Proportion, Scale, Similarity And Dilation.

Friday, March 5, 2010

Remove Formaldehyde From Trailer

Problem Jeep (Exploration Problem)

Description
Maximize the distance of a Jeep can get into the wilderness with a given amount of fuel. The jeep is allowed to go ahead, download a little fuel, and then return to base with fuel left in the tank. At its base, refuel and be off again. When it reaches the fuel saved earlier, then it can be used to partially fill your tank.

Problem
There are n units fuel stored in a fixed base. The jeep can carry at most 1 unit of fuel at any time, and can travel 1 unit of distance in 1 unit of fuel (fuel consumption of the jeep is supposed to be constant.) At any point in a jeep tour can leave any amount of fuel being performed on a fuel dump, or you can catch any remaining fuel in a fuel dump on a previous trip, as long as his fuel load in no case exceed 1 unit.

There are two variants of the problem:
• Exploring the desert - The jeep should return to base after each trip.
• Crossing the desert - The jeep has to return to base after each trip, except for the final trip, when the jeep traveling as far as I can before running out of fuel.

In any case objective is to maximize the distance traveled by jeep on his final journey. Alternatively, the goal may be to find the least amount of fuel needed to produce a final travel a given distance.

In the classical problem of fuel in the jeep and fuel dumps is treated as a continuous quantity. More complex variations on the problem have been proposed in which the fuel can only be left or collected in discrete amounts.

Solution

A strategy that maximizes the distance covered on the final voyage of the "variant to explore the wilderness" is as follows:

• On each trip starts from the bottom with 1 unit of fuel.
• In the first jeep trip covers a distance of 1 / (2n) units and leaves (n-1) / n units of fuel in a fuel dump. The jeep still has 1 / (2n) units of fuel - just enough to return to base.
• In each of these n-1 lists the jeep trips 1 / (2n) units of fuel from the first unloading fuel at the outlet, so that leaves the fuel discharge of 1 unit of transportation fuel . It also collects 1 / (2n) Units of fuel from the first discharge of fuel on the way back, which is just enough fuel to return to base.
• On the second trip the jeep travels to the unloading and reloading of fuel cells first. It then travels a distance of 1 / (2n-2) units and leaves (n-2) / (n-1) units of fuel in a fuel dump second. The jeep still has 1 / (2n-2) units fuel, which is just enough to return to the discharge of fuel first. Here is collected 1 / (2n) units of fuel, which is just enough fuel to return to base.
• In each of these n-2 jeep tours collects 1 / (2n-2) units of fuel to the fuel dump to second exit, which exits the discharge of fuel of 1 unit of fuel transport . It also collects 1 / (2n-2) units of fuel from the second discharge of fuel on the way back, which is just enough fuel to return to the discharge of fuel first.
• The jeep is in this way, so the trip k k sets a new fuel or dump at a distance of 1 / (2n-2k + 2) units above the discharge of fuel and leaves (n - k) / (n - k + 1) units of fuel there. In each of these n - k trips collects 1 / (2n-2k + 2) units of fuel k download, or on their way out and another 1 / (2n-2k + 2) units of fuel in its path return.

Solution to "explore the desert" variant for n = 3, showing the jeep fuel content and fuel tanks at the beginning of each trip and at the point of turnround on each trip.

When the jeep started his last journey, there are n-1 fuel dumps. The furthest contains 1 / 2 of a unit of fuel, far below contain 1 / 3 of a unit of fuel, and so on, and the nearest trash fuel is barely 1 / n units of fuel left. Along with 1 unit of fuel that starts from the base, this means that the jeep can travel a total distance round of the units in its last trip (the maximum distance traveled in the desert is half this).

It collects half of the remaining fuel in each stream to the output, what fills your tank. After leaving the dump fuel farthest traveled to 1 / 2 of a unit in the desert and then back away further from the fuel. Remaining fuel accumulates each discharge of fuel into the way back, which is just enough to get to the next or the discharge of fuel in the final stage, returning to base.

The distance traveled on the last trip is not number of harmonics [1] , Hn. Since they are not bounded in harmonic numbers is possible to overcome any distance on the trip end, and along with enough fuel is available at the base. However, the amount of fuel required and the number of fuel tanks, both increase exponentially with the distance to be covered.

The "crossing the desert" variant can be solved with a similar strategy, except that there is no longer required to collect the fuel on the way back in the final journey. So the trip k jeep down a new k fuel or dump at a distance of 1 / (2n-2k + 1) units of the fuel discharge and previous sheets (2n-2k-1) / (2n- 2k + 1) units of fuel there. In each of the next n - k-1 trip that collects 1 / (2n-2k + 1) units of fuel k download, or on their way out and another 1 / (2n-2k + 1) units of fuel on their way back.

Now, when the jeep started his last journey, there are n-1 fuel dumps. The furthest contains 1 / 3 of a unit of fuel, the following further contain 1 / 5 of a fuel unit, and so on, and the nearest fuel dump just 1 / (2n-1) units of fuel left. Along with 1 unit of fuel that starts from the base, this means that the jeep can travel a total distance of the units in their final journey. Collect all the fuel remaining in each stream to the output, what fills your tank. After leaving the dump fuel farthest travels a distance of 1 unit.

Note that it is theoretically possible, cross a desert of any given size sufficient fuel in the base. As before, the amount of fuel required and the number of fuel tanks, both increase exponentially with the distance to be covered.

--------------------------

Maximum distance Ln Alcanzada, n combustible

Referencias

[1] http://en.wikipedia.org/wiki/Harmonic_number

http://mathworld.wolfram.com/JeepProblem.html

http://en.wikipedia.org/wiki/Jeep_problem

http://page.mi.fu-berlin.de/rote/Papers/pdf/Optimal+logistics+for+expeditions+-+the+jeep+problem+with+complete+refilling.pdf

Remove Formaldehyde From Trailer

--------------------------

Maximum distance Ln Alcanzada, n combustible

Referencias

[1] http://en.wikipedia.org/wiki/Harmonic_number

http://mathworld.wolfram.com/JeepProblem.html

http://en.wikipedia.org/wiki/Jeep_problem

http://page.mi.fu-berlin.de/rote/Papers/pdf/Optimal+logistics+for+expeditions+-+the+jeep+problem+with+complete+refilling.pdf

Friday, February 19, 2010

Veet Strips Wax Pubic Area

Project 1 Project 2. Telephone Directory. Average

For our first project, Gustavo Chavana my partner and I decided to do the "Phone Book", which we premitirá find the phone number to insert the first letter of the name, nustro program after several numbered options will give us that by inserting the option number is going to display another list, which contains all the names together, and that number is stored in the person we are looking for.

Here is part of the flow chart we use to make the program then.

Then they left the show we did in Dev c + +, and I'll explain briefly what we are doing.

In this case look for the phone number for Garza de Leon Ricardo Francisco.

# include # include

int x = 0;

int main () {

int letter;
system ("CLS");
printf ("\\ t \\ t Directory Monterrey \\ n \\ n ");

/ * Here we start running our program, the first step is choosing a letter, in the example we set, as we remember to look Garza called Ricardo Leon Fransisco, so when you run the program begin by introducing the number 7, which corresponds to the letter G, which is the first letter of the name in question * /
printf ("Choose one of the following letters to start looking \\ n \\ n ");

printf (" A = 1 \\ tE = 5 \\ tI = 9 \\ tM = 13 \\ tP = 17 \\ tT = 21 \\ tX = 25 \\ nB = 2 \\ tF = 6 \\ tJ = 10 \\ tN = 14 \\ tQ = 18 \\ tU = 22 \\ tY = 26 \\ nC = 3 \\ tG = 7 \\ tK = 11 \\ tn = 15 \\ tR = 19 \\ tV = 23 \\ tZ = 27 \\ nD = 4 \\ tH = 8 \\ tL = 12 \\ tO = 16 \\ tS = 20 \\ tW = 24 \\ n \\ nOptions: ");
scanf ("% d ", & letter);

switch (letter) {
case
1:
printf ("\\ n \\ nYou can select the letter A to empzar to find, choose one \\ n of suguientes Surnames: \\ n1 = Avendaño \\ n \\ nOptions:");

scanf ("% d", & , letter);
if (x = 1) {
printf ("\\ n \\ nA selected the name Avendaño, there was a match of 5 names, choose an option \\ n \\ n");
printf ("1 = Alejandro Avendaño \\ n \\ n option: ");
scanf ("% d ", & letter);

if (x = 1) {
printf (" \\ nThe number is 87674535 Avendaño Alejadro ");

}}

case 7:
printf ("\\ n \\ nYou can select the letter G to empzar to find, choose one \\ nof the suguientes surname: \\ n \\ n1 = Galvan \\ t6 = Gorrion \\ N2 = Garcia \\ t7 = Guitierrez \\ n3 = Garza \\ n4 = Guevara \\ n5 = War \\ n \\ nOptions: ");

/ * When you insert the number corresponding to the letter "G", we see a list of surnames, which also is listed in the same way as letters, now we have to choose one of the names, so that we deploy another list, which contains the names of all persons who agree on the name, now by typing the number 3, which contains the surname Garza * /

scanf ("% d", & letter);
if (x = 3) {
printf ("\\ n \\ nA Garza selected the name, there was a match of 22 Surname \\ ncompuestos, choose an option \\ n \\ n ");
printf (" 1 = Garza Alvarado \\ n2 = Garza Alvarez \\ n3 = Garza Arellano \\ n4 = Garza Barba \\ n6 = Garza Barreda \\ n7 = Garza Benavides \\ n8 = Garza Blasques \\ n9 = Buentello Garza \\ n10 = Garza Castillo Martha \\ n11 = Garza Cavazos \\ n12 = Garza Casares \\ n13 = Garza Cantu \\ n14 = Garza Garza \\ n15 = Garza de Leon \\ n16 = Garza Delgado \\ n17 = Garza Maldonado \\ n18 = Garza Rodriguez \\ n19 = Garza Salinas \\ n20 = Garza Villarreal \\ n21 = Garza Zamora \\ n22 = Garza Zapata \\ n \\ nOptions: ");
scanf ("% d ", & lyrics);

/ * Finally choose the enumerated list, match the name on the person we are looking for, in this case is number 15 which corresponds to Garza de León, and typing this we get a list which contains the name and number of people whose last name Garza de León, there we will find that we are looking for, which is Ricardo Garza de León Francisco. * /

if (x = 15)
{printf ("\\ nhube 12 matches \\ n \\ n");
printf ("Garza de Leon Eustolia \\ t \\ t83588512 \\ nGarza Adrian Leon \\ t \\ t83608647 \\ nGarza de Leon Diana \\ t \\ t83492749 \\ nGarza Gloria Hilda Leon \\ t81340204 \\ nGarza de Leon Jose Angel \\ t83101866 \\ nGarza de Leon Jose Regino \\ t83262824 \\ nGarza of Leon Marcelino \\ t \\ t83820750 \\ nGarza de Leon Ramon Gerardo \\ t83144447 \\ nGarza de Leon Ricardo Francisco \\ t83430367 \\ nGarza de Leon Romulo \\ t \\ t83797724 \\ nGarza de Leon Silvia \\ t \\ t83347144 \\ nGarza de Leon Yolanda \\ t \\ t83350393) ;

} getch ();}

In our second case we are looking for your Romo Jose Luis Viramontes.

Just as in the first case, we first introduce the number of the initial letter of the name, who is 23, corresponding to the letter V, next to the list that appears we'll pick the 10 which corresponds to the name Viramontes, then typing the number 6, which is what lead us to the list that contains people with the surname Viramontes Romo, and then get the number of Jose Luis Viramontes Romo, that your phone is 83774291

# include # include

int x = 0;

int main () {

int letter;
system ("CLS");
printf ("\\ t \\ t Monterrey Directory \\ n \\ n ");
printf (" Choose one of the following letters to start looking \\ n \\ n ");
printf (" A = 1 \\ tE = 5 \\ tI = 9 \\ tM = 13 \\ tP = 17 \\ tT = 21 \\ tX = 25 \\ nB = 2 \\ tF = 6 \\ tJ = 10 \\ tN = 14 \\ tQ = 18 \\ tU = 22 \\ tY = 26 \\ nC = 3 \\ tG = 7 \\ tK = 11 \\ tn = 15 \\ tR = 19 \\ tV = 23 \\ tZ = 27 \\ nD = 4 \\ tH = 8 \\ tL = 12 \\ tO = 16 \\ tS = 20 \\ tW = 24 \\ n \\ nOptions: ");
scanf ("% d", & letter);

switch (letter)

{case 23:
printf ("\\ n \\ nYou can select the letter V to begin searching, choose one \\ nof the suguientes Surname : \\ n \\ n ");
printf (" 1 = Valles \\ n2 = Vara \\ n3 = Varela \\ n4 = Vargas \\ n5 = Vazquez \\ n6 = Velasco \\ n7 = Viera \\ n8 = Villanueva \\ n10 = Viramontes \\ n11 = Rate \\ n \\ nOptions: ");

scanf ("% d ", & letter);
if (x = 10)
{printf (" \\ n \\ nA selected the name Viramontes, there was a match of 8 Surnames \\ nCompuestos, choose an option \\ n \\ n ");
printf (" 1 = Viramontes Brown \\ n2 = Flores Viramontes \\ n3 = Viramontes Sources \\ n4 = Viramontes Garcia \\ n5 = Viramontes Gutierrez \\ n6 = Viramontes Romo \\ n7 = Viramontes Sierra \\ n8 = Viramontes Velazquez \\ n \\ n option: ");
scanf ("% d ", & letter);

if (x = 6)
{printf ("\\ nhube 4 matches \\ n");
printf ("Romo Jesus Viramontes \\ t \\ t83431255 \\ nViramontes Romo Jose Luis \\ t83774291 \\ nViramontes Romo Maria Eugenia \\ t83343068 \\ nViramontes Romo Silvia \\ t \\ t83772352 ");

}}}

getch ();}