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CO 250 - Winter 2014
Assignment 07
Due date : Friday, March 14th, 2014, at 2pm (sharp)
Submission Guidelines:
- Use a cover page to submit your solutions (available on the course webpage).
- Assignments submissions are exclusively accepted in the following dropbox slots [Section 001] Math DropBox #12; Slot #1 A-J, Slot #2 K-S, Slot #3 T-Z [Section 002] Math DropBox #12; Slot #4 A-J, Slot #5 K-S, Slot #6 T-Z
- You answers need to be fully justified, unless specified otherwise. Always remember the WHAT-WHY-HOW rule, namely explain in full detail what you are doing, why are you doing it, and how are you doing it. Dry yes/no or numerical answers will get 0 marks. Assignment policies: While it is acceptable to discuss the course material and the assignments, you are expected to do the assignments on your own. For example, copying or paraphrasing a solution from some fellow student or old solutions from previous offerings of related courses qualifies as cheating and we will instruct the TAs to actively look for suspicious similarities and evidence of academic offenses when grading. All students found to be cheating will automatically be given a mark of 0 on the assignment (where that grade will not be ignored as the lowest assignment). In addition, there will be a 5/100 penalty to their final mark, as well as all academic offenses will be reported to the Associate Dean for Undergraduate Studies and recorded in the student’s file (this may lead to further, more severe consequences). Remarking policies: If you have any complaints about the marking of assignments, then you should first check your solutions against the posted solutions. After that, if you see any marking error, then you should return your assignment paper to the head-TA within one week and with written notes on all the marking errors; please write the notes on a new sheet and attach it to your assignment paper. Marking: From every assignment only some of the questions will be marked (the number of them will vary). Still students are expected to submit answers to all of the questions.
COPYRIGHT: Reproduction, sharing or online posting of this document is forbidden.
Question 1 Consider the instance of the s-t path problem as it is depicted in Figure 1.
s^1
c
a b
e d
f g t
2
5 4
7 8
3
31
10
19
16
6
Figure 1: Graph G = (V, E). Edges e above are labeled by their costs ce.
(a) Guess an optimal solution to the s-t path problem.
Solution: The optimal s-t path is P := sabcdef gt with cost 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36. It is shown in Figure 2 in bold edges.
(b) Find appropriate feasible cut widths that certify that your guess in part (a) is indeed optimal. Prove your claims. (Do not run the algorithm for finding the shortest path).
Solution: We define positive widths for the s-t cuts U 1 ,... , U 8 , as they are shown in Fig- ure 2. The widths are set to be yUi = i, i = 1,... , 8. The rest of the widths are defined to be 0. Now we verify that the widths are indeed feasible. To that end we observe that for every edge in P , there is only one cut-width with positive value that the edge belongs to, and it has the exact same value. It remains to check edges
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to (P ). In other words, define the dual of (P ) analogously to the statement of Theorem 3.2 in your coursenotes. Solution:
h(x) := max{bT^ y : Ay ≤ b, y ≤ 0 } (D)
(b) For the LP (D) you defined in part (a), denote by h(y) the value of the objective function. Prove that for all x, y feasible to (P ), (D) respectively we have z(x) ≤ h(y). Conclude that if the last inequality is satisfied with equality for two feasible solutions to (P ), (D), then they must be optimal solutions. Solution: Let x, y feasible to (P ), (D). We have that
z(x) = cT^ x ≤ (AT^ y)T^ x = yT^ Ax ≤ yT^ b = h(y).
Therefore, no feasible to solution to (P ) evaluates the objective to more than h(y) for any feasible y to (D). So, if z(x) = h(y), then x evaluates the objective to the maximum possible value, hence x is optimal. Similarly we argue that y is optimal to (D).
Question 3 The United Nations Space Association (UNSA) wants to launch a new satellite-based wifi coverage program for the whole planet. There are M available orbits in the atmosphere which could accommodate any number of satellites each. The cost for building, launching and maintaining one satellite in orbit i is denoted by ci, i = 1,... , M. At the same time, each orbit covers only a specific subset of the N countries that have expressed interest in receiving service from the program, and these are the countries above which the orbit passes (if a satellite is active in some orbit, then it serves all countries that it passes as it travels in the orbit). Due to different populations, countries request to be covered by different numbers of satellites. The satellite coverage request of country j is denoted by dj , j = 1,... , N (with the understanding that if a smaller number of satellites covers country j, then the country will veto, and the whole wifi coverage program will be abandoned). UNSA wants to decide which orbits to use and how many satellites to build so as to satisfy the demands of all M countries, and with the minimum possible cost.
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In that direction UNSA proposes the following IP.
min
∑^ M
i=
ciyi (U N SA − IP )
s.t.
i: orbit i passes from country j
yi ≥ dj , j = 1,... , N
y ≥ 0 y ∈ ZM
(a) Briefly explain why (U N SA − IP ) solves the optimization problem of UNSA (start by giving some meaning to variables yi). Then introduce the LP relaxation (U N SA − LP ) of (U N SA − IP ). Solution: Variable yi is meant to denote the number of satellites that are to be built for orbit i. Then
∑M
i=1 ciyi^ measures the cost for building/launching/maintaining the satellites. Constraints
i: orbit i passes from country j yi^ ≥^ dj^ say that country^ j^ needs to be covered by at least dj many satellites, since
i: orbit i passes from country j yi counts the satellites that are built and cover country j. Finally, we require to build any non negative integer number of satellites (constraints y ≥ 0 , y ∈ ZM^ ). The LP relaxation of (U N SA − IP ) is
min
∑^ M
i=
ciyi (U N SA − LP )
s.t.
i: orbit i passes from country j
yi ≥ dj , j = 1,... , N
y ≥ 0
(b) Find the dual of (U N SA − LP ). Your answer needs to be expressed analogously to the dual of the LP relaxation of the IP of the shortest st-path formulation (see (3.15), p. 131 of your coursenotes). Explain what the dual LP represents, and prove that its objective function lower bounds the objective of (U N SA − LP ) (similarly to proposition 3.1, p. 122 of your coursenotes).
Solution:
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enough. Hence, the optimal solution to the IP is exactly two. However, consider the feasible solution to (U N SA − LP ), y 1 = y 2 = y 3 = 1/ 2. This evaluates the objective to 3/2.
(d) Consider four countries A, B, C, D with demands 2,3,4,1 respectively. There are three available orbits, that cover countries AB, BC and CD respectively. The cost for building one satellite for any of the three orbits costs 7,4,5 (million dollars) respectively. Prove or disprove that for this specific instance, the optimal solutions to (U N SA − IP ) and (U N SA − LP ) evaluate the objectives to the same value. Hint: Show that in an optimal solution to (U N SA−LP ), some inequalities must be satisfied with equalities.
Solution: The claim is true. To prove this, we write all constraints of (U N SA − LP ) (except from the non-negativity constraints):
y 1 ≥ 2 , y 1 + y 2 ≥ 3 , y 2 + y 3 ≥ 4 , y 3 ≥ 1.
Note that in any optimal solution y, we must have y 1 = 2, as otherwise we would be able to reduce the value of y 1 , effectively reducing the value of the objective. This means that for all optimal solutions to the LP we have
y 1 = 2, y 2 ≥ 1 , y 2 + y 3 ≥ 4 , y 3 ≥ 1.
The next observation is that also y 2 + y 3 ≥ 4 has to be satisfied with equality for any optimal solution. Otherwise, at least one among y 2 , y 3 would be more than 1, and again, we would be able to reduce the objective. Therefore y 2 = 4 − y 3. So, for any optimal solution y, the constraints can be equivalently written as
y 1 = 2, y 2 + y 3 = 4, 1 ≤ y 3 ≤ 3.
But then, the objective becomes 7 y 1 + 4y 2 + 5y 3 = 14 + 16 − 4 y 3 + 5y 3 = 30 + y 3. Since we are minimizing, it is best to take y 3 = 1, which also gives y 2 = 3. We just proved that the optimal solution to the LP is integral.
Question 4 Consider the graph G of Figure 3.
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10
s
t
e
f
c d
a
b
1
3
1
8
2
5
(^241)
2
11
Figure 3: Graph G = (V, E). Edges e above are labeled by their costs ce.
(a) Run the shortest path algorithm of section 3.1.4 of your coursenotes, in order to find the shortest route from vertex a to vertex t. Clearly indicate how the algorithm chooses edges. At the end, present the certificate of optimality, that proves that the a − t path that you found is indeed optimal. Solution: The execution of the algorithm is summarized in Figure (4). At every step we indicate the current cut we are considering, along with the edge that is chosen (the one with the smallest slack). The seven iteration of the algorithm, along with the associated slack-computations can be seen below.
Iteration 1 slack of af : 11-0= slack of as: 1-0=1 (smallest slack) slack of ab: 2-0= y{a} = 1 Iteration 2 slack of af : 11-1= slack of se: 10-0= slack of sc: 4-0= slack of ab: 2-0=1 (smallest slack) y{a,s} = 1 Iteration 3
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slack of dt: 5-0= slack of bt: 8-2-3= y{a,s,b,c,d} = 1 Iteration 6 slack of af : 11-1-1-2-3-1= slack of ef : 1-0=1 (smallest slack) slack of dt: 5-1= slack of bt: 8-2-3-1= y{a,s,b,c,d,e} = 1 Iteration 7 slack of f t: 2-0= slack of dt: 5-1-1= slack of bt: 8-2-3-1-1=1 (smallest slack) y{a,s,b,c,d,e,f } = 1
At this step, we reach t, and we stop. Note that we have found a path P = abt, from a to t, of cost 10 and it is equal to
U yU^ = 10. Hence, the path we found is indeed optimal.
(b) Using the shortest path algorithm of section 3.1.4 of your coursenotes (and maybe your calculations from part (a)), find the shortest path from vertex a to any other vertex of G. Solution: At the end of our algorithm, we have computed a number of widths yU. At the same time, we have reached all vertices (starting from vertex a) with paths of length at most
U yU^. Since in part (a) we have reached all vertices, we have already computed the shortest paths to all vertices, from vertex a.
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