Lab 5- Traveling Salesperson Problem with Depth First Search

We will explore two algorithms for solving the very famous traveling salesperson problem. 

Consider a graph of cities. The cost on an edge represents the cost to travel between two cities. The goal of the salesperson is to visit each city once and only once, and returning to the city he/she originally started at. Such a path is called a “tour” of the cities. 

How many tours are there? If there are N cities, then there can be N! possible tours. If we tell you which city to start at, there are (N-1)! Possible tours. For example, if there are 4 cities (A, B, C, D), and we always start at city A, then there are 3! possible tours: 

(A, B, C, D) (A, B, D, C) (A, C, B, D) (A, C, D, B) (A, D, B, C) (A, D, C, B) 

It is understood that one travels back to A at the end of the tour. 

The traveling salesperson problem consists of finding the tour with the lowest cost. The cost includes the trip back to the starting city. Clearly this is a horrendously difficult problem, since there are potentially (N-1)! possible solutions that need to be examined. We will consider DFS algorithms for finding the solution. 

The DFS algorithm is exhaustive – it will attempt to examine all (N-1)! possible solutions. This can be accomplished via a recursive algorithm (call it recTSP). This function is passed a “partial tour” (a sequence of M cities (M <= N) which is initially empty) and a “remaining cities” (sequence of N cities). There are clearly M-N cities not in this partial tour. Thus the function recTSP will have to call itself recursively M-N times, adding each of the M-N cities to the current partial tour out of the remaining cities. If M=N we have a complete tour. 

For example, we start with recTSP ({A}). This will have to call recTSP ({A, B}), recTSP ({A, C}, and recTSP ({A, D}). Here is a partial picture of how the sequence of function calls is done. This tree is not something you build explicitly – it arises from your function calls. You traverse this tree in a “depth-first” manner, The numbers tell you the order in which the nodes are processed. 

Each leaf node is a complete tour, which you will compute the cost of. Note that each non-leaf node is an incomplete tour, which you can also compute the cost of. If the cost of an incomplete tour is greater than the best complete tour that you have found thus far, you clearly do not have to continue working on that incomplete tour. Thus you can “prune” your search. 

Just how hard are these problems? For example, if there are 29 cities, how many possible tours are there? If you can check 1,000,000 tours per second, how many years would it take to check all possible tours? Has the universe been around that long? 

Since this program may take too long to complete, be sure to output the tour and its cost when it finds a new best tour.  

We have to first develop the distance matrix, also called adjacency matrix. This adjacency matrix is populated using a given data file. You will run your program to find the best tours for 12, 14, 16, 19, and 29 cities.

Here is a sample code to populate the distance matrix

 public void populateMatrix(int[][] adjacency){

int value, i, j; 

for (i = 0; i < CITI && input.hasNext(); i++) { //CITI is a constant  

  for (j = i; j < CITI && input.hasNext(); j++){ 

    if (i == j) { 

          adjacency[i][j] = 0;

    }

    else {

          value = input.nextInt();

           adjacency[i][j] = value;

           adjacency[j][i] = value;

    }

  }

}

}

Here is an algorithm to compute tour cost

Algorithm computeCost (ArrayList<Integer> tour)

           Set totalcost = 0

      For (all cities in this tour)

          totalcost += adjacency [tour.get(i)][ tour.get(i+1)]

           EndFor

           If (tour is a complete tour)

                       totalcost += adjacency [tour.get(tour.size()-1)][0]

           EndIf

           return totalcost

 End computeCost

Here is the DFS algorithm

Use ArrayList for “partialTour” and “remainingCities” – This implementation is inefficient due to higher space complexity. 

/* requies : partialTour = <0>, remainingCities = <1,2, 3, ….N-2, N-1>

  ensures: partialTour = <0,…..n> where n E <1,2,3, …, N-1> &&

                Cost(partialTour) is the absolute minimum cost possible.

*/

Algorithm recDFS (ArrayList<Integer> partialTour, ArrayList<Integer> remainingCities )

           If (remainingCities is empty)

                       Compute tour cost for partialTour

                      If (tour cost is less than best known cost)

                                  Set best known cost with tour cost

                                   Output this tour and its cost

                       EndIf

           Else

                       For (all cities in remainingCities)

                                   Create a newpartialTour with partialTour

                                   Add the i_th city of remainingCities to newpartialTour

                                   Compute the cost of newpartialTour

                                   If (newpartialTour cost is less than the best known cost) // pruning

                                               Create newRemainingCities with remainingCities

                                               Remove the i_th city from newRemainingCities

                                               Call recDFS with newpartialTour and newRemainingCities

                                  EndIf

                       EndFor

           EndIf              

 End recDFS

The minimal cost path for 12 cities is 821, and the minimal cost path for 29 cities is 1610, but 29! = 8841761993739700772720181510144 (!!!!!)

Well, Sorry to disappoint you. We will have to wait until the midterm to implement the second algorithm!!! 

Turn in your source program and outputs as an attachment of this assignment. You should turn in your outputs in a PDF. Do not turn in PDF with source code!!! 

 

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Label each of the following statements as True or False:

a. Isoelectric focusing of proteins separates by size.

b. After protein separation by SDS-PAGE each band on the gel represents only one protein.

c. SDS only binds to polar proteins.

 

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Label each of the following scenarios in which there are problems enacting and applying fiscal policy as being an example of either recognition lag, administrative lag, or operational lag.

a. To fight a recession, Parliament has passed a bill to increase infrastructure spending—but the legally required environmental-impact statement for each new project will take at least two years to complete before any building can begin.

b. Distracted by a conflict that is going badly, inflation reaches 8 percent before politicians take notice.

     (Click to select)Recognition lagAdministrative lagOperational lag

c. A sudden recession is recognized by politicians, but it takes many months of political deal making before a stimulus bill is finally approved. 

     (Click to select)Recognition lagAdministrative lagOperational lag

d. To fight a recession, the president issues an executive order requiring Federal agencies to get rid of petty regulations that burden private businesses—but the Federal agencies begin by spending a year developing a set of regulations on how to remove petty regulations.

 

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Label each as TRUE or FALSE:

1. Centrifugation can be used to separate cellular components effectively.
2. Centrifugation is an outdated technology and has been replaced by other methods.
3. Centrifugation can only be used to separate large objects, but small particles like microsomes and ribosomes cannot be separated.

 

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Lab Transshipment Problem (100 Points)

Problem 1 Transshipment Problem

Courtesy example from Spreadsheet Modeling and Decision Analysis By Cliff Ragsdale)

The Bavarian Motor Company (BMC) manufactures expensive luxury cars in Hamburg, Germany and exports cars to sell in the United States. The exported cars are shipped from Hamburg to ports in Newark, New Jersey, and Jacksonville, Florida. From these ports, the cars are transported by rail or truck to distributors located in Boston, Massachusetts; Columbus, Ohio; Atlanta, Georgia; Richmond, Virginia; and Mobile, Alabama. Figure 1 shows the possible shipping routes available to the company along with the transportation cost for shipping each car along the indicated path.

Currently, 200 cars are available at the port in Newark, NJ and 300 are available in Jacksonville, FL. The numbers of cars needed by the distributors in Boston, Columbus, Atlanta, Richmond, and Mobile are 100, 60, 170, 80, and 70, respectively. BMC wants to determine the least costly way of transporting cars from the ports in Newark and Jacksonville to the cities where are needed.

Problem 2 Maximum Flow Problem

Consider the north-south interstate highway system passing through Cincinnati, Ohio. The north-south vehicle flow reaches a level of 15,000 vehicles per hour at peak times. Due to a summer highway maintenance program, which calls a for the temporary closing of lanes and lower speed limits, a network alternative routes through Cincinnati has been proposed by a transportation planner committee. The alternative routes include other highways as well as city streets.

Because of differences in speed limits and traffic patterns, flow capacities vary, depending on the particular streets and roads used. The proposed network with arc flow capacities is shown in Figure 2.

The direction of flow for each arc is indicated, and the arc capacity is shown next to each arc. Note that most of the streets are one-way. However, a two-way street can be found between nodes 2 and 3 and between node 5 and 6. In both cases, the capacity is the same in each direction.

 

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