CDB 4333Z – PROCESS OPTIMIZATION

LINEAR PROGRAMMING

ZULFAN ADI PUTRA, PDENG

OFFICE: 05-03-08TELP: 05 368 7562

EMAIL: Zulfan.adiputra@utp.edu.myDISCUSSION TIME: FRIDAY, 15.00-18.00

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mailto:zulfan.adiputra@utp.edu.my

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Objective and Constraints formulation

Multi variable

One variable

Unconstrained optimization

Constrained Optimization

Methods:• Function value,

• Bracketing,• Newton,• Secant

Methods:• Simplex,• Newton,• Steepest

descentLagrange multiplier

and Karush Kuhn Tucker algorithm

Linear programming via Simplex method

Non Linear programming

Methods:• SQP,• GRG,

• PenaltyInteger programming via Branch and Bound

COURSE OVERVIEW

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COURSE OUTCOME

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1. Refinery case: Problem formulation

2. Graphical solution for two variable problems

3. Degenerate linear programming (LP) problems

4. Simplex method

5. Sensitivity analysis

6. Duality in LP

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CONTENTS

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GENERAL FEATURES

▪Linear Programming (LP)→most widely used in chemical industry

▪LP implies that objective function and all constraints are linear

▪LP problems are convex programming, where:

•the objective function is convex (linear) and

•the constraints form a convex region or solution space

The solution isglobal optimum

The basic steps involved in formulating an LP model are:

1) Identify decision variables

2) Express constraints as linear equalities or inequalities

3) Write the objective function as a linear function

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Product Volume Percent Yield Max allowable production

Crude 1 Crude 2 (bbl/day)

Gasoline 80 44 24000

Kerosene 5 10 2000

Fuel Oil 10 36 6000

Residues 5 10

Processing cost (USD/bbl) 0.5 1

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REFINERY PRODUCTION PLANNING

Refinery

Crude 1Price = 24 USD/bbl

Crude 2Price = 15 USD/bbl

Gasoline, Price = 36 USD/bblKerosene, Price = 24 USD/bblFuel Oil, Price = 21 USD/bblResidues, Price = 10 USD/bbl

Two different crudes are available to be refined. The objective is to maximize the profit

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▪ Your company has two plants, which are in Barcelona and Antwerp,and three market hubs, located in Rotterdam, Frankfurt, and Porto. Thedistances between the plants and the markets, together with thecapacities and demands of each hub are shown in TABLE Q1. Thetransportation cost is 10 RM/ton∙km.

▪ Determine the amount of goods that has to be transported fromeach plant to each market hub to minimize the total cost oftransportation

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MARKET AND DEMAND PROBLEM

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▪ Your company just bought two lubricant factories as subsidiary companies.Until now, they are only selling the lubricants via historical partnershipwith their customers. Apparently, they have two same customers.

▪ This situation does not make the new supply chain manager happy. So, hecalls you as the recommended senior process engineer within the company.

▪ He tells you that the 1st factory has 400 tons of lubricants in their storage,while the other has 300 tons. The 1st customer needs 200 tons oflubricants, while the other requires 300 tons. The supply chain managershows you the cost of shipments ($/ton) as shown below.

▪ Then, he asks for your advice on how much lubricant from each factoryshould be sent to each customer. What are you going to say?

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CUSTOMER PROBLEM (AND MANY MORE LP…)

Customer 1 Customer 2

Factory 1 $ 30 $ 25

Factory 2 $ 36 $ 30

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Product Volume Percent Yield Max allowable production

Crude 1 Crude 2 (bbl/day)

Gasoline 80 44 24000

Kerosene 5 10 2000

Fuel Oil 10 36 6000

Residues 5 10

Processing cost (USD/bbl) 0.5 1

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OPENING QUESTION: FORMULATE THE OPTIMIZATION PROBLEM

Refinery

Crude 1Price = 24 USD/bbl

Crude 2Price = 15 USD/bbl

Gasoline, Price = 36 USD/bblKerosene, Price = 24 USD/bblFuel Oil, Price = 21 USD/bblResidues, Price = 10 USD/bbl

Two different crudes are available to be refined. The objective is to maximize the profit

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STEP 1. DEFINE VARIABLES

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STEP 2. FORMULATE THE CONSTRAINTS

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STEP 3. FORMULATE THE OBJECTIVE FUNCTION

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FULL VERSION OF THE LP PROBLEM FORMULATION

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THE REDUCED VERSION

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GRAPHICAL SOLUTION

Graphical solution shows that the optimum value of 8.1x1+10.8x2 is at:X1 = 26206 bbl/dayX2 = 6896 bbl/dayWith the maximum profit of 286758 $/day

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SIMPLEX METHOD (STANDARD MAXIMIZATION)

Standard form of simplex maximization problem

Maximize z = c1x1 + c2x2 + … + cnxn

Subject to: a11x1 + a12x2 + … a1nxn ≤ b1

… …

am1x1 + am2x2 + … amnxn ≤ bm

x1 ≥ 0, … xn ≥ 0

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SIMPLEX METHOD (STANDARD MAXIMIZATION)

Steps in Simplex Method:1. If the problem is min z, convert it to max y = –z

2. Introduceslack variables and convert the inequalities into equality constraints:ai1x1 + ai2x2 + … ainxn ≤ bi → ai1x1 + ai2x2 + … ainxn + si = bi

if larger than and equal to, convert to less than and equal to, then introduce the slack variableai1x1 + ai2x2 + … ainxn ≥ bi → -ai1x1 - ai2x2 - … ainxn + si = -bi

si ≥ 0

3. Construct the problem into matrix of coefficients, vector of variables, and vector of the right handsideof the equations.

4. Check if basic variables (variable that only appear in ONE (1) equation) are all positive. If not, these variables do not make feasible basic solution. Go to step 6.

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SIMPLEX METHOD (STANDARD MAXIMIZATION)

5. Perform below calculations:a) Find pivot column based on the most negative coefficient of the objective functionb) Find pivot row, the smallest positive ratio between the righthandside numbers and the

coefficients in the pivot column. The intersection between the pivot column and pivot row is the pivot point.

c) Make the pivot point into 1d) Apply Gauss-Jordan elimination to make the rest of the numbers in the pivot column zero.e) Back to (a) until there is no negative coefficient

6. Perform below calculations to remove all negative coefficients on the righthand side:a) Find pivot row based on the most negative coefficient on the righthand side numbersb) Find pivot column based on the most negative coefficient in the pivot row. The intersection

between the pivot column and pivot row is the pivot point.c) Make the pivot point into 1d) Apply Gauss-Jordan elimination to make the rest of the numbers in the pivot column zero.e) Back to (a) until there is no negative coefficient

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SIMPLE EXAMPLE

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

Objective function, f = 2x1 + x2A(0,4), f = 4B(0,1), f = 1 → minimum valueC(1,0), f = 2D(16/7, 9/7), f = 41/7 → maximum valueE(1,2), f = 4

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MAXIMIZATION

Maximize f = 2x1 + x2Standardize the constraints: x1 + x2 ≥ 1 → -x1 - x2 ≤ -1 → -x1 - x2 +s1 = -13x1 + 4x2 ≤ 12 → 3x1 + 4x2 +s2 = 12x1 – x2 ≤ 1 → x1 – x2 + s3 = 1

Make table in standard simplex LP maximization problem

x1 x2 s1 s2 s3 f

R1 -1 -1 1 0 0 0 -1

R2 3 4 0 1 0 0 12

R3 1 -1 0 0 1 0 1

R4 -2 -1 0 0 0 1 0

Check basic solution:x1 = 0 → not in feasible regionx2 = 0 → not in feasible regions1 = -1 → not feasibles2 = 12, s3 = 1 → OK (positive)

Move to the feasible region by removing all negatives on the right-hand side

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1ST ITERATION (MAXIMIZATION)

x1 x2 s1 s2 s3 f

R1 -1 -1 1 0 0 0 -1

R2 3 4 0 1 0 0 12

R3 1 -1 0 0 1 0 1

R4 -2 -1 0 0 0 1 0

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

Pick arbitrary between these two negatives

1

2

Pivot selected is at step (3), which is -1Hence, R1 = R1/-1R2 = R2 -4R1R3 = R3 + R1R4 = R4 + R1

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1ST ITERATION (MAXIMIZATION)

x1 x2 s1 s2 s3 f

R1 1 1 -1 0 0 0 1

R2 -1 0 4 1 0 0 8

R3 2 0 -1 0 1 0 2

R4 -1 0 -1 0 0 1 1

Check basic solution:x1 = 0 → OKx2 = 1 → OKs1 = 0 → OKs2 = 8 → OKs3 = 2 → OK

All are feasiblex1 and x2 are now at point B(0,1). Moved from previously (0,0)Continue with standard simplex, removing all negatives in R4

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

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x1 x2 s1 s2 s3 f

R1 1 1 -1 0 0 0 1

R2 -1 0 4 1 0 0 8

R3 2 0 -1 0 1 0 2

R4 -1 0 -1 0 0 1 1

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2ND ITERATION (MAXIMIZATION)

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

Pick arbitrary between these two negatives

2

1

Pivot selected is at step (3), which is 2Hence, R3 = R3/2R1 = R1 – R3R2 = R2 + R3R4 = R4 + R3

3Both gives the same value, 1.Pick one

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2ND ITERATION (MAXIMIZATION)

x1 x2 s1 s2 s3 f

R1 0 1 -1/2 0 -1/2 0 0

R2 0 0 7/2 1 ½ 0 9

R3 1 0 -1/2 0 ½ 0 1

R4 0 0 -3/2 0 ½ 1 2

Check basic solution:x1 = 1 → OKx2 = 0 → OKs1 = 0 → OKs2 = 9 → OKs3 = 0 → OK

All are feasiblex1 and x2 are now at point C(1,0). Moved from previously B(0,1)Continue with standard simplex, removing all negatives in R4

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

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x1 x2 s1 s2 s3 f

R1 0 1 -1/2 0 -1/2 0 0

R2 0 0 7/2 1 ½ 0 9

R3 1 0 -1/2 0 ½ 0 1

R4 0 0 -3/2 0 ½ 1 2

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3RD ITERATION (MAXIMIZATION)

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

The only negative

2

1

Pivot selected is at step (3), which is 7/2Hence, R2 = R2/(7/2)R1 = R1 + 1/2R2R3 = R3 + 1/2R2R4 = R4 + 3/2R2

3

The only positive

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3RD ITERATION (MAXIMIZATION)

x1 x2 s1 s2 s3 f

R1 0 1 0 1/7 -3/7 0 9/7

R2 0 0 1 2/7 1/7 0 18/7

R3 1 0 0 1/7 4/7 0 16/7

R4 0 0 0 3/14 5/7 1 41/7

Check basic solution:x1 = 16/7 → OKx2 = 9/7 → OKs1 = 18/7 → OKs2 = 0 → OKs3 = 0 → OK

All are feasiblex1 and x2 are now at point D(16/7,9/7). Moved from previously C(0,1)Maximum value of f(16/7,9/7) = 41/7

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

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MINIMIZATION

Minimize f = 2x1 + x2 → transform to maximize, y = -f = - 2x1 - x2 Standardize the constraints: x1 + x2 ≥ 1 → -x1 - x2 ≤ -1 → -x1 - x2 +s1 = -13x1 + 4x2 ≤ 12 → 3x1 + 4x2 +s2 = 12x1 – x2 ≤ 1 → x1 – x2 + s3 = 1

Make table in standard simplex LP maximization problem

x1 x2 s1 s2 s3 y

R1 -1 -1 1 0 0 0 -1

R2 3 4 0 1 0 0 12

R3 1 -1 0 0 1 0 1

R4 2 1 0 0 0 1 0

Check basic solution:x1 = 0 → not in feasible regionx2 = 0 → not in feasible regions1 = -1 → not feasibles2 = 12, s3 = 1 → OK (positive)

Move to the feasible region by removing all negatives on the right-hand side

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1ST ITERATION (MINIMIZATION)

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

Pick arbitrary between these two negatives

1

2

Pivot selected is at step (3), which is -1Hence, R1 = R1/-1R2 = R2 -4R1R3 = R3 + R1R4 = R4 - R1

3x1 x2 s1 s2 s3 y

R1 -1 -1 1 0 0 0 -1

R2 3 4 0 1 0 0 12

R3 1 -1 0 0 1 0 1

R4 2 1 0 0 0 1 0

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x1 x2 s1 s2 s3 y

R1 1 1 -1 0 0 0 1

R2 -1 0 4 1 0 0 8

R3 2 0 -1 0 1 0 2

R4 1 0 1 0 0 1 -1

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1ST ITERATION (MINIMIZATION)

A (0,4)

B (0,1)

C (1,0)

D (16/7,9/7)

E (1,2)

x1-x2 ≤ 1

3x1+4x2 ≤ 12

x1+x2 ≥ 1

x1

x2

Check basic solution:x1 = 0 → in feasible regionx2 = 1 → in feasible regions1 = 0 → feasibles2 = 8, s3 = 2 → OK (positive)

All are feasibleX1 and x2 are now at B(1,0). Moved from (0,0)R4 are all positive. No further calculationMaximum value of y = -1 = -f Minimum value of f = -(-f) = 1

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LET’S SOLVE THE REFINERY PROBLEM

Step 1:0.8x1 + 0.44x2 ≤ 240000.05x1 + 0.1x2 ≤ 20000.1x1 + 0.36x2 ≤ 6000

Step 2:0.8x1 + 0.44x2 + s1 = 24000 (a)0.05x1 + 0.1x2 + s2 = 2000 (b)0.1x1 + 0.36x2 + s3 = 6000 (c)f – 8.1x1 – 10.8x2 = 0

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Make table in standard simplex LP maximization problem

x1 x2 s1 s2 s3 f

R1 0.8 0.44 1 0 0 0 24000

R2 0.05 0.1 0 1 0 0 2000

R3 0.1 0.36 0 0 1 0 6000

R4 -8.1 -10.8 0 0 0 1 0

Check basic solution:x1 = 0 → In feasible regionx2 = 0 → In feasible regions1, s2, s3 = 0 → feasible

Non Basic Basic

1ST ITERATION

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x1 x2 s1 s2 s3 f

R1 0.8 0.44 1 0 0 0 24000

R2 0.05 0.1 0 1 0 0 2000

R3 0.1 0.36 0 0 1 0 6000

R4 -8.1 -10.8 0 0 0 1 0

Gauss Jordan elimination:R1 = R1 – 0.44R3R2 = R2 - 0.1R3 R3 = R3/0.36R4 = R4 + 10.8R3

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1

Most negative column

Smallest positive ratio

Pivot point3

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x1 x2 s1 s2 s3 f

R1 61/90 0 1 0 -11/9 0 50000/3

R2 1/45 0 0 1 -5/18 0 1000/3

R3 5/18 1 0 0 25/9 0 50000/3

R4 -5.1 0 0 0 0 1 180000

Check solution:x1 = 0 → feasiblex2 = 50000/3 = 16666.67 → feasibles1, s2 ≥ 0 → feasibles3 = 0 → feasiblef = 180000 → point (1)

Basic

2ND ITERATION

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x1 x2 s1 s2 s3 f

R1 61/90 0 1 0 -11/9 0 50000/3

R2 1/45 0 0 1 -5/18 0 1000/3

R3 5/18 1 0 0 25/9 0 50000/3

R4 -5.1 0 0 0 0 1 180000

2

1

The only negative column

Smallest positive ratio

Pivot point3

Gauss Jordan elimination:R1 = R1 – 61/90R2R2 = R2*45 R3 = R3 – 5/18R2R4 = R4 + 5.1R2

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x1 x2 s1 s2 s3 f

R1 0 0 1 -30.5 7.25 0 6500

R2 1 0 0 45 -12.5 0 15000

R3 0 1 0 -12.5 6.25 0 12500

R4 0 0 0 229.5 -33.75 1 256500

Check solution:x1 = 15000 → feasiblex2 = 12500 → feasibles1 = 0 → feasibles2, s3 ≥ 0 → feasiblef = 256500 → point (2)

Basic

3RD ITERATION

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x1 x2 s1 s2 s3 f

R1 0 0 1 -30.5 7.25 0 6500

R2 1 0 0 45 -12.5 0 15000

R3 0 1 0 -12.5 6.25 0 12500

R4 0 0 0 229.5 -33.75 1 256500

2

1

The only negative column

Smallest positive ratio

Pivot point3

Gauss Jordan elimination:R1 = R1/7.25R2 = R2 + 12.5R1 R3 = R3 – 6.25R1R4 = R4 + 33.75R1

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x1 x2 s1 s2 s3 f

R1 0 0 4/29 -4.21 1 0 26000/29

R2 1 0 50/29 2830/9 0 0 26206

R3 0 1 -0.86 -38.8 0 0 6896

R4 0 0 4.65 87.41 0 1 286758

Check solution:x1 = 26206 → feasiblex2 = 6896→ feasibles1, s3 = 0 → feasibles2 ≥ 0 → feasiblef = 286758 → point (3)No more negative values in R4. Iteration stops

Basic

4TH ITERATION

Basic

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x1 x2 s1 s2 s3 f

R1 0 0 4/29 -4.21 1 0 26000/29

R2 1 0 50/29 2830/9 0 0 26206

R3 0 1 -0.86 -38.8 0 0 6896

R4 0 0 4.65 87.41 0 1 286758

Basic

4TH ITERATION

Basic

Maximum value is 286758 $/dayx1 = 26206 bbl/dayx2 = 6896 bbl/day

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EXCEL SOLVER

We’ll do the same exercise in Excel Solver

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SENSITIVITY ANALYSIS

• In all LP problems, all coefficients are supplied as input data to the model

• These coefficients are not 100% certain. There are uncontrollable situations

• Hence, only finding the optimal solution does not solve a practical problem

• Thus, sensitivity analysis is done by varying the coefficients and check theresults

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SENSITIVITY ANALYSIS FOR SMALL PROBLEM

This final table from the refinery case, has the following equation:f = -4.65s1 – 87.41s2 + 286758

See it in a different perspective:At s1 and s2 = 0, we get the maximum value of f, which is 286758

x1 x2 s1 s2 s3 y

R1 0 0 4/29 -4.21 1 0 26000/29

R2 1 0 50/29 2830/9 0 0 26206

R3 0 1 -0.86 -38.8 0 0 6896

R4 0 0 4.65 87.41 0 1 286758

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SENSITIVITY ANALYSIS FOR SMALL PROBLEM

In the optimal solution (point (3)), both gasoline and kerosene are constrained. But not fuel oil!

(Example) constraint for gasoline: 0.8x1 + 0.44x2 ≤ 24000, x1 and x2 are the amount of crude 1 and crude 2 (bbl/day), or0.8x1 + 0.44x2 + s1 = 24000 → s1 (slack for gasoline produced) is 0 in the optimal solution

If we relax the production of gasoline by 1 unit (s1 = -1 bbl/day), then Δf = 4.65 $/day.This 4.65$/bbl is the marginal price for constraint A (gasoline).

A = constraint for gasolineB = constraint for keroseneC = constraint for fuel oil

Equation from the final table:f = -4.65s1 – 87.41s2 + 286758

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SENSITIVITY ANALYSIS

• Similarly, we can do the same thing with the kerosene constraint. Its marginal price is 87.41 $/bbl.

• Based on the final equation (f = -4.65s1 – 87.41s2 + 286758), small change in fuel oil does not make any difference.

• If crude oil price change, the coefficients in the objective function will change.

• In the end, iterative calculations are necessary!

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DEGENERATE LP PROBLEMS

Degenerate LP problems are LP problems that:- does not have a feasible region, or- does not have a finite solution

Minimize f(x) = -x1-x2Subject to: 3x1-x2 ≥ 0 (A)

x2 ≤ 3 (B)x1, x2 ≥ 0

Minimize f(x) = -x1-x2Subject to: x1+x2 ≤ -2 (A)

x1 +2x2 ≤ 0 (B)x1, x2 ≥ 0

The bigger the number, the minimum the objective function → has no finite solution

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DUALITY IN LINEAR PROGRAMMING

• Marginal prices of:• gasoline = $ 4.65/bbl• kerosene = $ 87.41/bbl, while • fuel oil = $ 0/bbl

• If maximum capacity of these products are multiplied by their marginal prices, we get

4.65(24000) + 87.41(2000) + 0(6000) = 286758

This is the same as the optimum profit calculated for this problem

• This equivalence is an important property of LP, known as “duality”

• It implies that original LP (called “primal” problem) could be formulated in an alternative way called “dual” problem

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DUAL PROBLEM FORMULATION

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SIMPLEX METHOD

(MINIMIZATION VIA DUAL PROBLEM)

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SIMPLEX METHOD

(MINIMIZATION VIA DUAL PROBLEM)

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Original problem:

Minimize f = 3x1 + 5x2 f = cTx

Subject to x1 + 3x2 ≥ 10 Ax ≥ b

2x1 – x2 ≥ 4

-x1 + 4x2 ≥ -2

-x1 – x2 ≥ -20

x1, x2 ≥ 0

Dual problem: f = bTy

ATy ≤ c

Maximize f = 10y1 + 4y2 – 2y3 – 20y4y1 + 2y2 – y3 – y4 ≤ 1

3y1 – y2 + 4y3 – y4 ≤ 5

• Dual problem is advised for problems with many constraints and few variables.

• Minimization problems are solved by converting them to the maximization problem via dual problem.

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EXAMPLE

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▪ Please minimize w = 29y1 + 10y2▪ Subject to: 3y1 + 2y2 ≥ 2

5y1 + y2 ≥ 3

y1, y2 ≥ 0

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PAIR TEST: DUALITY PROBLEM

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▪ Standard linear programming problem:

• Maximization

• Inequality is less than or equal to (≤)

▪ Non-standard linear programming problem:

• Minimization

• Inequality is combination of both ≤ and ≥

▪ How to solve non-standard linear programming problem:

• Transform into standard maximization problem

• Change minimize to maximize

• Change inequality from ≥ to ≤

• If some of the constraints are substituted into the other constraints and the objective function, make sure that the respective variables have to be positive (all variables are positive) by adding necessary additional constraints

• Use dual problem, if applicable

• JUST USE MS EXCEL or any other LP software! ☺

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NON-STANDARD LINEAR PROGRAMMING PROBLEM

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▪ Your company just bought two lubricant factories as subsidiary companies.Until now, they are only selling the lubricants via historical partnershipwith their customers. Apparently, they have two same customers.

▪ This situation does not make the new supply chain manager happy. So, hecalls you as the recommended senior process engineer within the company.

▪ He tells you that the 1st factory has 400 tons of lubricants in their storage,while the other has 300 tons. The 1st customer needs 200 tons oflubricants, while the other requires 300 tons. The supply chain managershows you the cost of shipments ($/ton) as shown below.

▪ Then, he asks for your advice on how much lubricant from each factoryshould be sent to each customer. What are you going to say?

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FINAL QUESTION: CASE STUDY

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Objective and Constraints formulation

Multi variable

One variable

Unconstrained optimization

Constrained Optimization

Methods:• Function value,

• Bracketing,• Newton,• Secant

Methods:• Simplex,• Newton,• Steepest

descentLagrange multiplier

and Karush Kuhn Tucker algorithm

Linear programming via Simplex method

Non Linear programming

Methods:• SQP,• GRG,

• PenaltyInteger programming via Branch and Bound

COURSE OVERVIEW

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