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Goal Programming

Goal Programming - Definition

Goal programming (GP) is a branch of Multi Objective Optimization, which in turn is abranch of Multi-Criteria Decision Analysis (MCDA). This is an optimization program. Itcan be thought of as an extension or generalization of linear programming to handlemultiple, normally conflicting objective measures. Each of these measures is given agoal or target value to be achieved. Unwanted deviations from this set of targetvalues are then minimized in an achievement function. This can be a vector ora weighted sum dependent on the goal programming variant used. As satisfaction ofthe target is deemed to satisfy the decision maker(s), an underlying satisficingphilosophy is assumed.

Goal programming is used to perform three types of analysis:

• Determine the required resources to achieve a desired set of objectives.• Determine the degree of attainment of the goals with the available resources.• Providing the best satisfying solution under a varying amount of resources and

priorities of the goals.

Basic DefinitionsDecision Maker

: The decision maker(s) refer to the person(s), organization(s), orstakeholder(s) to whom the decision problem under consideration belongs.

Criterion : A criterion is a single measure by which the goodness of any solution to adecision problem can be measured. There are many possible criteria arisingfrom different fields of application but some of the most commonly arisingrelate at the highest level to:• Cost• Profit• Time• Distance• Performance of a system• Company or organizational strategy• Personal preferences of the decision maker(s)• Safety considerations

A decision problem which has more than one criterion is therefore referredto as a multi-criteria decision making (MCDM) or multi-criteria decision aid(MCDA) problem. The space formed by the set of criteria is known ascriteria space.

Decision Variable

: A decision variable is defined as a factor over which the decision maker hascontrol. An example is a manufacturing company which has to decide howmany of a certain product to make in the next month.The set of decision variables fully describe the problem and form thedecision to be made. The purpose of the goal programming model can beviewed as a search of all the possible combinations of decision variablevalues (known as decision space) in order to determine the point which bestsatisfies the decision maker’s goals and constraints.

Objective : An objective is be referred to as a criterion with the additional informationof the direction (maximize or minimize) in which the decision maker(s)prefer on the criterion scale, for example minimize cost or maximize theperformance of a system. A decision problem with a set of objectives to bemaximized or minimized is referred to as a multi-objective optimizationproblem. In practice, these objectives will be conflicting, that is they cannotreach their optimal values simultaneously. If they could, then the model canbe solved as a single-objective problem for any of the objectives. The spaceformed by the values of the set of objectives is known as objective space.

Aspiration Level

: The numerical value specified by the decision maker that reflects his/herdesire or satisfactory level with regard to the objective function underconsideration

Goal : An objective function along with its aspiration level

Goal Deviation : The difference between what we actually achieve and what we desire toachieve.If the Goal Deviation is positive, it reflects over-achievement of a goalas the actual achieved value is more than the aspiration level.On the other hand, a negative value of Goal Deviation denotes under -achievement of the goal as the aspired level is more than what we couldactually manage to achieve.

➢ Desirable vs. Undesirable Deviations: (depend on the objectives)• Max goals (≥) - the more the better - 𝑑𝑖

+ or 𝑝𝑖 desirable.• Min goals (≤) - the less the better - 𝑑𝑖

− or 𝑛𝑖 desirable.• Exact goals (=) - exactly equal - both 𝑑𝑖

+ (or 𝑝𝑖) and 𝑑𝑖− (or 𝑛𝑖) undesirable

➢ In all the situations, we first identify the undesirable deviation of the expression inthe goal and then attempt to minimize the same.

➢ In GP, the objective is to minimize the (weighted) sum of undesirable deviations(all undesirable 𝑑𝑖

+ (or 𝑝𝑖) and 𝑑𝑖− (or 𝑛𝑖) →→ 0 ).

➢ For each goal, at least, one of 𝑑𝑖+ (or 𝑝𝑖) and 𝑑𝑖

− (or 𝑛𝑖) must be equal to "0“

➢ The actual constraints of the problem are called RIGID goals or HARD goalswhereas, the goals representing the given objective functions are called SOFTgoals. When a set of criteria has been determined it is necessary to distinguishbetween goals and hard constraints. Hard constraints are in variable space and areconditions that must be satisfied in order for the solution to be implementable.Any condition that does not fulfil this requirement should be included as a goaland not a hard constraint.

Formulation of GP Problems

➢ Goals are prioritized in some sense, and their level of aspiration is stated.

➢ An optimal solution is attained when all the goals are reached as close as possible to their aspiration level, while satisfying a set of constraints.

➢ There are two types of goal programming models:

Formulation of GP Problems

Non- preemptivegoal programming

: No goal is pre-determined to dominate any other goal

Pre-emptive goal programming

: Goals are assigned different priority levels. Level 1goal dominates level 2 goal, and so on.

Examples

1. Beaver Creek Pottery Company Example

2. Nicolo Investment Advisors Problem

Beaver Creek Pottery Company Example

Beaver Creek Pottery Company is a small crafts operation run by a Native Americantribal council. The company employs skilled artisans to produce clay bowls and mugswith authentic Native American designs and colors. The two primary resources usedby the company are special pottery clay and skilled labor. Given these limitedresources, the company desires to know how many bowls and mugs to produce eachday in order to maximize profit.

The Problem (1 of 2)

The two products have the following resource requirements for production and profitper item produced (i.e., the model parameters):

Resource Requirements

ProductLabour

( Hr./Unit )Clay

( Lb./Unit)Profit

( $/Unit )

Bowl 1 4 40

Mug 2 3 50

There are 40 hours of labor and 120 pounds of clay available each day for production.

Now, the company (in order of importance, i.e. priorities):

Does not want to use fewer than 40 hours of labor per day.

Would like to achieve a satisfactory profit level of $1,600 per day.

Prefers not to keep more than 120 pounds of clay on hand each day.

Would like to minimize the amount of overtime.

The Problem (2 of 2)

➢ Labor goals constraint (1, less than 40 hours labor; 4,minimum overtime):

• Minimize {P1d1-, P4d1

+}

➢ Add profit goal constraint (2, achieve profit of $1,600):


-, P4d1+}

➢ Add material goal constraint (3, avoid keeping more than 120pounds of clay on hand):


-, P3d3+, P4d1

+}

Goal Constraints and Objective Function (1 of 2)

Complete Goal Programming Model:

Minimize {P1d1-, P2d2

-, P3d3+, P4d1

+}

subject to:x1 + 2x2 + d1

- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0

Goal Constraints and Objective Function (2 of 2)


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0

Goal Constraints

Graphical Interpretation (1 of 6)

The First-Priority Goal: Minimize


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0


The Second-Priority Goal: Minimize


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0


The Third-Priority Goal: Minimize


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0


The Fourth-Priority Goal: Minimize


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0


Goal programming solutions do not always achieve all goals andthey are not optimal, they achieve the best or most satisfactorysolution possible.


-, P3d3+, P4d1

+}


- - d1+ = 40

40x1 + 50 x2 + d2- - d2

+ = 1,6004x1 + 3x2 + d3

- - d3+ = 120

x1, x2, d1-, d1

+, d2-, d2

+, d3-, d3

+ 0

Solution: x1 = 15 bowlsx2 = 20 mugsd1

- = 15 hours


Additional NotesThe graphical solution procedure can also be worked out as mentioned:

1. Rank the goals of the problem in order of importance (i.e., priority wise).

2. Identify the feasible solution points that satisfy the problem constraints.

3. The solution procedure considers one goal at a time, starting with the highestpriority goal and ending with the lowest. The process is carried out such that thesolution obtained from a lower – priority goal never degrades any higher –priority solutions. Identify all feasible solutions that achieve the highest – prioritygoal; if no feasible solutions achieve the highest – priority goal, identify thesolution(s) that comes closest to achieving it. Let the highest priority goal, G1,attain a value G1 = G1*.

4. Move down one priority level. Add the constraint G1 = G1* to the existingconstraints of the problem and determine the “best” solution.

5. Repeat Step 4 until all priority levels have been considered.

The following example illustrates this procedure.

Nicolo Investment Advisors Problem

Nicolo Investment Advisors Problem

A client has $80,000 to invest and, as an initial strategy, would like the investmentportfolio restricted to two stocks:

U. S. Oil, which has a return of $3 on a $25 share price, provides an annual rate ofreturn of 12%, whereas Hub Properties provides an annual rate of return of 10%. Therisk index per share, 0.50 for U. S. Oil and 0.25 for Hub Properties, is a rating Nicoloassigned to measure the relative risk of the two investments. Higher risk index valuesimply greater risk; hence, Nicolo judged U. S. Oil to be the riskier investment. Byspecifying a maximum portfolio risk index, Nicolo will avoid placing too much of theportfolio in high risk investments.

Stock Price/ShareEstimated Annual

Return / ShareRisk Index / Share

U. S. Oil $25 $3 0.50

Hub Properties $50 $5 0.25

To illustrate how to use the risk index per share to measure the total portfolio risk,suppose that Nicolo chooses a portfolio that invests all $80,000 in U. S. Oil, the higherrisk but higher return, investment. Nicolo could purchase $80,000/$25 = 3200 sharesof U. S. Oil, and the portfolio would have a risk index of 3200(0.50) = 1600.Conversely, if Nicolo purchases no shares of either stock, the portfolio will have norisk, but also no return. Thus, the portfolio risk index would vary from 0 (least risk) to1600 (most risk).

Nicolo’s client would like to avoid a high risk portfolio; thus, investing all funds inU. S. Oil would not be desirable. However, the client agreed that an acceptable levelof risk would correspond to portfolios with a maximum total risk index of 700 or less.

Another goal of the client is to obtain an annual return of at least $9000. This goalcan be achieved with a portfolio consisting of 2000 shares of U. S. Oil [at a cost of2000($25) = $50,000] and 600 shares of Hub Properties [at a cost of 600($50) =$30,000]; the annual return in this case would be 2000($3) + 600($5) = $9,000. Note,however, that the portfolio risk index for this investment strategy would be2000(0.50) + 600(0.25) = 1150; thus, this portfolio achieves the annual return goalbut does not satisfy the portfolio risk index goal.

Suppose that the client’s top – priority goal is to restrict the risk; that is, keeping theportfolio risk index at 700 or less is so important that the client is not willing to tradethe achievement of this goal for any amount of an increase in annual return. As longas the portfolio risk index does not exceed 700, the client seeks the best possiblereturn. Based on this statement of priorities, the goals for the problem are as follows:

Primary Goal (Priority Level 1)Goal 1: Find a portfolio that has a risk index of 700 or less.

Secondary Goal (Priority Level 2)Goal 2: Find a portfolio that will provide an annual return of at least $9,000

Lex Min z = {𝑃1 d1+, 𝑃2 d2

-}

subject to:25U + 50H ≤ 80,000 (Funds available)

0.50U + 0.25H – d1+ + d1

- = 700 (Goal 1)3U + 5H – d2

+ + d2- = 9,000 (Goal 2)

U, H, d1+, d1

-, d2+, d2

- ≥ 0

where:U: number of shares of U. S. Oil purchasedH: number of shares of Hub Properties purchased

d- : negative deviational variabled+: positive deviational variable

Problem Formulation

Portfolios that satisfy the Available Funds Constraint

25U + 50H ≤ 80,000 (Funds available)U, H ≥ 0

Feasible Portfolios

Portfolios that satisfy the P1 GoalMin d1

+

subject to:25U + 50H ≤ 80,000

0.50U + 0.25H – d1+ + d1

- = 7003U + 5H – d2

+ + d2- = 9,000

U, H, d1+, d1

-, d2+, d2

- ≥ 0

d1+ = 0

Feasible Portfolios that will achieve Priority Level 1 Goal

Best Solution with respect to Both Goals

Min d2-

subject to:25U + 50H ≤ 80,0000.50U + 0.25H – d1

+ + d1- = 700

3U + 5H – d2+ + d2

- = 9,000 d1

+ = 0U, H, d1

+, d1-, d2

+, d2- ≥ 0

Final SolutionThus, the solution recommends that :

U = 800 sharesH = 1200 shares

Note that the priority level 1 goal of a portfolio risk index of 700 or less has beenachieved. However, the priority level 2 goal of at least a $9,000 annual return is notachievable. The annual return for the recommended portfolio is $8,400.

Simplex Method for

Goal Programming

Always remember

A lower priority goal can never be achieved at the expense of a

higher priority goal”

Solve the following GPP using Simplex Method:

𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒆 𝒛 = 𝑷𝟏𝒅𝟏− + 𝑷𝟐𝒅𝟒

+ + 𝟓𝑷𝟑𝒅𝟐− + 𝟑𝑷𝟑𝒅𝟑

− + 𝑷𝟒𝒅𝟏+

𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

𝑥1 + 𝑥2 + 𝑑1− − 𝑑1

+ = 80𝑥1 + 𝑑2

− = 70𝑥2 + 𝑑3

− = 45𝑑1+ + 𝑑4

− − 𝑑4+ = 10

𝑥1, 𝑥2, 𝑑1−, 𝑑1

+, 𝑑2−, 𝑑3

−, 𝑑4−, 𝑑4

+ ≥ 0

Example 1 (1 of 9)

Solution:The first four rows of the simplex table are set up in the same way as for theSimplex Method. The next four rows stand for priority goal levels. The goallevels 𝑃1, 𝑃2, 𝑃3 and 𝑃4 are arranged in descending order.

Example 1 (2 of 9)

Table 1 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2

𝐶𝐵 𝐵 𝑥1 𝑥2 𝑑1− 𝑑2

− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

𝑃1 𝑑1− 1 1 1 0 0 0 -1 0 80

5𝑃3 𝑑2− 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 1 0 0 1 0 0 0 45

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗(fill the

coefficient of 𝑃𝑖 in the respective

rows)

𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 5 3 0 0 0 0 0 0 485

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 1 1 0 0 0 0 -1 0 80

Key columnThe key column can be determined by selecting the largest positive element in𝑧𝑗 − 𝑐𝑗 row at the highest priority goal level.

Example 1 (3 of 9)

Table 1 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

𝑃1 𝑑1− 1 1 1 0 0 0 -1 0 80

5𝑃3 𝑑2− 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 1 0 0 1 0 0 0 45

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 5 3 0 0 0 0 0 0 485

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 1 1 0 0 0 0 -1 0 80

In the table, the largest positive element 1 in the 𝑃1 row occurs at two places..

Key columnIn order to break this tie, check the next lower priority goal levels.

Example 1 (4 of 9)

Table 1 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

𝑃1 𝑑1− 1 1 1 0 0 0 -1 0 80

5𝑃3 𝑑2− 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 1 0 0 1 0 0 0 45

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 5 3 0 0 0 0 0 0 485

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 1 1 0 0 0 0 -1 0 80

Since the largest positive element is 5 in 𝑃3 row, therefore, column under 𝑥1 becomes the key column.

Key rowMinimum positive value = ( 80/1, 70/1) = 70So 𝑑2

− row is the key row.

Example 1 (5 of 9)

Table 1 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

𝑃1 𝑑1− 1 1 1 0 0 0 -1 0 80

5𝑃3 𝑑2− 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 1 0 0 1 0 0 0 45

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 5 3 0 0 0 0 0 0 485

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 1 1 0 0 0 0 -1 0 80

Pivot element = 1

Therefore, 𝑑2− departs and 𝑥1 enters.

Key column = 𝑥2 columnMinimum positive value = Min(10/1, 45/1) = 10So 𝑑1

− row is the key row.𝑑1− departs & 𝑥2 enters.

Example 1 (6 of 9)

Table 2 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

𝑃1 𝑑1− 0 1 1 -1 0 0 -1 0 10

0 𝑥1 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 1 0 0 1 0 0 0 45

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 0 3 0 -5 0 0 0 0 135

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 0 1 0 -1 0 0 -1 0 10

Key column = 𝑑1+column

Minimum positive value = Min(35/1, 10/1) = 10So 𝑑4

− row is the key row.𝑑4− departs & 𝑑1

+ enters.

Example 1 (7 of 9)

Table 3 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

0 𝑥2 0 1 1 -1 0 0 -1 0 10

0 𝑥1 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 0 -1 1 1 0 1 0 35

0 𝑑4− 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 0 -1 0 0

𝑃3 0 0 -3 -2 0 0 3 0 105

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 0 0 -1 0 0 0 0 0 10

In the above table, since all values in 𝑃1 and 𝑃2 row are either zero or negative, the first goal andthe second goal are completely achieved. The third goal is not completely attained, becausethere is a positive value, i.e., 3 in the 𝑃3 row. Since element 3 in 𝑃3 row is above the element 1 inthe 𝑃2 row, therefore, the rule is that if there is a positive element at a lower priority level in𝑧𝑗 − 𝑐𝑗 , the variable in that column cannot be introduced into the solution, as long as there is a

negative element at a higher priority level. Likewise, the positive element 1 in the 𝑃4 row will notbe considered.

Example 1 (8 of 9)Table 4 𝑐𝑗 0 0 𝑃1 5𝑃3 3𝑃3 0 𝑃4 𝑃2


− 𝑑3− 𝑑4

− 𝑑1+ 𝑑4

+ 𝑥𝐵

0 𝑥2 0 1 1 -1 0 1 0 -1 20

0 𝑥1 1 0 0 1 0 0 0 0 70

3𝑃3 𝑑3− 0 0 -1 1 1 -1 0 1 25

𝑃4 𝑑1+ 0 0 0 0 0 1 1 -1 10

𝑧𝑗 − 𝑐𝑗 𝑃4 0 0 0 0 0 1 0 -1 10

𝑃3 0 0 -3 -2 0 -3 0 3 75

𝑃2 0 0 0 0 0 0 0 -1 0

𝑃1 0 0 -1 0 0 0 0 0 0

𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒆 𝒛 = 𝑷𝟏𝒅𝟏− + 𝑷𝟐𝒅𝟒

+ + 𝟓𝑷𝟑𝒅𝟐− + 𝟑𝑷𝟑𝒅𝟑

− + 𝑷𝟒𝒅𝟏+


𝑥1 + 𝑥2 + 𝑑1− − 𝑑1

+ = 80𝑥1 + 𝑑2

− = 70𝑥2 + 𝑑3

− = 45𝑑1+ + 𝑑4

− − 𝑑4+ = 10

𝑥1, 𝑥2, 𝑑1−, 𝑑1

+, 𝑑2−, 𝑑3

−, 𝑑4−, 𝑑4

+ ≥ 0

The optimal solution is:𝑥1 = 70, 𝑥2 = 20, 𝑑1

+ = 10, 𝑑3− = 25, 𝑑1

− = 0, 𝑑2− = 0

Example 1 (9 of 9)

Solve the following GPP using Simplex Method:

𝑴𝒊𝒏𝒊𝒎𝒊𝒛𝒆 𝒛 = 𝑷𝟏𝒅𝟏− + 𝟐𝑷𝟐𝒅𝟐

− + 𝑷𝟐𝒅𝟑− + 𝑷𝟑𝒅𝟏

+


𝑥1 + 𝑥2 + 𝑑1− − 𝑑1

+ = 350𝑥1 + 𝑑2

− = 200𝑥2 + 𝑑3

− = 300𝑑1+ + 𝑑4

− − 𝑑4+ = 10

𝑥1, 𝑥2, 𝑑1−, 𝑑1

+, 𝑑2−, 𝑑3

− ≥ 0

Example 2 (1 of 5)

Example 2 (2 of 5)

Table 1 𝑐𝑗 0 0 𝑃1 2𝑃2 𝑃2 𝑃3


− 𝑑3− 𝑑1

+ 𝑥𝐵

𝑃1 𝑑1− 1 1 1 0 0 -1 350

2𝑃2 𝑑2− 1 0 0 1 0 0 200

𝑃2 𝑑3− 0 1 0 0 1 0 300

𝑧𝑗 − 𝑐𝑗 𝑃3 0 0 0 0 0 -1 0

𝑃2 2 1 0 0 0 0 700

𝑃1 1 1 0 0 0 -1 350

Key column = 𝑥1columnMinimum positive value = Min(350/1, 200/1) = 200So 𝑑2

− row is the key row.𝑑2− departs & 𝑥1enters.

Example 2 (3 of 5)



− 𝑑3− 𝑑1

+ 𝑥𝐵

𝑃1 𝑑1− 0 1 1 -1 0 -1 150

0 𝑥1 1 0 0 1 0 0 200

𝑃2 𝑑3− 0 1 0 0 1 0 300

𝑧𝑗 − 𝑐𝑗 𝑃3 0 0 0 0 0 -1 0

𝑃2 0 1 0 -2 0 0 300

𝑃1 0 1 0 -1 0 -1 150

Key column = 𝑥2columnMinimum positive value = Min(150/1, 300/1) = 150So 𝑑1

− row is the key row.𝑑1− departs & 𝑥2enters.

Example 2 (4 of 5)



− 𝑑3− 𝑑1

+ 𝑥𝐵

0 𝑥2 0 1 1 -1 0 -1 150

0 𝑥1 1 0 0 1 0 0 200

𝑃2 𝑑3− 0 0 -1 1 1 1 150

𝑧𝑗 − 𝑐𝑗 𝑃3 0 0 0 0 0 -1 0

𝑃2 0 0 -1 -1 0 1 150

𝑃1 0 0 -1 0 0 0 0

Key column = 𝑑1+column

Minimum positive value = Min(150/1) = 150So 𝑑3

− row is the key row.𝑑3− departs & 𝑑1

+ enters.

Example 2 (5 of 5)



− 𝑑3− 𝑑1

+ 𝑥𝐵

0 𝑥2 0 1 0 0 1 0 300

0 𝑥1 1 0 0 1 0 0 200

𝑃3 𝑑1+ 0 0 -1 1 1 1 150

𝑧𝑗 − 𝑐𝑗 𝑃3 0 0 -1 1 1 0 150

𝑃2 0 0 0 -2 -1 0 0

𝑃1 0 0 -1 0 0 0 0

The optimal solution is:

𝑥1 = 200, 𝑥2 = 300, 𝑑1− = 0, 𝑑2

− = 0, 𝑑3− = 0, 𝑑1

+ = 150.

Goal Programming - University of Delhidu.ac.in/du/uploads/departments/Operational...Goal Programming...

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