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3 years ago

The four main structural elements of a computer system are:

Computers and Technology

1 answer:

Tom [10]3 years ago

4 0

Processor, Main Memory, I/O Modules, System Bus

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You are an interior decorator, confronted with a dark living room. To lighten the room up, you have n candles and want to build

user100 [1]

Answer:

Algorithm to find a solution of min. cost

Function:cost(Graph G,Graph G1);

GC --> empty graph

for i in edges E:

if E(i,j) in G:

c(i,j)=c(i,j)

else if E(i,j) in G1:

c(i,j)=-c(i,j)

Function:Mincost(Graph G):

GC=Cost(G,G1)

while(negativecycle(GC)):

Update residal graph(G1)

GC=Cost(G,G1)

mincost=sum of Cij*F(i,j)

return mincost;

Explanation:

1) Start the program

2) Read the no. of edges and vertices and also read the cost of the two nodes.

3) Find the min cost by travelling to the destination i.e.. finding all possible min. cost values.

4) Compare the all possible min.cost values

5) And display the least min. cost

6) Stop the program

Correctness of algorithm

1)Here in these algorithm we are calculating all the possible cases.

2)These algorithm also supports the negative cost.

3)These algorithm occupies more space.

4)Takes less time

5)so,these algorithm provides the cost efficient solution very effectively.

Run Time Analysis

1) While reading the values during the run time the program execution will stop until user provides the values.

2) Based on the User input Output vary.

3) Time consumption and space consumption is depends on the no. of inputs the user is given.

6 0

3 years ago

In which generation microprocessor was developed short answer of computer science

3241004551 [841]

Explanation:

in fourth generation microprocessor was developed.

may it helped u

5 0

3 years ago

It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.

Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

For optimal 4-way merging, we need one dummy run with size 0.

Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

$L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000$

The resulting run has length 6000 records.

Part b

For step 1

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec$

So the input/output time is 46 seconds for step 01.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec$

So the CPU time is 23 seconds for step 01.

Total time in step 01

$T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\$

Total time in step 01 is 69 seconds.

For step 2

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec$

So the input/output time is 120 seconds for step 02.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec$

So the CPU time is 60 seconds for step 02.

Total time in step 02

$T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\$

Total time in step 02 is 180 seconds

Merging Time (Total)

Now the total time for merging is given as

$T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\$

Total time in merging is 249 seconds seconds

5 0

3 years ago

You are configuring two switches of different vendors such that they connect to each other via a single link that will carry mul

Jet001 [13]

Answer:

The recommended type of trunk for interoperability is an IEEE 802.1Q trunk.

Explanation:

IEEE 802.1Q is an open industry standard and is the most commonly implemented on layer 2 switches of different vendors, assuring interoperability.

Commonly know as dot1q, is the networking standard that supports virtual LANs (VLANs) on an IEEE 802.3 Ethernet network. It specifies the mechanisms for tagging frames with VLAN data and the procedures for handling this data by switches and bridges.

8 0

3 years ago

This is a graded practice activity. This is not an actual quiz.

Lerok [7]

Answer:

Explanation:

x=27

y=5

22+(27-5)

22+(22)

8 0

1 year ago