2012 Fall Final Exam+Solution
Sample Exam # 2 solutions
ESI 6314 Deterministic Methods in Operations Research
1. Valdosta Tile Co. has two clay mining pits, P1 and P2, each of which supplies 15 tons of clay per month to three tile making kilns, K1, K2 and K3, each of which require 10 tons of clay per month. The shipping costs ($ per ton) are given in the table below:
[pic]
a) Formulate a transportation linear program to minimize shipping costs
b) Determine a “Northwest Corner Rule” basic feasible solution for this problem. Calculate the total transportation cost associated with this solution.
c) Perform one iteration of the transportation simplex method. What do you observe?
a) min 1.2 x11 + 0.6 x12 +1.5 x13 +3.6 x21 +2.4 x22 +1.2 x23
s.t. x11 + x12 +x13 =15 x21 + x22 +x23 =15 x11 + x21 =10 x12 + x22 =10 x13 + x23 =10 xij>=0
b)
| |1.2 |0.6 |1.5 | |
| |10 |5 | |15 |
| |3.6 |2.4 |1.2 | |
| | |5 |10 |15 |
10 10 10
The total cost is: 10*1.2 + 5*0.6 + 5*2.4 + 10*1.2 = 39
c) We have: u1=0, u2=1.8, v1=1.2, v2=0.6, v3= -0.6
Using these values, we have:
_
c13 = 0 + (-0.6) – 1.5 = -2.1
_
c21 = 1.8 + 1.2 – 3.6 = -0.6
So, this is the optimal solution.
2. Think of the network shown in the figure below as a highway map, and the number recorded next to each arc as the travel time for each arc. A traveler plans to drive from node 1 to node 12 on this highway. Find the best path for this traveler using Dijkstra’s algorithm.
[pic]
Node d(j) pred (j)
1 0 0
2 1 1
3 10 2
4 4 1
5 5 2
6 12 5
7 10
ESI 6314 Deterministic Methods in Operations Research
1. Valdosta Tile Co. has two clay mining pits, P1 and P2, each of which supplies 15 tons of clay per month to three tile making kilns, K1, K2 and K3, each of which require 10 tons of clay per month. The shipping costs ($ per ton) are given in the table below:
[pic]
a) Formulate a transportation linear program to minimize shipping costs
b) Determine a “Northwest Corner Rule” basic feasible solution for this problem. Calculate the total transportation cost associated with this solution.
c) Perform one iteration of the transportation simplex method. What do you observe?
a) min 1.2 x11 + 0.6 x12 +1.5 x13 +3.6 x21 +2.4 x22 +1.2 x23
s.t. x11 + x12 +x13 =15 x21 + x22 +x23 =15 x11 + x21 =10 x12 + x22 =10 x13 + x23 =10 xij>=0
b)
| |1.2 |0.6 |1.5 | |
| |10 |5 | |15 |
| |3.6 |2.4 |1.2 | |
| | |5 |10 |15 |
10 10 10
The total cost is: 10*1.2 + 5*0.6 + 5*2.4 + 10*1.2 = 39
c) We have: u1=0, u2=1.8, v1=1.2, v2=0.6, v3= -0.6
Using these values, we have:
_
c13 = 0 + (-0.6) – 1.5 = -2.1
_
c21 = 1.8 + 1.2 – 3.6 = -0.6
So, this is the optimal solution.
2. Think of the network shown in the figure below as a highway map, and the number recorded next to each arc as the travel time for each arc. A traveler plans to drive from node 1 to node 12 on this highway. Find the best path for this traveler using Dijkstra’s algorithm.
[pic]
Node d(j) pred (j)
1 0 0
2 1 1
3 10 2
4 4 1
5 5 2
6 12 5
7 10