TRAN 650 HW04 solution
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TRAN 650 Urban Systems Engineering HW Assignment 4: Transportation and Assignment Problems, Transportation Simplex, and Network Models Problem 1. W. Winston textbook, Chapter 7 Review Problems, Group A, Problem 1, page 407 Televco produces TV picture tubes at three plants. Plant 1 can produce 50 tubes per week; plant 2, 100 tubes per week; and plant 3, 50 tubes per week. Tubes are shipped to three customers. The profit earned per tube depends on the site where the tube was produced and on the customer who purchased the tube (see Table 1 below). Customer 1 is willing to purchase as many as 80 tubes per week; customer 2, as many as 90; and customer 3, as many as 100. Televco wants to find a shipping and production plan that will maximize profits. a.Formulate a balanced transportation problem that can be used to maximize Televco’s profits.b.Use the northwest corner method to find a bfs to the problem.c.Use the transportation simplex to find an optimal solution to the problem.Table 1. To ($) From Customer 1 Customer 2 Customer 3 Plant 1 75 60 69 Plant 2 79 73 68 Plant 3 85 76 70 Sol:Given, 1.Supply at Plants:•Plant 1 can produce = 50 tubes per week•Pant 2 can produce = 100 tubes per week•Pant 3can produce = 50 tubes per week2.Demand at customers:•Customer 1 requires 80 tubes per week•Customer 2 requires 90 tubes per week
•Customer 3 requires 100 tubes per week 3.Profit per tube shipped from each plant to each customer:Customer 1Customer 2Customer 3Plant 1756069Plant 2797368Plant 3857670a)Formulate a balanced transportation problem:Decision VariablesLet xijrepresent the number of tubes shipped from Plant i to Customer j. Formulation: 1.Objective Function: •Maximize the total profit Z:Z= 71x11+ 60x12+ 69x13+ 79x21 + 73x22+ 68x23+ 85x31+ 76x32+ 70x33 2.Constrains: •Supply Constrains: Plant 1 : x11+ x12 +x13 ≤50 Plant 2 : x21 + x22+ x23 ≤100 Plant 3: x31+ x32+ x33≤50 •Demand Constrains:Customer 1: x11+ x12 +x13 = 80 Customer 2: x21 + x22+ x23= 90 Customer 3: x31+ x32+ x33= 100 b)Basic Feasible Solution Using the Northwest Corner Method: The northwest corner solution is used to find an initial solution by starting from the top- left corner of the matrix and allocate as much as possible without exceeding supply or demand , then moves right or down as needed:
1.Start at x11: (Plant 1 to customer 1) •Allocate x11= 50, which exhaust Plant 1 supply •Customer 1’s demand => 80-50 = 30 2.At x21: (Plant 2 to customer 1) •Allocate x21= 30, fulfilling the remaining demand for customer 1 •Plant 2 new supply = 100- 30 = 70 3.At x22:(Plant 2 to customer 2 ): •Allocate x22= 70, using up Plant 2’s remaining supply for customer 2. •Plant 2’s remaining supply : 90 –70 = 20 4.At x32 (Plant 3 to customer 2): •Allocate x32= 20, meeting the remaining demand for customer 2. •Plant’s 3 remaining supply is now 50 –20 = 30 5.Allocate Remaining demand to x23and x33: •Assign x23= 50, meeting part of Customer 3’s demand using Plant 2’s supply •Assign x33= 50 , meeting rest of Customer 3’s demand using Plant 3s supply The resulting BFS table is : Customer 1 Customer 2 Customer 3 Supply Plant 1 50 0 0 50 Plant 2 30 70 50 100 Plant 3 0 20 50 50 Demand 80 90 100 c)Optimization Using the Transportation Simplex Method:To optimize this solution, we need to apply the Transportation Simplex Method by calculating the dual variables and reduced costs for non-basic cells and adjusting allocations to increase total profit: Step 1 : Calculate Dual Variable: 1)Set u1= 0 2)For each basic cell , solve ui+ vi=cijto determine uiand vjvalues : •For x11: u1+ v1= 75 => v1= 75 •For x21: u2+ v1= 79 => u2= 4
•For x22: u2+ v2 = 73 => v2= 73 –4 = 69 •For x23: u2+ v3= 68 => v3= 68 –4 = 64 •For x32 : u3+ v2= 76 => u3= 76 –69 = 7 •For x33: u3+ v3= 70 = > 7 + 64 = 70 •We have : •u1=0 , u2= 4, u3= 7 •v1= 75 , v2= 69, v3= 64 Step 2 : Calculate Reduce Costs: The reduced cost for each non- basic cell xijis calculated as: Reduced Cost = Cij–(ui+ vj) For each non- basic cell: 1.For x12: c12- (u1+ v2) = 60 - (0 + 69) = -9 2. For x13: c13–(u1+ v3) = 69 –(0 + 64) = 5 3. For x31: c31–(u3+ v1) = 85 –(7 + 75 ) = 3 Since x12 has the most negative reduced cost (-9), it will enter the basis. Step 3: Pivot to Include x121.Loop for Adjustments: We create a loop with X12, X22, X21, and x11alternating additions and subtractions. 2.Adjustments: oAdd 50 units to X12and X21oSubtract 50 units from X11and X22The updated solution is: Customer 1 Customer 2 Customer 3 Supply Plant 1 0 50 0 50 Plant 2 80 20 50 100 Plant 3 0 20 50 50 Plant 4 80 90 100 Final Profit Calculation: 1.Profit Calculation: •X12= 50: 50* 60 = 3000•X21= 80 : 80 * 79 = 6320
•X22= 20 : 20 * 73 = 1460•X23= 50 : 50 * 68 = 3400•X32= 20 : 20 * 76 = 1520•X33= 50 : 50 * 70 = 3500The optimal Profit: Z = 3000 + 6320 + 1460 + 3400 + 1520 + 3500 = 22200 Problem 2. W. Winston textbook, Chapter 7 Review Problems, Group A, Problem 6, page 408 The Gotham City Police have just received three emergency calls. Five cars are available to respond. The distance (in city blocks) of each car from each call location is given in Table 2. Gotham City Police wants to minimize the total distance cars must travel to respond to the three emergency calls. Use the Hungarian Method to determine which car should respond to which call. Table 2. Distance (city blocks) Car Call 1 Call 2 Call 3 1 10 11 18 2 6 7 7 3 7 8 5 4 5 6 4 5 9 4 7 Sol:Step 1: Construct the Cost MatrixSince there are five cars and only three calls, we need to create a 5x5 matrix to apply the Hungarian Method. We do this by adding two "dummy" columns representing additional "dummy calls" with zero distance (cost) so that each car can potentially be assigned to one of these dummy columns. This will help balance the matrix for the algorithm.The modified cost matrix is:
Call 1Call 2Call 3Dummy 1Dummy 2Car 110111800Car 267700Car 378500Car 456400Car 594700Step 2: Row Reduction :For each row, subtract the smallest element in that row from all elements in the row. This step creates a row-reduced matrix with at least one zero in each row.1.Row 1 : Minimum = 0 -> [10 , 11 , 18, 0 , 0]2.Row 2 : Minimum = 0 -> [6, 7 ,7 ,0, 0]3.Row 3 : Minimum = 0 -> [ 7 ,8 , 5 , 0 , 0]4.Row 4: Minimum = 0 -> [5, 6, 4, 0, 0]5.Row 5 : Minimum = 0 -> [9 , 4 ,7 ,0, 0 ]Since all rows already contain a zero due to the dummy columns, there is no need to adjust further. The matrix after row reduction remains:Call 1Call 2Call 3Dummy 1Dummy 2Car 110111800Car 267700Car 378500Car 456400Car 594700Step 3: Column Reduction:Next, subtract the smallest element in each column from all elements in that column. This ensures that each column also contains at least one zero.1. Column 1: Minimum = 5 => [5, 1, 2, 0, 4]2.Column 2: Minimum= 4 =>[7, 3, 4, 2, 0 ]3.Column 3: Minimum= 4 =>[14, 3, 1 , 0, 3]4.Column 4: Minimum= 0 =>[0, 0, 0, 0, 0]5.Column 5: Minimum= 0 =>[0, 0, 0, 0, 0]
The matrix after Column Reduction is:Call1Call2Call3Dummy1Dummy 2Car1571400Car213300Car324100Car402000Car540300Step 4: Cover All Zeros with the Minimum Number of LinesTo determine if the solution is optimal, we need to cover all zeros in the matrix with the minimum number of lines. If the number of lines equals the matrix size (5 in this case), then we have an optimal solution. If not, we need to adjust the matrix.Covering Procedure1.Cover Column 4and Column 5since they contain only zeros.2.Cover Row 4to cover the zero in Column 1.This gives us 3 lines, which is less than 5, so we need further adjustments.Step 5: Adjust the Matrix1.Find the smallest uncovered element in the matrix, which is 1.2.Subtract 1 from all uncovered elements and add 1 to elements covered twice.The adjusted matrix becomes:Call1Call 2Call 3Dummy 1Dummy 2Car461300Car02200Car13000Car01000Car30200Step 6: Test for Optimality and Make AssignmentsAfter the adjustment, we repeat the covering process. Now we can cover all zeros using exactly 5 lines (equal to the matrix size), indicating that we have reached an optimal solution. Now, we can assign each car to a unique call using only the zero-cost cells.Assignments1.Car 1 to Dummy 1(0 in Row 1, Column 4)2.Car 2 to Call 1(0 in Row 2, Column 1)3.Car 3 to Call 3(0 in Row 3, Column 3)4.Car 4 to Call 2(0 in Row 4, Column 2)5.Car 5 to Dummy 2(0 in Row 5, Column 5)
Final AnswerThe optimal assignment of cars to calls (excluding dummy columns) is:•Car 2should respond to Call 1(6 blocks).•Car 4should respond to Call 2(6 blocks).•Car 3should respond to Call 3(5 blocks).Total Minimum Distance: 6+6+5=176 + 6 + 5 = 176+6+5=17 blocks.Problem 3. W. Winston textbook, Chapter 8 Review Problems, Group A, Problem 1, page 471 A truck must travel from New York to Los Angeles. As shown in the Figure below, a variety of routes are available. The number associated with each arc is the number of gallons of fuel required by the truck to traverse the arc: a.Use Dijkstra’s algorithm to find the route from New York to Los Angeles that uses the minimum amount of gas.b.Formulate a balanced transportation problem that could be used to find the route from New York to Los Angeles that uses the minimum amount of gas.c.Find the minimum spanning tree for the network shown in the Figure below.St. LouisClevelandNashvilleNew YorkDallasPhoenixSalt Lake CityLos Angeles40095060013001800120080060090011004006009001000
Sol:Part (a): Use Dijkstra’s Algorithm to Find the Route from New York to Los Angeles with Minimum FuelDijkstra’s Algorithm is a classic algorithm used to find the shortest path in a weighted graph. Here, we want to minimize fuel usage from New Yorkto Los Angeles.Steps for Dijkstra’s Algorithm1.Initialize Distances:oSet the starting node (New York) with a distance of 0.oSet the distances to all other nodes as infinity (∞\infty∞).2.Iterate and Update Distances:oStarting from the node with the smallest known distance, update the tentative distances of its neighbors.oSelect the unvisited node with the smallest tentative distance as the next node to explore.oRepeat until the destination node (Los Angeles) is reached.Solution Using Dijkstra’s Algorithm1.New York(0 gallons)oCleveland: 0+400=4000 + 400 = 4000+400=400oSt. Louis: 0+950=9500 + 950 = 9500+950=950oNashville: 0+800=8000 + 800 = 8000+800=800Current Minimum Distances:oNew York: 0 (start)oCleveland: 400oSt. Louis: 950oNashville: 800
2.Cleveland(400 gallons)oPhoenix: 400+1800=2200400 + 1800 = 2200400+1800=2200oDallas: 400+900=1300400 + 900 = 1300400+900=1300Current Minimum Distances:oNew York: 0oCleveland: 400oSt. Louis: 950oNashville: 800oPhoenix: 2200oDallas: 13003.Nashville(800 gallons)oDallas: 800+600=1400800 + 600 = 1400800+600=1400 (existing distance to Dallas is shorter, so we keep 1300)oSalt Lake City: 800+1200=2000800 + 1200 = 2000800+1200=2000Current Minimum Distances:oNew York: 0oCleveland: 400oSt. Louis: 950oNashville: 800oPhoenix: 2200oDallas: 1300oSalt Lake City: 20004.St. Louis(950 gallons)
oDallas: 950+600=1550950 + 600 = 1550950+600=1550 (existing distance to Dallas is shorter, so we keep 1300)Current Minimum Distances:oNew York: 0oCleveland: 400oSt. Louis: 950oNashville: 800oPhoenix: 2200oDallas: 1300oSalt Lake City: 20005.Dallas(1300 gallons)oLos Angeles: 1300+1300=26001300 + 1300 = 26001300+1300=2600oSalt Lake City: 1300+1000=23001300 + 1000 = 23001300+1000=2300Current Minimum Distances:oNew York: 0oCleveland: 400oSt. Louis: 950oNashville: 800oPhoenix: 2200oDallas: 1300oSalt Lake City: 2000oLos Angeles: 2600
6.Salt Lake City(2000 gallons)oLos Angeles: 2000+600=26002000 + 600 = 26002000+600=2600 (no update, as the distance to Los Angeles is already 2600)7.Los Angeles(2600 gallons) - Destination Reached.Final Answer for Part (a)The minimum-fuel route from New Yorkto Los Angelesrequires 2600 gallons.(b) balanced transportation problem that could be used to find the route from New York to Los Angeles that uses the minimum amount of gas:1.Supplyand Demandpoints need to be defined.2.The transportation problem is balanced by ensuring that the total supply equals the total demand.For this problem:•Supply: We have 1 unit of supply at New York(representing one truck trip).•Demand: We need 1 unit of demand at Los Angeles.Since intermediate cities serve as transshipment points (where inflow equals outflow), we can model this as a minimum cost flow problem, setting up a network where each route's fuel usage represents a transportation cost.By solving this transportation problem using a minimum-cost flow approach, we obtain the same result as in Part (a), confirming that 2600 gallonsis the minimum fuel required to travel from New Yorkto Los Ange
Part (c): Find the Minimum Spanning Tree (MST) for the NetworkThe Minimum Spanning Tree (MST) connects all nodes with the minimum sum of edge weights, without forming cycles. We can use Prim’s or Kruskal’s Algorithm.MST Solution Using Prim’s AlgorithmStarting with any node, we connect the nearest unconnected node and continue until all nodes are connected:1.Start with New York:oConnect to Cleveland (400 gallons)oConnect to Nashville (800 gallons)2.Expand from Cleveland and Nashville:oConnect Nashville to Dallas (600 gallons)oConnect Dallas to Los Angeles (1300 gallons)oConnect Los Angeles to Salt Lake City (600 gallons)3.Continue connecting the nearest nodes:oConnect Cleveland to Dallas (900 gallons)oConnect Phoenix to Los Angeles (400 gallons)The MST total cost (in gallons) for connecting all cities is:400+800+600+1300+600+900+400=5000gallons400 + 800 + 600 + 1300 + 600 + 900 + 400 = 5000 \text{ gallons}400+800+600+1300+600+900+400=5000gallonsResult for Part (c): The MST connects all cities with a minimum total fuel of 5000 gallons.