Dongxu Wu Sining GuHong 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 projectparam 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 Spot1 s2 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 s18s1 12 5 4 8 7 0 0 0 0 6 0 0 9 ,

10 1 ,2 0 3 , 11 0 0

S2 6 11 12 7 5 0 0 0 4 0 0 0 0 0 0 3 , 8 1 ,10 2,9

S3 0 2 5 12 3 7 0 0 6 0 0 0 0 0 0 4, 8 1 , 10 9,11

S4 0 11 5 9 12 8 0 0 2 0 0 0 0 0 0 4,6 7,10 1 , 3S5 0 12 5 9 10 3 0 0 11 0 0 0 0 0 0 4, 6 7 , 8 1, 2S6 0 0 11 12 1 10 3 0 9 0 0 0 0 0 4, 6 0 7 ,8 2 ,5S7 0 0 0 9 10 5 3 11 2 0 0 0 0 4 ,6 1,

12 0 0 7 ,8S8 0 0 0 0 0 10 11 12 0 3 1 5 4 , 8 6 ,7 2,9 0 0 0S9 0 0 2 3 1 11 5 0 9 0 0 0 8 4,7 0 0 6 2 ,

10S10 6 0 0 0 0 0 4 9 0 12 2 7 0 1,5 10,

11 3, 8 0 0

S11 0 0 0 1 0 5 0 3 10 11 9 0 0 4,8 2, 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 61 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 62.5 15.4 18.4 9.0 11.0 18.18.0 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=92 2,1,4,3 p22=17, p12=7, p42=11,

p32=133 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 0H10 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-air

craft-carriers/all/*http://www.jmp.com/community/ *http://www.columbia.edu/~

cs2035/courses/ieor4405.S14/index.html

*Thank you!