Homework solutions, Chapter 8

Transcription

1 Homework solutions, Chapter 8 NOTE: We might think of 8.1 as being a section devoted to setting up the networks and 8.2 as solving them, but only 8.2 has a homework section. Section Use Dijkstra s algorithm to find the shortest path from Figure 4. I ll type asterisks in place of the boxes we used in class (easier to typeset) The shortest path is 14, and the path is Repeat problem 2 as a transshipment problem. SOLUTION: Node 1 is the supply node, and node 5 is the demand node, and the other nodes are transshipment nodes. We set the supply equal to 1, which is also our s. Node 2 Node 3 Node 4 Node 5 Supply 2 8 M M Node Node 2 1 M 0 5 M Node 3 1 M M 0 10 Node 4 Demand

2 4. If we use Dijkstra s algorithm, we get the path 1 3 4, but there is a shorter path, Dijkstra s algorithm assumes that because node 3 is not connected to node 1 by an arc, node 3 would have to go through node 2, which should make the distance greater than 2. With a negative path, it made the distance zero- Dijkstra s algorithm only works for networks with positive edges. 5. Just as in Example 2, we let node 1 be the beginning of year 1, and node 7 to be the end of year 6/beginning of year 7. Therefore, a table of our edges would be the following, where position (i, j) is C ij : Using Dijkstra s algorithm, we have the following (in hundreds): We see the best solution is with a value of = Problem 6 is identical in flavor to 5 (and Example 2). Here are the costs involved (in tens). We have nodes 1-7 for years beginning 1-6. Because we may only keep the phone 2

3 for at most 5 years, then node 1 is not connected to node / This tells us that is the shortest path at $260. That is, we buy the phone and keep it for three years, then buy another phone and keep it for three years to the end of year 6 (beginning of year 7). 7. Done in class (and solution posted to the class website). 8. Done in class. 9. The box factory. For this one, notice that there is a fixed cost of $1000 to produce any particular box. This is, in essence, a penalty on the number of different sizes we will use. We will assume that, for j > i, then x ij will be the situation where we use box type i in place of box types i, i + 1, i + 2,, j 1. Therefore, for example, x 47 = Box type 4 in place of types 4 and 5 Here are a couple of sample computations: C 13 = = C 47 = = And here is the cost matrix (in hundreds):

4 Therefore, the shortest path is , with costs C 13 +C 34 +C 45 +C 58 = This means we use box type 1 in place of box types 1 and 2, box type 3 for its own demand, box type 4 for its own, then box type 5 for the remainder (types 5, 6, 7). Section

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6 Section 8.4 5(a) 6. (a) One big thing to note. Once we separated activities C and D branching off of node 3, we had to bring them together again for activity E. The only way to do that (without violating the rule about two nodes being connected by at most one edge) is to create a dummy edge so that the two activities can be brought together again at node 5. (b) The linear program may be formulated a couple of different ways. We ll show both, although only the first one was used in class. SOLUTION 1: Let x ij = {0, 1}. Although this isn t strictly a linear program, it turns out that the x ij will take on the values either 0 or 1 in the optimal solution. 6

7 We ll have seven variables (one for each edge): max z = 2x x x x x x x 56 such that (each node has outflow-inflow expressions): x 12 = 1 Node 1 (Supply) x 23 x 12 = 0 Node 2 x 34 + x 35 + x 36 x 23 = 0 Node 3 x 45 x 34 = 0 Node 4 x 56 x 45 x 35 = 0 Node 5 x 56 x 36 = 1 Node 6 (Demand) with x ij 0. For this problem, here is the output from LINDO. You might notice something interesting with the Reduced Cost column. LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE 1) VARIABLE VALUE REDUCED COST X X X X X X X ROW SLACK OR SURPLUS DUAL PRICES 2) ) ) ) ) ) SOLUTION 2: Our textbook author suggests the following setup. Let x j be the time at which event j will take place. These are relative times, so x j is URS. Then: 7

8 min w = x 6 x 1 st x 2 x x 3 x x 4 x x 5 x x 5 x 4 x 6 x x 6 x 3 + x 4 Here is the LINDO output: min x6-x1 st -x1+x2>2 -x2+x3>4 -x3+x4>2 -x3+x5>3 -x4+x5>0 x6-x5>10 -x3+x6>4 end free x1 free x2 free x3 free x4 free x5 free x6 LP OPTIMUM FOUND AT STEP 5 OBJECTIVE FUNCTION VALUE 1) VARIABLE VALUE REDUCED COST X X X X X X ROW SLACK OR SURPLUS DUAL PRICES 2) ) ) ) ) ) ) For exercise 7, use the same techniques as exercise 6. What s really new is building the network model, so you might focus on that aspect, then build the linear program. 8

9 The linear program is much like the one for Exercise 6 (Solution 1): max z = 3x x x x x x 67 st x 12 = 1 x 12 + x 25 + x 23 = 0 x 23 + x 34 = 0 x 34 + x 45 = 0 x 25 x 45 + x 56 = 0 with x ij 0 (in the optimal path, they will be either 0 or 1). x 56 + x 67 = 0 x Explain why the total float will always be bigger than the free float. SOLUTION: Looking at how these are computed on activity (i, j): = 1 F F (i, j) = ET (j) ET (i) t ij T F (i, j) = LT (j) ET (i) t ij And by definition, LT (j) is always larger than (or equal to) ET (j). 9

10 10. Diagram is below. We needed a dummy edge to make it all work. 16. Basic set of computations. See below. 17. Same techniques as 16, you should find that the critical path has length