M=20 (total capacity of bag) N=4 (no of items) Let’s find Pi/Wi (profit/weight) for each items: P1/W1=3/5 = 0.60 (O1) P2/W2=10/13= 0.76(O2) P3/W3= 15/12 = 1.25 (O3) P4/W4 = 5/8 = 0.62 (O4) Where O is object no Let’s arrange them in order: 1.25 >> 0.76 >> 0.62 >> 0.60 Object takRead more
M=20 (total capacity of bag)
N=4 (no of items)
Let’s find Pi/Wi (profit/weight) for each items:
P1/W1=3/5 = 0.60 (O1)
P2/W2=10/13= 0.76(O2)
P3/W3= 15/12 = 1.25 (O3)
P4/W4 = 5/8 = 0.62 (O4)
Where O is object no
Let’s arrange them in order: 1.25 >> 0.76 >> 0.62 >> 0.60
Object taken
profit
weight
Remaining weight
O3
15
12
8
O2
10
13
0
O4
5
8
0
O1
3
5
0
Now,
Knapsack weight left to be filled is 8 kg but item-2 has a weight of 13 kg.
Since in fractional knapsack problem, even the fraction of any item can be taken.
1.D7 bit of control word register is responsible for mode selection. 2.D5, D6 are used for mode selection for group A .The format is as follow: D6 D5 Mode 0 0 0 0 1 1 1 x 2 3.Value of CWR for setting PC3 is 07H. 4. D3=1,D2=0,D1=0 to select PC4. 5 . D3=0,D2=1,D1=0 to select PC2.
1.D7 bit of control word register is responsible for mode selection.
2.D5, D6 are used for mode selection for group A .The format is as follow:
ANSWER 1: the various Applications of Stack in Data Structure: Evaluation of Arithmetic Expressions Backtracking Delimiter Checking Reverse a Data Processing Function Calls answer 2: A queue may have a limited space depending on the implementation. We must implement check conditions to see if we arRead more
ANSWER 1:
the various Applications of Stack in Data Structure:
Evaluation of Arithmetic Expressions
Backtracking
Delimiter Checking
Reverse a Data
Processing Function Calls
answer 2:
A queue may have a limited space depending on the implementation. We must implement check conditions to see if we are not adding or deleting elements more than it can maximum support.
The underflow condition checks if there exists any item before popping from the queue. An empty one cannot be dequeued further.
if(front == rear) // underflow condition
The overflow condition checks if the queue is full (or more memory is available) before enqueueing any element. This prevents any error if more space cannot be allocated for the next item.
Solve using fractional knapsack: M=20, n=4 P= (3, 10, 15, …
M=20 (total capacity of bag) N=4 (no of items) Let’s find Pi/Wi (profit/weight) for each items: P1/W1=3/5 = 0.60 (O1) P2/W2=10/13= 0.76(O2) P3/W3= 15/12 = 1.25 (O3) P4/W4 = 5/8 = 0.62 (O4) Where O is object no Let’s arrange them in order: 1.25 >> 0.76 >> 0.62 >> 0.60 Object takRead more
M=20 (total capacity of bag)
N=4 (no of items)
Let’s find Pi/Wi (profit/weight) for each items:
P1/W1=3/5 = 0.60 (O1)
P2/W2=10/13= 0.76(O2)
P3/W3= 15/12 = 1.25 (O3)
P4/W4 = 5/8 = 0.62 (O4)
Where O is object no
Let’s arrange them in order: 1.25 >> 0.76 >> 0.62 >> 0.60
Object taken
profit
weight
Remaining weight
O3
15
12
8
O2
10
13
0
O4
5
8
0
O1
3
5
0
Now,
< O3,(8/13)O2 >
See lessWrite down the steps to test ACUNETIX website for SQL …
For Answer Download below given attachment:
For Answer Download below given attachment:
See lessIntelligent test were given to two groups of boys and girls with mean , Standard deviation and Size 75,8,60 for …
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See lessThe 8255 operates in 2 modes,Answer the following wrt 8255 modes Which bit of control word register is responsible for …
1.D7 bit of control word register is responsible for mode selection. 2.D5, D6 are used for mode selection for group A .The format is as follow: D6 D5 Mode 0 0 0 0 1 1 1 x 2 3.Value of CWR for setting PC3 is 07H. 4. D3=1,D2=0,D1=0 to select PC4. 5 . D3=0,D2=1,D1=0 to select PC2.
1.D7 bit of control word register is responsible for mode selection.
2.D5, D6 are used for mode selection for group A .The format is as follow:
3.Value of CWR for setting PC3 is 07H.
4. D3=1,D2=0,D1=0 to select PC4.
5 . D3=0,D2=1,D1=0 to select PC2.
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question 1:
question 1:
See lessQUESTION 1. What are the various applications of a stack data structure? QUESTION 2. When do you get error message …
ANSWER 1: the various Applications of Stack in Data Structure: Evaluation of Arithmetic Expressions Backtracking Delimiter Checking Reverse a Data Processing Function Calls answer 2: A queue may have a limited space depending on the implementation. We must implement check conditions to see if we arRead more
ANSWER 1:
the various Applications of Stack in Data Structure:
answer 2:
A queue may have a limited space depending on the implementation. We must implement check conditions to see if we are not adding or deleting elements more than it can maximum support.
The underflow condition checks if there exists any item before popping from the queue. An empty one cannot be dequeued further.
if(front == rear)
// underflow condition
The overflow condition checks if the queue is full (or more memory is available) before enqueueing any element. This prevents any error if more space cannot be allocated for the next item.
if(rear == SIZE-1)
See less// overflow condition