Analisis Regresi

TUGAS INDIVIDU ANALISIS REKAYASA

SI-5101

Dosen :

Ir. Biemo W. Soemardi, Ph.D.

Oleh :

Davin Yuan Kermite (25014003)

MANAJEMEN REKAYASA KONSTRUKSI

SEKOLAH PASCA SARJANA

INSTITUT TEKNOLOGI BANDUNG

2014

1

7-14. The Electrocomp Corp manufactures 2 electrical products: Air Conditioner and Large Fans. The

assembly process for each is similar in the both require a certain amount of wiring and drilling. Each

air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of

wiring and 1 hour of drilling. During the next production period, 240 hours of wiring time are

available and up to 140 hours of drilling time may be used. Each air conditioner sold yields a profit of

$25. Each fan assembled may be sold for a $15 profit. Formulate and solve this LP production mix

situation to find the best combination of air conditioners and fans that yields the highest profit. Use

the corner point graphical approach.

= Air Conditioner

= Large Fan

Objective Function: max profit = 25 + 15

s.t. 3 + 2 240 (hours of wiring time available)

2 + 140 (hours of drilling time available)

, 0

Dengan menggunakan metode corner point, perlu dicari titik-titik setiap persamaan sehingga dapat dibuat

garis dari persamaan tersebut, yaitu:

a. wiring time constraint

(0, 3(0) + 2 = 240

= 120 (0, 120)

( 0) 3 + 2(0) = 240

= 80 (80, 0)

b. drilling time constraint

(0, ) 2(0) + = 140

= 140 (0, 140)

(, 0) 2 + (0) = 140

= 70 (70, 0)

2

Titik perpotongan antara garis 1 & garis 2 adalah:

3 + 2 240 = 2 (2 + 140)

31 + 2 240 = 4 + 2 280

40 =

= 140 2(40) = 60 (40,60)

Langkah selanjutnya adalah dengan menghitung nilai dari fungsi tujuan di setiap titik:

AC () Fan () Profit = 25 + 15

0 0 0 0 120 1800

70 0 1750 40 60 1900

Jadi kombinasi yang memberikan profit paling besar adalah 40 AC dan 60 Fan.

0

20

40

60

80

100

120

140

160

0 20 40 60 80 100

Fan

Air Conditioner

Series1

series 2

Linear (Series1)

Linear (series 2)

3

7-15. Electrocomps management realizes that it forgot to include 2 critical constraints (see Problem 7-14).

In particular, management decides that there should be a minimum number of air conditioners

produced in order to fulfill a contract. Also, due to an oversupply of fans in the preceding period, a

limit should be placed on the total number of fans produced.

(a) If Electrocomp decides that at least 20 AC should be produced but no more than 80 fans should

be produced, what would be the optimal solutions? How much slack is there for each of the

four constraints?

tambahan constraint:

20

80

solusi optimal (40, 60) tidak berubah karena masih memenuhi tambahan constraint.

Constraint Slack (Amount of resource available) (Amount of resource used)

Surplus (Actual amount) (Minimum amount)

Wiring Hours 0 Drilling Hours 0

Fan Production 20 AC 20

(b) If Electrocomp decides that at least 30 AC should be produced but no more than 50 fans should be

produced, what would be the optimal solution? How much slack is there for each of the four constraints

at the optimal solution?

Karena jumlah maksimal kipas dikurangi menjadi lebih kecil daripada jumlah kipas pada solusi optimal

sebelumnya, maka solusi optimal juga berubah, yaitu:

Air Conditioner (X) = 45

Large Fan (Y) = 50

profit = 25(45) +15(50) = $1875

Air Conditioner

X

Large Fan

Y

jumlah 40 60 profit

max 25 15 1900

Hours Used Constraint

wiring 3 2 240

4

Constraint Slack (Amount of resource available) (Amount of resource used)

Surplus (Actual amount) (Minimum amount)

Wiring Hours 0 Drilling Hours 0

Fan Production 0 AC 15

7-19. MSA Computer Corporation manufactures 2 models of minicomputers, the Alpha 4 and the Beta 5.

The firm employs 5 technicians, working 160 hours each per month, on its assembly line.

Management insists that full employment (i.e., all 160 hours of time) be maintained for each worker

during next months operations. It requires 20 labor hours to assemble each Alpha 4 computer and

25 labor hours to assemble each Beta 5 model. MSA wants to see at least 10 Alpha 4s and at least 15

Beta 5s produced during the production period. Alpha 4s generate $1,200 profit per unit, and Beta

5s yield $1,800 each. Determine the most profitable number of each model of minicomputer during

the coming month.

= Alpha 4

= Beta 5

Objective Function: max profit = 1200A + 1800B

s.t. 20A + 25B = 800 (semua labor hours digunakan)

A 10, B 15

Dengan menggunakan Excel Solver, dicari solusi optimal dari model yang dibuat.

Dengan memproduksi jumlah Alpha 4 minimal (10 unit) dan memaksimalkan jumlah Beta 5 (24

unit) akan didapatkan profit sebesar $55.200

7-24. The stock brokerage firm of Blank, Leibowitz, and Weinberger has analyzed and recommended two

stocks to an investors club of college professors. The professors were interested in factors such as

short-term growth, intermediate growth, and dividend rate. These data on each stock are as follows:

Factor Stock ($)

Louisiana Gas and Power Trimex Insulation Company Short-term growth potential Per dollar invested

.36 .24

Intermediate growth potential (over next 3 years) Per dollar invested

1.67 1.50

Dividend rate potential 4% 8%

Alpha 4

A

Beta 5

B

jumlah 10 24 profit

max 1200 1800 55200

Resource used Constraint

labor hours 20 25 800 = 800

Alpha qty 1 10 >= 10

Beta qty 1 24 >= 15

5

Each member of the club has an investment goal of:

(1) an appreciation of no less than $720 in the short term,

(2) an appreciation of at least $5,000 in the next 3 years, and

(3) a dividend income of at least $200 per year.

What is the smallest investment that a professor can make to meet these three goals?

Ketiga goal diatas, dijadikan sebagai constraint yang harus dipenuhi dengan fungsi objektif minimasi

biaya investasi:

= Besarnya investasi di Louisiana Gas and Power

= Besarnya investasi di Trimex Insulation Company

Karena investasi menggunakan uang yang mempunyai satuan terkecil tertentu (dalam sen), maka untuk mempermudah

satuan terkecil dijadikan 1$ (integer)

Objective Function: min investment = X + Y

s.t. 0.36X + 0.24Y 720 (short-term growth)

1.67X + 1.50Y 5000 (intermediate growth)

0.04X + 0.08Y 200 (dividend per year)

X, Y 0 (integer)

Dengan menggunakan perhitungan melalui Excel Solver didapatkan bahwa nilai investasi minimum

yang dapat diberikan dengan memenuhi ketiga syarat adalah sebesar $3.180

7-27. Consider the following 4 LP formulations. Using a graphical approach, determine

(a) which formulation has more than 1 optimal solution

(b) which formulation is unbounded

(c) which formulation has no feasible solution

(d) which formulation is correct as is

Persamaan 1

Maximize 101 + 102

s.t. 1 5

1 + 22 8 (8,0) dan (0,4)

2 2

1 = 6

Lousiana Gas and Power

(X)

Trimex Insulation Company

(Y)

jumlah 1360 1820 investment

min 1 1 3180

1360 1820

Resource used Constraint

short 489,6 436,8 926,4 >= 720

intermediate 2271,2 2730 5001,2 >= 5000

Dividend 54,4 145,6 200 >= 200

6

Tidak ada daerah solusi (c)

Persamaan 2

Maximize 1 + 22

s.t. 1 1

2 1

1 + 22 2 (2,0) dan (0,1)

Karena fungsi objektif parallel dengan constraint 1 + 22 2 , maka model ini memiliki lebih

dari 1 solusi optimal. (a)

Persamaan 3

Maximize 31 + 22

s.t. 1 + 22 5 (5,0) dan (0,5

2 )

1 2

2 4

X =6

Feasible

Region

Redundant constraint 2 1

1 + 22

7

Dari grafik dapat dilihat bahwa tidak ada yang membatasi jumlah maksimal 1 dan 2 sehingga

nilai solusi dari fungsi objektif dapat mencapai tak terhingga (b)

Persamaan 4

Maximize 31 + 32

s.t. 21 + 32 24 (12,0) dan (0,8)

21 + 2 6 . (3,0) dan (0,6)

2 1

1 1

Dari grafik di atas, meskipun ada constraint yang redundan, namun hasil solusinya memenuhi

syarat-syarat/constrain (d)

7-28. Graph the following LP problem and indicate the optimal solution point:

Maximize profit = $3X + $2Y

s.t. 2X + Y 150

2X + 3Y 300

(a) Does the optimal solution change if the profit per unit of X changes to $ 4.50?

(b) What happens if the profit function should have been $3X + $3Y?

Persamaan Titik yg dilewati

2X + Y 150 (75, 0) dan (0, 150) 2X + 3Y 300 (150, 0) dan (0, 100)

Feasible

Region

8

Dari gambar di atas, titik ekstrim yang ada pada daerah yang diarsis adalah pada potongan antara 2

garis constraint, yaitu:

2X + Y - 150 = 2X + 3Y 300

150 = 2Y

Y = 75 (X = 37.5)

Titik 3 + 2

(0, 0) 0 (75, 0) 225

(0, 100) 200 (37.5, 75) 262.5

Dari semua titik yang ada, nilai optimal terdapat pada titik (37.5, 75)

Adjustable Cells

Final Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease

$B$2 jumlah X 37,5 0 3 1 1,666666667

$C$2 jumlah Y 75 0 2 2,5 0,5

Constraints

Final Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$D$5 a Resource used 150 1,25 150 1E+30 1E+30

$D$6 b Resource used 300 0,25 300 1E+30 1E+30

(a) apabila fungsi tujuan berubah menjadi $4.5X + $2Y, apakah solusi optimal berubah?

Tanda merah pada tabel di atas menunjukkan besar perubahan yang dilakukan tanpa

mengubah solusi optimal.

batas maksimum perubahan pada profit X diubah adalah = $3+$1 = $4 (lebih kecil dari

$4.5)

Berdasarkan hasil Excel Solver didapatkan bahwa solusi optimal adalah X = 75, Y = 0

(b) apabila fungsi tujuan berubah menjadi $3X + $3Y, apa yang akan terjadi?

batas maksimum perubahan pada profit Y diubah adalah = $2+$2.5 = $4.5 , sehingga

perubahan koefisien pada Y menjadi 3 (masih di bawah batas maksium) tidak mengubah

solusi optimal. Hanya nilai dari profit saja yang berubah, yaitu:

profit = $3(37.5) + $3(75) = $337,5

X Y

75 0 profit

o.f. max 4,5 2 337,5

constrain

a 2 1 150

9

7-31. Consider the following LP problem:

Maximize profit = 5X +6Y

s.t. 2X + Y 120

2X + 3Y 240

X,Y 0

(a) What is the optimal solution to this problem? Solve it graphically.

(b) If a technical breakthrough occurred that raised the profit per unit of X to $8, would this affect

the optimal solution?

(c) Instead of an increase in the profit coefficient X to $8, suppose that profit was overestimated

and should only have been $3. Does this change the optimal solution?

(a) Grafik dari model di atas adalah sebagai berikut:

Titik 5X + 6Y

(0, 0) 0 (60, 0) 300 (0, 80) 480

(30, 60) 510

Dari hasil perhitungan nilai fungsi objektif masing-masing titik, solusi optimalnya adalah X =

30 dan Y = 60

(b) Dengan menggunakan QM, didapatkan analisis sensitivitas sebagai berikut:

10

Perubahan koefisien X pada fungsi tujuan menjadi 8 tidak akan mempengaruhi solusi optimal

karena nilai 8 masih di antara lower bound (4) dan upper bound (12), hanya akan mengubah

nilai dari profit yang didapat, yaitu:

Profit = 8(30) + 6(60) =600

(c) Apabila ternyata koefisien X berubah menjadi 3, maka solusi optimalnya akan berubah

karena sudah keluar dari batas lower bound (4), maka solusi optimal yang baru berubah

menjadi:

X = 0, Y = 80

Profit = 0 + 6(80) = 480

7-37. Bhavika Investments, a group of financial advisors and retirement planners, has been requested to

provide advice on how to invest $200,000 for one of its clients. The client has stipulated that the

money must be put into either a stock fund or a money market fund, and the annual return should

be at least $14,000. Other conditions related to risk have also been specified, and the following

linear program was developed to help with this investment decision:

Minimize risk = 12S + 5M

s.t. S + M = 200,000 (total investment is $200,000)

0.1S + 0.05M 14,000 (return must be at least $14,000)

M 40,000 (at least $40,000 must be in money market fund)

S, M 0

where

S = dollars invested in stock fund

M = dollars invested in money market fund

11

(a) How much money should be invested in money market fund (M) and the stock fund(S)? What

is the total risk?

Dari hasil yang ditampilkan QM di atas, solusi optimalnya adalah:

S = $ 80,000

M = $ 120,000

Risk = 12(80,000) + 5(120,000) = $ 1,560,000

(b) What is the total return? What rate of return is this?

Total return yang didapat adalah sebesar $ 14,000

Rate of return = 14,000

200 ,000 = 0.07

(c) Would the solution change if risk measure for each dollar in the stock fund(S) were 14 instead

of 12?

Karena upper bound dari variabel S adalah tak terhingga, maka kenaikan koefisien ke 14 tidak

mengubah solusi, hanya besarnya risk yang berubah, yaitu:

risk = 14(80,000) + 5(120,000) = $ 1,720,000 (d)

(d) For each additional dollar that is available, how much does the risk change?

Perubahan besarnya risk adalah $ 1,720,000 (perhitungan ada di (c))

(e) Would the solution change if the amount that must be invested in the money market fund

were changed from $40,000 to $50,000?

Perubahan nilai pada konstrain ketiga menjadi M $50,000 tidak mengubah solusi karena

belum melebihi upper bound (120,000)

12

8-15. (Agricultural production planning problem) Margaret Blacks family owns 5 parcels of farmland

broken into a southeast sector, north sector, northwest sector, west sector, and southwest sector.

Margaret is involved primarily in growing wheat, alfalfa, and barley crops and is currently preparing

her production plan for next year. The Pennsylvania Water Authority has just announced its yearly

water allotment, with the Black farm receiving 7,4000 acre-feet, Each parcel can only tolerate a

specified amount of irrigation per growing season, as specified in the following table:

Parcel Area (Acres) Water Irrigation Limit (Acre-feet)

Southeast 2,000 3,200 North 2,300 3,400 Northwest 600 800 West 1,100 500 Southwest 500 600

Each of Margarets crops needs a minimum amount of water per acre, and there is a projected limit

on sales of each crop. Crop data follow:

Crop Maximum Sales Water Needed per Acre

Wheat 110,000 bushels 1.6 Alfalfa 1,800 tons 2.9 Barley 2,200 tons 3.5

Margarets best estimate is that she can sell wheat at a net profit of $2 per bushel, alfalfa at $40 per

ton, and barley at $50 per ton. One acre of land yields an average of 1.5 tons of alfalfa and 2.2 tons

of barley. The wheat yield is approximately 50 bushels per acre.

(a) Formulate Margarets production plan

misal: wheat (W), Alfalfa (A), Barley (B)

asumsi: W, A, B adalah bilangan bulat, karena satuan terkecil yang digunakan adalah 1 acre

objective function: max profit = 2W + 40A + 50B

s.t. W 110,000

A 1,800

B 2,200

W + A + B 6,500 (luas lahan tersedia)

1.6W + 2.9A + 3.5B 8,500 (jumlah air tersedia)

W, A, B 0

(b) What should the crop plan be, and what profit will it yield?

dengan menggunakan Excel Solver didapatkan:

Wheat = 0

Alfalfa = 277

Barley = 2199

Solusi = $ 121,030

(gambar dari Excel Solver terdapat di halaman berikut)

13

(c) The Water Authority informs Margaret that for a special fee of $6,000 this year, her farm

will qualify for an additional allotment of 600 acre-feet of water. How should she respond?

Dengan penambahan biaya $ 6,000 dolar akan memberikan tambahan air sebesar 600 acre-

feet. Maka dari itu perlu dihitung apakah keuntungan yang didapat lebih besar daripada biaya

tambahan yang dikeluarkan atau tidak.

profit = ($ 129,300 - $ 121,030 =$ 8,270

Berarti dengan penambahan $ 6,000, keuntungan bersih yang didapatkan adalah:

$ 8,270 - $ 6,000 = $ 2,270

Karena profit yang didapatkan lebih besar dari biaya tambahan yang dikeluarkan maka

sebaiknya Margaret mengambil kesempatan tersebut.

variabel wheat alfalfa barley

solution 0 277 2199 Result

maximize 2 40 50 121.030

constraints LHS RHS

wheat max sale limit 1 0

14

8-16. (Material blending problem) Amalgamated Products has just received a contract to construct steel body frames for automobiles that are to be produced at the new Japanese factory in Tennessee. The Japanese auto manufacturer has strict quality control standards for all of its component subcontractors and has informed Amalgamated that each frame must have the following steel content:

Amalgamated mixes batches of eight different available materials to produce one ton of steel used in the body frames. The table on this page details these materials. Formulate and solve the LP model that will indicate how much each of the eight materials should be blended into a 1-ton load of steel so that Amalgamated meets its requirements while minimizing costs.

Jawaban :

Objective Function:

minimize f($) = 0.121 + 0.132 + 0.153 +0.091 + 0.072 +0.11 +0.122 +0.093

s.t.

0.021 0.71 + 0.552 + 0.123 +0.011 + 0.052

2000 0.023

0.043 .151 + 0.32 + 0.263 +0.11 + 0.032 +0.241 +0.252 +0.233

2000 0.046

0.0505 .151 + 0.32 + 0.263 +0.11 + 0.032 +0.241 +0.252 +0.233

2000 0.054

2 300 ; 1 50; 2 200; 3 300

1,2,3,1 ,2 ,1 ,2 ,3 0

15

Hasil perhitungan linear programming dengan menggunakan excel solve adalah sebagai berikut:

Hasil yang didapatkan adalah:

A1 : 60 pounds

C1 : 41.67 pounds

C2 : 200 pounds (resource max)

C3 : 100 pounds (resource max)

Berdasarkan hasil dari perhitungan excel, model tidak dapat memberikan solusi yang memenuhi konstrain spesifikasi (kandungan karbon di

bawah spesifikasi), dimana:

kandungan minimal carbon adalah 5.05%, sedangkan hasil dari solusi yang didapat hanya 3,715%

Apabila dilakukan penambahan campuran C1 sampai mencapai kandungan carbon minimum, maka akan berakibat pada kandungan silicon

melebihi batas maksimal. Oleh karena itu batasan spesifikasi tidak dapat dipenuhi.

variable A1 A2 A3 I1 I2 C1 C2 C3

(pounds) produced 60 0 0 0 0 41.66666048 200.000001 100.000001 cost

objective min cost 0.12 0.13 0.15 0.09 0.07 0.1 0.12 0.09 44.36666636

s.t. resource used RHS

manganese 0.70 0.55 0.12 0.01 0.05 - - - 2.100% min 2.1%

max 2.3%

silicon 0.15 0.30 0.26 0.10 0.03 0.24 0.25 0.23 4.600% min 4.3%

max 4.6%

carbon 0.03 0.01 - 0.03 - 0.18 0.20 0.25 3.715% min 5.05%

max 5.4%

A1 (pounds)

A2 (pounds) 1 0 max 300

A3 (pounds)

I1 (pounds)

I2 (pounds)

C1 (pounds) 1 41.66666048 max 50

C2 (pounds) 1 200.000001 max 200

C3 (pounds) 1 100.000001 max 100

16

10-13. An airline owns an aging fleet of Boeing 737 jet airplanes. It is considering a major purchase of up to 17 new Boeing model 757 and 767 jets. The decision must take into account numerous cost and capability factors, including the following: (1) the airline can finance up to $1.6 billion in purchases (2) each Boeing 757 will cost $80 million, and each Boeing 767 will cost $110 million (3) at least one-third of the planes purchased should be the longer-range 757 (4) the annual maintenance budget is to be no more than $8 million (5) the annual maintenance cost per 757 is estimated to be $800,000 and it is $500,000 for each

767 purchased (6) each 757 can carry 125,000 passengers per year, whereas each 767 can fly 81,000 passengers

annually.

Formulate this as an integer programming problem to maximize the annual passenger-carrying capability. What category of integer programming problem is this? Solve this problem

misal: - Boeing 757 =

- Boeing 767 = - jumlah pesawat adalah bilangan bulat (integer) karena satuan terkecil adalah 1 pesawat

objective function max capacity = 125,000X + 81,000Y

s.t. 80X + 110Y 1600 (in million).. financial

X 1 3 (X+ Y) menjadi 2X - Y 0 ............... jumlah pesawat

0.8X + 0.5Y 8 (in million) maintenance cost

karena yang jumlah pesawat harus dalam bentuk integer (semua variabel harus dalam bentuk integer),

maka kasus ini termasuk dalam pure integer programming.

Dengan menggunakan excel Solver didapatkan:

dengan X=5 dan Y=8, maka kapasitas totalnya adalah 1,273,000 penumpang

Boeing 757 Boeing 767

variable X Y

solution 5 8 result

max capacity 125.000 81.000 1.273.000

constraint LHS RHS

financial 80 110 1.280 1600

maintenance 0,8 0,5 8 8

plane qty 2 -1 2 0

17

10-16. Innis Construction Company specializes in building moderately priced homes in Cincinnati, Ohio. Tom Innis has identified eight potential locations to construct new single-family dwellings, but he cannot put up homes on all of the sites because he has only $300,000 to invest in all projects. The accompanying table shows the cost of constructiong homes in each area and the expected profit to be made from the sale of each home. Note that the home-building costs differ considerably due to lot costs, site preparation, and differences in the models to be built. Note also that a fraction of a home cannot be built.

Location Cost of building ($) Expected Profit ($)

Clofton 1 60,000 5,000 Mt. Auburn 2 50,000 6,000 Mt. Adams 3 82,000 10,000 Amberly 4 103,000 12,000 Norwood 5 50,000 8,000 Covington 6 41,000 3,000 Roselawn 7 80,000 9,000 Eden Park 8 69,000 10,000

(a) Formulate Innis problem using 0-1 integer programming

1,2, 3, 4, 5, 6, 7, 8 dalam bilangan biner

objective function max profit = 51 + 62 + 103 + 124 + 85 + 36 + 97 + 108

s.t. 601 + 502 + 823 + 1034 + 505 + 416 + 807 + 698 300

1,2, 3, 4, 5, 6, 7, 8 0

(b) Solve with QM for Windows or Excel

dari hasil perhitungan di atas, maka solusi optimalnya adalah dengan mengerjakan proyek di: - Mt. Aubum - Mt. Adams - Norwood - Covington - Eden Park Maka Innis akan mendapatkan profit sebesar $37,000

variable X1 X2 X3 X4 X5 X6 X7 X8

solution 0 1 1 0 1 1 0 1 result

maximize profit 5 6 10 12 8 3 9 10 37

constraint LHS RHS

financial 60 50 82 103 50 41 80 69 292 300

18

10-17. A real estate developer is considering 3 possible projects: a small apartment complex, a small shopping center, and a mini-warehouse. Each of these requires different funding over the next 2 years, and the NPV of the investments also varies. The following table provides the required investment amounts (in $1,000s) and the NPV of each (also expressed in $1,000s):

NPV

INVESTMENT

Year 1 Year 2 Apartment 1 18 40 30 Shopping center 2 15 30 20 Mini-warehouse 3 14 20 20

The company has $80,000 to invest in year 1 and $50,000 to invest in year 2.

(a) develop an integer programming model to maximize the NPV

1,2, 3, dalam bilangan biner

objective function maximize profit = 181 + 152 + 143

s.t. 401 + 302 + 203 80

301 + 202 + 203 50

1,2, 3 0

(b) solve using computer software. Which of the 3 projects would be undertaken if NPV is

maximized? How much money would be used each year?

Untuk memaksimasi NPV, maka proyek yang akan dikerjakan adalah Apartment dan Shopping Center dengan biaya investasi sebesar: year 1 = $70,000 year 2 = $50,000

variable X1 X2 X3

solution 1 1 0 result

maximize profit 18 15 14 33

constraint LHS RHS

year 1 40 30 20 70 80

year 2 30 20 20 50 50

19

10-19. Triangle Utilities provides electricity for 3 cities. The company has 4 electric generators that are used to provide electricity. The main generator operates 24 hours per day, with an occasional shutdown for routine maintenance. Three other generators (1,2,3) are available to provide additional power when needed. A startup cost is incurred each time one of these generators is started. The startup costs are $6,000 for 1; $5,000 for 2; and $4,000 for 3. These generators are used in the following ways: A generator may be started at 6:00 AM and run for either 8 hours or 16 hours, or it may be started at 2:00 PM and run for 8 hours (until 10:00 PM). All generators except the main generator are shut down at 10:00 PM. Forecasts indicate the need for 3,200 megawatts more than provided by the main generator before 2:00 PM, and this goes up to 5,700 megawatts between 2:00 and 10:00 PM. Generator 1 may provide up to 2,400 megawatts, generator 2 may provide up to 2,100 megawatts, and generator 3 may provide up to 3,300 megawatts. The cost per megawatt used per 8 hour period is $8 for 1, $9 for 2, and $7 for 3.

(a) Formulate this problem as an integer programming problem to determine the least-cost way to

meet the needs of the area

Power (megawatts)

Startup Cost ($)

Cost per megawatt per 8 hour($)

Total Cost per 8 hour ($)

Generator 1 1 2,400 6,000 8 19,200 Generator 2 2 2,100 5,000 9 18,900 Generator 3 3 3,300 4,000 7 23,100

1 = 1 jika generator 1 dinyalakan pada periode-1, 0 jika tidak






minimize cost = 61 + 52 + 43 + 19,21 + 18,92 + 21,73 (menggunakan C dan T untuk menyederhanakan fungsi tujuan)

dimana:

1= (1+4) 1= (1 + 42 -14)

2= (2+5) 2= (2 + 52 -25)

3= (3+6) 3= (3 + 62 -36)

s.t. 2,4001 + 2,1002 + 3,3003 3,200 (8 jam pertama)

2,4002 + 2,1003 + 3,3004 5,700 (8 jam kedua)

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(b) Solve using computer software

Dari hasil perhitungan melalui excel solver pada NLP objective function di atas, didapatkan solusi optimal berupa: Generator-1 dinyalakan mulai 2:00 PM (selama 8 jam) dan Generator-3 dinyalakan sejak 6:00 AM (selama 16 jam) sehingga total biaya operasionalnya adalah $72,600

variable x1 x2 x3 x4 x5 x6

values 0 0 1 1 0 1

terms C1 C2 C3 T1 T2 T3

calculated values 1 0 1 1 0 2 result

minimize cost 6 5 4 19,2 18,9 21,7 72,60

constraints LHS RHS

watt period 1 2400 2100 3300 3300 3200

watt period 2 2400 2100 3300 5700 5700

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10-24. An Oklahoma manufacturer makes 2 products speaker telephones (1) and pushbutton telephones (2). The following goal programming model has been formulated to find the number of each to produce each day to meet the firms goals:

Minimize 11 + 22

+ 33+ + 41

+

subject to 21 + 42 + 1 - 1

+ = 80

81 + 102 + 2 - 2

+ = 320

81 + 62 + 3 - 3

+ = 240 all , 0 Find the optimal solution using a computer

asumsi: setiap deviasi memiliki bobot yang sama yang bernilai 1. Dari hasil perhitungan pada QM

didapat 1 = 40, 2 = 0. Dari solusi tersebut dapat dilihat bahwa constrain 1 & 2 terpenuhi, sedangkan constrain 3

menunjukkan kelebihan nilai sebesar 80, sehingga total deviasi untuk model ini adalah 80.

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10-32. The integer programming problem in the box below has been developed to help First National Bank

decide where, out of 10 possible sites, to locate 4 new branch offices: where represents Winter Park, Maitland, Osceola, Downtown, South Orlando, Airport, Winter Garden, Apopka, Lake Mary, Cocoa Beach, for i equals 1 to 10, respectively.

(a) Where should the 4 new sites be located, and what will be the expected return?

dari hasil perhitungan di atas, 4 lokasi yang baru adalah: - South Orlando (5) - Apopka (8) - Lake Mary (9) - Cocoa (10)

expected return = 655 (b) If at least 1 new branch must be opened in Maitland () or Osceola (), will this change the

answers? Add the new constraint and re-run dengan kondisi seperti soal (b), tambahan constraint adalah X2 + X3 1

variable X1 X2 X3 X4 X5 X6 X7 X8 X9 X10

solution 0 0 0 0 1 0 0 1 1 1 result

maximize profit 120 100 110 140 155 128 145 190 170 150 665

constraint LHS RHS

1 20 30 20 25 30 30 25 20 25 30 105 110

2 15 5 20 20 5 5 10 20 5 20 50 50

3 1 1 1 1 1 2 3

6 1 1 1 1 1 1 1 1 1 1 4 4

4 1 1 1 1 1 3 2

5 1 1 1 1 1

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dengan tambahan batasan di atas, maka perubahan lokasi adalah sebagai berikut: - Osceola (3) - South Orlando (5) - Apopka (8) - Lake Mary (9)

expected return = 625 (c) The expected return at Apopka was overestimated. The correct value is $160,000 per year (that

is, 160). Using the original assumptions (namely, ignoring (b)), does your answer to part (a) change?

dari perubahan koefisien pada 8 dari 190 menjadi 160, mengubah tidak solusi optimal sebelumnya, hanya besarnya expected return yang berubah menjadi 635



maximize profit 120 100 110 140 155 128 145 190 170 150 625

constraint LHS RHS

a 20 30 20 25 30 30 25 20 25 30 95 110

b 15 5 20 20 5 5 10 20 5 20 50 50

c 1 1 1 1 1 1 3

d 1 1 1 1 1 1 1 1 1 1 4 4

e 1 1 1 1 1 4 2

f 1 1 1 1 1

g 1 1 1 1



maximize profit 120 100 110 140 155 128 145 160 170 150 635

constraint LHS RHS

a 20 30 20 25 30 30 25 20 25 30 105 ? 110

b 15 5 20 20 5 5 10 20 5 20 50 ? 50

c 1 1 1 1 1 2 ? 3

d 1 1 1 1 1 1 1 1 1 1 4 ? 4

e 1 1 1 1 1 3 ? 2

f 1 1 1 1 ? 1

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