 = 00000001

Converting the result (00000001) to decimal gives 1

11 or 11111111 is - 1 in decimal.

01010101

Its MSB is 0 showing that it is also positive. Now, convert to decimal

01010101 = 85

01010101 is + 85 in decimal.

11111111 + 01010101 = 01010100

The result (01010100), is also positive as the MSB is 0. Now, convert to decimal.

01010100 = 84

Therefore the result (01010100) is + 84 in decimal.

===================================================

(c) 0101 + 110

Convert the two numbers into their 4-bit representations.

=> 0101 = 0101

=> 110 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 1 0 1 ------------------> + 5 (in decimal)

1 1 1 0 -----------------> - 2 (in decimal)

1 0 0 1 1 -------------------> + 3 (in decimal)

Dicard the carry-out bit (leftmost bit) to get 0011 as result.

Explanation for (c)

0101

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now, convert to decimal.

0101 = 5

0101 is +5 in decimal.

110 or 1110

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 = 2

1110 is -2 in decimal

0101 + 110 = 0101+ 1110 = 0011

The result (0011), is a positive number as the MSB is 0.

Converting the result (0011) to decimal gives

0011 = 3

Therefore, the result, 0011 is 3 in decimal.

===================================================

(d) 01 + 10

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 10 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1 ------------------> + 1 (in decimal)

1 1 1 0 ------------------> - 2 (in decimal)

1 1 1 1 -------------------> - 1 (in decimal)

Explanation for (d)

01 or 0001

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now convert to decimal.

0001 = 1

0001 is + 1 in decimal.

1110

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 = 2

1110 is -2 in decimal

01 + 1011 or 0001+ 1110 = 1111

The result (1111), is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1111 => 0000 + 1 = 0001.

Converting the result (0001) to decimal gives

0001 = 1

Therefore, the result, 1111 is -1 in decimal.

7 0

2 years ago

Assume that two parallel arrays have been declared and initialized: healthOption an array of type char that contains letter code

eduard

/*

Since we have to check the first two options only as mentioned in last part of question the loop will work 2 times only and will compare the cost of first element and second and assign the healthoption accordingly

*/

for(int i =0;i<=1;i++){

if(annualCost[i]<annualCost[i+1]

best2 = healthOption[i]

else

best2 = healthOption[i+1]

}

6 0

2 years ago

Write a static method named countLastDigits that accepts an array of integers as a parameter and examines its elements to determ

Crazy boy [7]

Answer:

See explaination for the program code

Explanation:

code:

public static int[] countLastDigits(int[] list) {

int[] count = new int[10];

for (int i = 0; i < list.length; i++) {

int digit = list[i] % 10;

count[digit]++;

}

return count;

}

8 0

2 years ago

g Suppose the information content of a packet is the bit pattern 1110 0110 1001 1101 and an even parity scheme is being used. Wh

erma4kov [3.2K]

Answer: see description

Explanation:

first we accommodate the bit pattern in a matrix of 4x4 which is the minimum length checksum field, now with even parity two-dimensional scheme we need to complete this matrix by adding one row and one column by adding at the end of each row a 1 or a 0 to complete pairs of 1's:

we have

$\left[\begin{array}{cccc}1&1&1&0\\0&1&1&0\\1&0&0&1\\1&1&0&1\end{array}\right]$

so we complete with this, adding a row at the end which matches a pair number of 1's

$\left[\begin{array}{ccccc}1&1&1&0&1\\0&1&1&0&0\\1&0&0&1&0\\1&1&0&1&1\\1&1&0&0&0\end{array}\right]$

4 0

2 years ago

Create a program named Auction that allows a user to enter an amount bid on an online auction item. Include three overloaded met

Olin [163]

Answer:

Explanation:

The following code is written in Java and creates the overloaded methods as requested. It has been tested and is working without bugs. The test cases are shown in the first red square within the main method and the output results are shown in the bottom red square...

class Auction {

public static void main(String[] args) {

bid(10);

bid(10.00);

bid("10 dollars");

bid("$10");

bid("150 bills");

}

public static void bid(int bid) {

if(bid >= 10) {

System.out.println("Bid Accepted");

} else {

System.out.println("Bid not high enough");

}

public static void bid(double bid) {

if(bid >= 10) {

System.out.println("Bid Accepted");

} else {

System.out.println("Bid not high enough");

}

public static void bid(String bid) {

if (bid.charAt(0) == '$') {

if (Integer.parseInt(bid.substring(1, bid.length())) > 0) {

System.out.println("Bid Accepted");

return;

} else {

System.out.println("Bid not in correct Format");

return;

}

int dollarStartingPoint = 0;

for (int x = 0; x < bid.length(); x++) {

if (bid.charAt(x) == 'd') {

if (bid.substring(x, x + 7).equals("dollars")) {

dollarStartingPoint = x;

} else {

break;

}

if (dollarStartingPoint > 1) {

if (Integer.parseInt(bid.substring(0, dollarStartingPoint-1)) > 0) {

System.out.println("Bid Accepted");

return;

} else {

System.out.println("Bid not in correct format");

return;

}

} else {

System.out.println("Bid not in correct format");

return;

}

3 0

1 year ago