Dongxu Wu

Sining Gu

Hong Ouyang

* U.S Aircrafts Carriers Arrangement Scheduling

*The Map

*Distance between Dif. Spots

1. Alaska

2. Korea Strait

3 Makassar Strait

4  Sunda Strait

5  Malacca

6  Red Sea, the southern tip

7   Suez Canal

8  Gibraltar

9 Strait of Hormuz

10The Straits of Florida

11  South Atlantic Ocean

12      Iceland - British Waterway

13  WA

14    San Diego

15   Virginia State

16    Hawaii

17     Japan Yokosuka

18     Guam Island

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*S1. Goal: Minimize

Response Time

*Basic Stand-by Strategy

*Each base can accommodate at most 2 carriers

*Each carrier can only stay on one hot-spot\

*12 hotspots. Gen. 12 different strategies that minimize the total response time of aircraft carriers for each hot post.

*Compare and analyze, then get the best solution.

*St1. Formulation

* For each strategy: ||

* Let Tij be the time needed for aircraft carrier i to travel from spot j to destination.

* Let Cij be the placement of aircraft carrier i at spot j (binary).

* Min

S.T.

, All variables

*S1. ampl mod. Code

# File Project.mod

# model for final project

param T{i in 1..12, j in 1..18};

var c{i in 1..12, j in 1..18} integer >=0;

minimize Response:

sum{i in 1..12, j in 1..18} T[i,j]*c[i,j];

subject to Sixbase:

sum{i in 1..12, j in 1..12} c[i,j] = 6;

subject to Sixhotspot:

sum{i in 1..12, j in 13..18} c[i,j] = 6;

subject to maxbase {j in 13..18}:

sum{i in 1..12} c[i,j] <= 2;

subject to maxhotspot {j in 1..12}:

sum{i in 1..12} c[i,j] <= 1;

subject to oneship {i in 1..12}:

sum{j in 1..18} c[i,j]=1;

*St1. Data file

V Spot

1s2 s3 s4 s5 s6 s7 s8 s9 s10 s11 s12 s13 s14 s15 s16 s17 s18

56

Carrier1

0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

56

C2 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

56

C3 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

65

C4 0.0 4.5

5.8 7.1 7.1 9.6 12.8

10.3

12.2

8.3 15.4

15.4

1.3 2.9 10.6

3.2 3.8 5.1

56

C5 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

65

C6 0.0 4.5

5.8 7.1 7.1 9.6 12.8

10.3

12.2

8.3 15.4

15.4

1.3 2.9 10.6

3.2 3.8 5.1

60

C7 0.0 4.9

6.3 7.6 7.6 10.4

13.9

11.1

13.2

9.0 16.7

16.7

1.4 3.1 11.5

3.5 4.2 5.6

65

C8 0.0 4.5

5.8 7.1 7.1 9.6 12.8

10.3

12.2

8.3 15.4

15.4

1.3 2.9 10.6

3.2 3.8 5.1

56

C9 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

56

C10 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

56

C11 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

56

C12 0.0 5.2

6.7 8.2 8.2 11.2

14.9

11.9

14.1

9.7 17.9

17.9

1.5 3.3 12.3

3.7 4.5 6.0

*St1. Results

s1 s2 s3 s4 s5 s6 s7 s8 s9 s10 s11 s12 s13 s14 s15 s16 s17 s18

s1 12 5 4 8 7 0 0 0 0 6 0 09 , 10

1 ,2 03 , 11

0 0

S2 6 11 12 7 5 0 0 0 4 0 0 0 0 0 0 3 , 81 ,1

02,9

S3 0 2 5 12 3 7 0 0 6 0 0 0 0 0 0 4, 81 , 10

9,11

S4 0 11 5 9 12 8 0 0 2 0 0 0 0 0 0 4,6 7,10 1 , 3

S5 0 12 5 9 10 3 0 0 11 0 0 0 0 0 0 4, 6 7 , 8 1, 2

S6 0 0 11 12 1 10 3 0 9 0 0 0 0 0 4, 6 0 7 ,8 2 ,5

S7 0 0 0 9 10 5 3 11 2 0 0 0 0 4 ,61, 12

0 0 7 ,8

S8 0 0 0 0 0 10 11 12 0 3 1 5 4 , 8 6 ,7 2,9 0 0 0

S9 0 0 2 3 1 11 5 0 9 0 0 0 8 4,7 0 0 62 , 10

S10 6 0 0 0 0 0 4 9 0 12 2 7 0 1,510, 11

3, 8 0 0

S11 0 0 0 1 0 5 0 3 10 11 9 0 0 4,82, 12

6 ,7 0 0

s12 0 0 0 0 0 12 11 9 0 10 1 3 0 6,7 2, 5 4 , 8 0 0

*St1.Conclusion

*We get the U.S navy 12 carriers response ability is as right side

*(based on 6 existed bases and 12 hot spots)

Total Days Avg_days

51 4.3

41.7 3.5

38 3.2

45.3 3.8

52.2 4.4

78.2 6.5

87.3 7.3

78.1 6.5

85.7 7.1

69.7 5.8

106.5 8.9

114.1 9.5

*Strategy 2. Assumptions

*No 1-12 carriers are located at the 12 hotspots respectively. Each hotspot has one carrier.

*All the carriers are now performing on some tasks. Once they receive the order to return to bases, they should first accomplish the task and then return.

Goal:*Find the shortest time needed for all the 12 carriers to return to a base.

* St2. Formulation

*Qm|rj|Cmax.

*Qm = 12 carriers with different speed

*rj = the time each carrier spend on continue finishing the task

*Cmax = minimize the length of time needed for all the 12 carriers to return to the bases

*St2. Data

Carrier speed rj 1 2 3 4 5 6

1 56 1 1.5 3.3 12.3 3.7 4.5 6.0

2 56 2 6.0 6.0 13.4 5.2 1.5 2.2

3 56 3 8.9 8.2 13.4 6.0 3.0 2.2

4 65 4 9.0 8.3 12.8 6.4 3.8 3.2

5 56 5 11.2 10.4 13.4 8.9 6.0 5.2

6 65 6 13.5 12.8 9.6 11.5 9.0 8.3

7 60 7 12.5 10.8 8.3 12.5 11.8 11.1

8 65 8 9.0 7.4 5.1 9.0 12.8 12.8

9 56 9 12.6 11.9 14.1 13.4 12.6 9.7

10 56 10 8.2 6.3 1.5 7.4 11.9 11.2

11 56 11 16.4 14.5 7.4 11.9 16.4 15.6

12 56 12 16.4 14.5 7.1 15.6 20.1 23.8

*St2. cplexamp Code

param numcarriers = 12;

param numbases = 6;

param T {1..numcarriers, 1..numbases} >=0;

var X {1..numcarriers, 1..numbases} binary;

minimize Makespan {i in 1..numcarriers}:

max { sum {j in 1..numbases} T[i,j]*X[i,j]};

subject to Carrier_Constraints {i in 1..numcarriers} :

sum {j in 1..numbases} X[i,j] = 1;

subject to Base_Constraints {j in 1..numbases} :

sum {i in 1..numcarriers} X[i,j] = 2;

*St2. Result

*Cmax = 19.1

1 2 3 4 5 6

2.5 15.4 18.4 9.0 11.0 18.1

8.0 16.3 19.1 10.4 15.0 18.7

(1,2) (8,10) (11,12) (3,4) (5,6) (7,9)

*Strategy 3, Job Shop

*Job Shop Scheduling Problems

* 3 jobs (carriers), 4 machines(hotspots)

*Each carrier follows a predetermined route

*(a sequence of hotspots)

Strategy 3:

*Operation (i,j) : Processing of carrier j on hotspot i

*Processing time pij

*Assume that carriers do not recirculate

*Minimize Cmax

*St3. data

Carriers

Hotspot Sequence Processing Times

1 1,2,3 p11=21, p21=17, p31=9

2 2,1,4,3 p22=17, p12=7, p42=11, p32=13

3 1,2,4 p13=9, p23=15, p43=7

*St3. Code

* param numhotspots = 3;

* param carrier = 3;

* param M = 10000;

* param C >= 0;

* param T [i in 1..numhotspots, j in 1.. carrier];

* param P [i in 1..numhotspots, j in 1.. carrier];

* var X[ i in 1..numhotspots, j in 1..carrier, k in 1.. carrier] binary;

*  

* minimize Makespan: C;

*  

* subject to makespan1: C>=T[3,1] + P[3,1];

* subject to makespan2: C>=T[3,2] + P[3,2];

* subject to makespan3: C>=T[4,3] + P[4,3];

*  

* subject to job1:t21 ≥ t11 + p11; 

* subject to job2:t31 ≥ t21 + p21;

* subject to job3:t12 ≥ t22 + p22;

* subject to job4:t42 ≥ t12 + p12;

* subject to job5:t32 ≥ t42 + p42;

* subject to job6:t23 ≥ t13 + p13;

* subject to job7:t43 ≥ t23 + p23

*  

* subject to constraint {i in 1..numhotspots,}:

* {j in 1.. carrier, j<k} 

* {

*    T[i,1]+P[i,1] <= T[i,2]+M*(1-X[i,1,2]);

*    T[i,2]+P[i,2] <= T[i,1]+M*X[i,1,2];

*    T[i,1]+P[i,1] <= T[i,3]+M*(1-X[i,1,3]);

*    T[i,3]+P[i,3] <= T[i,1]+M*X[i,1,3];

*    T[i,2]+P[i,2] <= T[i,3]+M*(1-X[i,2,3]);

*    T[i,3]+P[i,3] <= T[i,2]+M*X[i,2,3];

* }

*St 3. Result

*Optimal solution with makespan 60

Hotspots Carrier Sequence

1 1,2,3 (21,7,9)2 2,1,3(17,17,1

5)3 1,2(9,13)4 2,3(11,7)

*Strategy 4. A Bigger Job Shop

*Assume there are 10 carriers (C0–C9) available and there are also 10 hotspots(H1–H10) need to be watched. Each time only one carrier watching one hotspot.

* Every hotspot must be processed on each of the 10 carriers in a predefined sequence.

* The objective is to minimize the completion time of the last Hotspot(job) to be processed (Makespan).

* The hotspots are described in the data set “arr” by using the following statements.

H1 C2 C3 C5 C4 C0 C7 C8 C9 C1 C6

H2 4 7 1 8 0 3 2 5 9 6

H3 9 6 4 3 1 0 8 2 7 5

H4 1 2 7 5 8 4 3 6 9 0

H5 6 1 3 0 2 8 4 7 9 5

H6 7 5 8 2 4 6 3 1 9 0

H7 6 1 4 5 2 3 7 8 9 0

H8 0 5 8 9 3 6 4 7 2 1

H9 5 2 3 6 4 7 8 9 1 0

H10

9 4 6 7 0 2 8 5 3 1

*Carriers Arrange Schedule for 10 Hotspots

*SAS/OR Code

* /* Hotspots specification */

data arr (drop=i mid);

do i=1 to 10;

input mid _DURATION_ @;

_RESOURCE_=compress(‘C'||put(mid,best.));

output;

end;

datalines;

2 44 3 5 5 58 4 97 0 9 7 84 8 77 9 96 1 58 6 89

4 15 7 31 1 87 8 57 0 77 3 85 2 81 5 39 9 73 6 21

9 82 6 22 4 10 3 70 1 49 0 40 8 34 2 48 7 80 5 71

1 91 2 17 7 62 5 75 8 47 4 11 3 7 6 72 9 35 0 55

6 71 1 90 3 75 0 64 2 94 8 15 4 12 7 67 9 20 5 50

7 70 5 93 8 77 2 29 4 58 6 93 3 68 1 57 9 7 0 52

6 87 1 63 4 26 5 6 2 82 3 27 7 56 8 48 9 36 0 95

0 36 5 15 8 41 9 78 3 76 6 84 4 30 7 76 2 36 1 8

5 88 2 81 3 13 6 82 4 54 7 13 8 29 9 40 1 78 0 75

9 88 4 54 6 64 7 32 0 52 2 6 8 54 5 82 3 6 1 26

*St4. Infeasible Schedule(Uncontrained)

/* create the Activity data set */

data act (drop= i j);

format _ACTIVITY_ $8. _SUCCESSOR_ $8.;

set arr;

_QTY_ = 1;

i=mod(_n_-1,10)+1;

j=int((_n_-1)/10)+1;

_ACTIVITY_ = compress('J'||put(j,best.)||'P'||put(i,

best.));

JOB=j;

TASK=i;

if

i LT 10 then _SUCCESSOR_ = compress('J'||put(j,best.)||'P'||put((i+1),best.));

else

_SUCCESSOR_ = ' ';

output;

run;

The minimum makespan would be 617—the time it takes to go to Hotspot 1.(finish hotspot 1 job)

*St 4. Add an Edge-finder algorithm

(Get a Contrained Schedule)

*proc const domain=[0,842]

actdata=actdata

schedout=sched_jobshop

dpr=50

restarts=150

showprogress;

schedule finish=842 edgefinder nf=1 nl=1;

run;

Result: 842 days

*Sources

*http://www.fas.org/sgp/crs/weapons/RS20643.pdf

*http://www.fas.org/programs/ssp/man/uswpns/navy/aircraftcarriers/cvn68nimitz.html

*http://www.wired.com/dangerroom/2013/03/replacing-aircraft-carriers/all/

*http://www.jmp.com/community/

*http://www.columbia.edu/~cs2035/courses/ieor4405.S14/index.html

*Thank you!