MATH 341 2002/2003

Assignment 5

1. 9.7-1 Without using your OR Courseware, consider the minimum costflow problem shown below, where the bi values (net lows generated) aregiven by the nodes, the cij values (costs per unit flow) are given by thearcs, and the uij vlalues (arc capacities) are given to the right of thenetwork.

A

DB

C

E

[0][20]

[-30]

[0][10]

5

2

3

6

5

3

4

Arc capacities

A C: 10B C: 25Others:

(a) Obtain an initial BF solution by solving the feasible spanning treewith basic arcs A → B, C → E, D → E, and C → A (a reversearc), where one of the nonbasic arcs (C → B) also is a reversearc. Show the resulting network (including bi, cij, and uij) in thesame format as the above one (except use dashed lines to drawthe nonbasic arcs), and add the flows in parentheses next to thebasic arcs.

(b) Use the optimality test to verify that this initial BF solution isoptimal and that there are multiple optimal solutions. Apply oneiteration of the network simplex method to find the other opti-mal BF solution, and then use these results to identify the otheroptimal solutions that are not BF solutions.

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(c) Now consider the following BF solution.

Basic Arc Flow Nonbasic ArcA→ D 20 A→ BB → C 10 A→ CC → E 10 B → DD→ E 20

Starting from this BF solution, apply one iteration of the networksimplex method. Identify the entering basic arc, the leaving basicarc, and the next BF solution, but do not proceed further.

Solution:

(a)

A

DB

C

E

[35][10]

[-30]

[0][-15]

5

2

3

-6

5

3

4

(15)(30)

(0)

Arc capacities

C A: 10C B: 25Others:

(5)

(b)

Nonbasic arc unique cycle ∆Z when θ = 1CB CB → BA→ AC 3− 2− (−6) = 7 > 0AD AD → DE → EC → CA 5 + 4− 3− 6 = 0BD BD→ DE → EC → CA→ AB 5 + 4− 3− 6 + 2 = 2 > 0

Since all ∆Z ≥ 0, the current solution is optimal. Because thearc AD had ∆Z = 0, we let AD be the entering BV to findanother optimal solution. To send a flow of θ unit along the cycleAD → DE → EC → CA, we need

AD θ <∞DE θ <∞EC θ ≤ 30CA θ ≤ 10− 5 = 5

2

The new diagram is

A

DB

C

E

[25][20]

[-30]

[0][-15]

5

2

3

6

5

3

4

(15)(25)

(5)

(5)

Arc capacities

A C: 10C B: 25Others:

All the other optimal solutions that are not BF solutions can beobtained by sending a flow of θ unit along the cycle AD→ DE →EC → CA for some 0 < θ < 5. That is, xAC = 5 − θ, xAD = θ,xAB = 15, xBC = 25, xBD = 0, xCE = 30− θ, xDE = θ.

(c)

Nonbasic arc unique cycle ∆Z when θ = 1AB AB → BC → CE → ED→ CB 2 + 3 + 3− 4− 5 = −1 < 0AC AC → CE → ED→ DA 6 + 3− 4− 5 = 0BD BD→ DE → EC → CB 5 + 4− 3− 3 = 3 > 0

AB is the entering BV. To send a flow of θ unit along the cycleAB → BC → CE → ED→ CB, we need

AB θ <∞BC θ ≤ 15CE θ ≤ ∞ED θ ≤ 20DA θ ≤ 20

The minimum is 10. Therefore BC is the the leaving BV (it be-

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comes a reversed arc) and the next BF solution is

A

DB

C

E

[25][20]

[-30]

[0][-15]

5

23

6

5

3

4

Arc capacities

A C: 10B C: 25Others:

(15)

(5)(5)

(25)

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2. 9.7-7 Consider the transportation problem having the following costand requirements table:

Destination1 2 3 Supply

Source 1 6 7 4 402 5 8 6 60

Demand 30 40 30

Formulate the network representation of this problem as a minimumcost flow problem. Use the northwest corner rule to obtain an initial BFsoluton. Then use the network simplex method to solve the problem.

Solution: The minimum cost flow problem:

S1

[30][40]

[40]

[30][60]

7

4

5

6

6

8

S2

D1

D2

D3

4

The initial BF solution corresponding to the solution obtained by usingthe northwest corner rule is

S1

[30][40]

[40]

[30][60]

7

4

5

6

6

8S2

D1

D2

D3

(30)

(10)

(30)

(30)

For S1 → D3, ∆Z = 4 − 6 + 8 − 7 = −1. For S2 → D1, ∆Z =5 − 6 + 7 − 8 = −2. S2 → D1 is the entering BV, S1 → D1 is theleaving BV. θ = 30. The new BF solution is

S1

[30][40]

[40]

[30][60]

7

4

5

6

6

8S2

D1

D2

D3

(30)

(40)

(0)

(30)

For S1 → D1, ∆Z = 6 − 5 + 8 − 7 = 2. For S1 → D3, ∆Z =4− 6 + 8− 7 = −1. S1 → D3 is the entering BV and S2 → D3 is the

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leaving BV. θ = 30. The new BF solution is

S1

[30][40]

[40]

[30][60]

7

4

5

6

6

8S2

D1

D2

D3

(30)

(10)

(30) (30)

This solution is optima with a total cost of 580. ¤

3. 9.7-8 Consider the minimum cost flow problem shown below, where thebi values are given by the nodes, the cij values are given by the arcs,and the finite uij values are given in parentheses by the arcs. Obtianan initial BF solution by solving the feasible spanning tree with basicarcs A→ C, B → A, C → D, and C → E, where one of the nonbasicarc (D → A) is a reverse arc. Then use the network simplex methodto solve this problem.

A D

B

C

E

[0]

[50] [-70]

[-60][80]

(uAD=40)

(uBE=40)

5

3

6

4

5

2

1

6

Solution: The initial BF solution is

A D

B

C

E

[0]

[10] [-30]

[-60][80]

(uDA=40)

(uBE=40)

5

3

-6

4

5

2

1

(60)

(30)

(80)

(90)

Nonbasic arc unique cycle ∆Z when θ = 1DA DA→ AC → CD −6 + 4 + 3 = 1BC BC → CA→ AB 2− 4− 1 = −3BE BE → EC → CA→ AB 5− 5− 4− 1 = −5

BE is the entering BV. It is the leaving BV as well (reversed). Thenext BF solution is

A D

B

C

E

[0]

[10] [-30]

[-20][40]

(uDA=40)

(uEB=40)

5

3

-6

4

-5

2

1

(20)

(30)

(40)

(50)

Nonbasic arc unique cycle ∆Z when θ = 1DA DA→ AC → CD −6 + 4 + 3 = 1BC BC → CA→ AB 2− 4− 1 = −3EB EB → BA→ AC → CE −5 + 1 + 4 + 5 = 5

7

BC is the entering BV. BA is the leaving BV. The next BF solution is

A D

B

C

E

[0]

[10] [-30]

[-20][40]

(uDA=40)

(uEB=40)

5

3

-6

4

-5

2

1

(20)

(30)

(40)

(10)

Nonbasic arc unique cycle ∆Z when θ = 1DA DA→ AC → CD −6 + 4 + 3 = 1BA BA→ AC → CB 1 + 4− 2 = 3EB EB → BC → CE −5 + 2 + 5 = 2

The current solution is optimal. It corresponds to the real flows witha cost of 750.

A D

B

C

E

[0]

[50] [-70]

[-60][80]

(uDA=40)

(uEB=40)

5

3

6

4

5

2

1

(20)

(30)

(40)

(10)

(40)

(40)

¤

4. 10.3-12 Consider the following integer nonlinear programming problem.

Maximize Z = 18x1 − x21 + 20x2 + 10x3

subject to2x1 + 4x2 + 3x3 ≤ 11

8

andx1, x2, x3 are nonnegative integers.

Use dynamic programming to solve this problem.

Solution: Number of stages = 3. For n = 1, 2, 3, at stage n, we deter-mine the value of xn. Let sn be the remaining slack in the functionalconstraint, i.e.,

sn =

11 n = 1s1 − 2x1 n = 2s2 − 4x2 n = 3

We havef∗3 (s3) = max

x3=0,1,...,bs3/3c10x3

s3 f ∗3 (s3) x∗30 0 01 0 02 0 03 10 14 10 15 10 16 20 27 20 28 20 29 30 310 30 311 30 3

f ∗2 (s2) = maxx2=0,1,...,bs2/4c

20x2 + f ∗3 (s2 − 4x2)

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x2s2 0 1 2 f∗2 (s2) x∗20 0 0 01 0 0 02 0 0 03 10 10 04 10 20 20 15 10 20 20 16 20 20 20 17 20 30 30 28 20 30 40 40 29 30 30 40 40 210 30 50 40 50 111 30 50 50 50 1, 2

f ∗1 (s1) = maxx1=0,1,...,5

18x1 − x21 + f∗2 (11− 2x1)

x1s1 0 1 2 3 4 5 f∗1 (s1) x∗111 50 57 62 65 66 65 66 4

The optimal solution is (x1, x2, x3) = (4, 0, 1) with Z = 66. ¤

5. 10.3-21 Consider the following nonlinear programming problem.

Maximize Z = x1 (1− x2)x3,

subject tox1 − x2 + x3 ≤ 1

andx1 ≥ 0, x2 ≥ 0, x3 ≥ 0.

Use dynamic programming to solve this problem.

Solution: (Based on the work of Sarah Jones) In stage i we will deter-mine the value of xi. The state of stage i is Si where Si is the slack inthe constraint.

S1 = 1

S2 = 1− x1

S3 = S2 + x2

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At stage 3 wemax f3 (S3, x3) = x3

s.t.x3 ≤ S3

The optimal solution is x∗3 = S3 and f ∗3 (S3) = S3.

At stage 2 we

max f2 (S2, x2) = (1− x2) f∗3 (S2 + x2)

= (1− x2) (S2 + x2)

s.t.−x2 ≤ S2 and x2 ≥ 0.

Since f2 = S2−S2x2+x2−x22, f 02 = −S2+1−2x2. f 02 = 0 =⇒ x2 =1−S22

.Now we consider the constraints. S2 = 1− x1 ≤ 1, therefore 1−S2

2≥ 0

is always true. The other constraint is −x2 ≤ S2 or x2 ≥ −S2.1− S22

< −S2 =⇒ S2 < −1

So in the case of S2 < −1 we will take the left endpoint (which is closerto 1−S2

2) −S2.

x∗2 =½

1−S22

if S2 ≥ −1;−S2 if S2 < −1

and

f ∗2 (S2) =½

14(1 + S2)

2 if S2 ≥ −1;0 if S2 < −1

At stage 1 wemax f1 (x1) = x1f

∗2 (1− x1)

s.t.x1 ≥ 0

Note here that it is not necessary for x1 to be less than or equal to 1for x2 is negative in the original constraint. S1 = 1 is not used. SinceS2 = 1− x1, S2 ≥ −1⇐⇒ x1 ≤ 2.

f1 (x1) =

½x1 · (2−x1)

2

4if 0 ≤ x1 ≤ 2

0 if x1 > 2

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For

f1 (x1) = x1 · (2− x1)2

4= x1 − x21 +

1

4x31

f 01 (x1) = 1− 2x1 + 34x21 =

1

4(3x1 − 2) (x1 − 2)

f 01 = 0 =⇒ x1 =2

3or x1 = 2

It is easy to check (just check x1 = 0,23and 2) that x∗1 =

23and f∗1 =

z∗ = 827. Since S2 = 1

3> −1, x∗2 = 1−S2

2= 1

3and x∗3 = S3 = S2+x∗2 =

23.

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6. 12.2-2 The research and development division of a company has beendeveloping four possible new product lines. Management must nowmake a decision as to ahich of these four products actually will beproduced and at what levels. Therefore, they have asked the OR de-partment to formulate a mathematical programming model to find themost profitable product mix.

A substantial cost is associated with beginning the production of anyproduct, as given in the first row of the following table. The marginalnet revenue from each unit produced is given in the second row of thetable.

Product1 2 3 4

Start-up cost, $ 50,000 40,000 70,000 60,000Marginal revenue, $ 70 60 90 80

Let the continuous decision variables x1, x2,x3 and x4 be the productionlevels of products 1, 2, 3, and 4, respectively. Management has imposedthe following policy constraints on these variables:

1. No more than two of the products can be produced.

2. Either product 3 or 4 can be produced only if either product 1 or 2is produced.

3. Either 5x1 + 3x2 + 6x3 + 4x4 ≤ 6000or 4x1 + 6x2 + 3x3 + 5x4 ≤ 6000.

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Introduce auxiliary binary variables to formulate an MIP model for thisproblem.

Solution: For i = 1, 2, 3, 4 let

yi =

½1 if product i is produced0 otherwise

Let M be a very large number. The problem can be formulated asfollows:

Maximize Z = 70x1 + 60x2 + 90x3 + 80x4

−50000y1 − 40000y2 − 70000y3 − 60000y4subject to

(1)4X

i=1

yi ≤ 2

xi ≤ Myi i = 1, 2, 3, 4

(2) y3 + y4 ≤ y1 + y2

(3) 5x1 + 3x2 + 6x3 + 4x4 ≤ 6000 + wM

4x1 + 6x2 + 3x3 + 5x4 ≤ 6000 + (1− w)M

xi ≥ 0 i = 1, 2, 3, 4. yi is binary for i = 1, 2, 3, 4. w is binary.

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7. 12.2-4 Consider the following mathematical model.

Maximize Z = 3x1 + 2f (x2) + 2x3 + 3g (x4) ,

subject to the restrictions

1. 2x1 − x2 + x3 + 3x4 ≤ 15.2. At least one of the following two inequalities holds:

x1 + x2 + x3 + x4 ≤ 4

3x1 − x2 − x3 + x4 ≤ 3.

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3. At least two of the following four inequalities holds:

5x1 + 3x2 + 3x3 − x4 ≤ 10

2x1 + 5x2 − x3 + 3x4 ≤ 10

−x1 + 3x2 + 5x3 + 3x4 ≤ 10

3x1 − x2 + 3x3 + 5x4 ≤ 10.

4. x3 = 1, or 2, or 3.

5. xj ≥ 0 (j = 1, 2, 3, 4),where

f (x2) =

½ −5 + 3x2 if x2 > 0,0 if x2 = 0,

and

g (x4) =

½ −3 + 5x4 if x4 > 0,0 if x4 = 0.

Formulate this problem as an MIP problem.

Solution:

Maximize 3x1 − 10y1 + 6x2 + 2x3 − 9y2 + 15x4subject to

x1 ≤ My1

x4 ≤ My2

(1) 2x1 − x2 + x3 + 3x4 ≤ 15

(2) x1 + x2 + x3 + x4 ≤ 4 +My3

3x1 − x2 − x3 + x4 ≤ 3 +M(1− y3)

(3) 5x1 + 3x2 + 3x3 − x4 ≤ 10 +My4

2x1 + 5x2 − x3 + 3x4 ≤ 10 +My5

−x1 + 3x2 + 5x3 + 3x4 ≤ 10 +My6

3x1 − x2 + 3x3 + 5x4 ≤ 10 +My7

y4 + y5 + y6 + y7 = 2

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(4) x3 = y8 + 2y9 + 3y10

y8 + y9 + y10 = 1

(5) xi ≥ 0 i = 1, 2, 3, 4 and yj binary j = 1, 2, ..., 10.

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