Linear Programming is a technique for determining an optimum schedule of interdependent activities in view of the available resources. 'linear' means that the relationships are represented by straight lines, i.e., the relationships are of the form

A = C + Bx.

The word 'programming' is just an other word for ‘planning’ and is concerned with optimal allocation of limited resources.

Basic TerminologyBasic Terminology Linear Function

A linear function contains terms, each of which is composed of only a single, continuous variable raised to (and only to) the power of 1.

Objective Function

It is a linear function of the decision variables expressing the objective of the decision-maker. The most typical forms of objective functions are:

maximize f(x) or minimize f(x). Decision Variables

These are economic or physical quantities whose numerical values indicate the solution of the linear programming problem. These variables are under the control of the decision-maker and could have an impact on the solution to the problem under consideration. The relationships among these variables should be linear.

Constraints

These are linear equations arising out of practical limitations. The mathematical forms of the constraints are:

f(x) ≥ b or f(x) ≤ b or f(x) = b

Feasible Solution

Any non-negative solution which satisfies all the constraints is known as a feasible solution.

Optimal Solution

The solution where the objective function is maximized or minimized is known as optimal solution.

General Linear Programming

Optimize ( maximize or minimize) z = c1x1 + c2x2 + c3x3 + .........+ cnxn

subject to (constraints) a11x1 + a12x2 + a13x3 + .........+ a1nxn (≥, ≤,

=) b1

a21x1 + a22x2 + a23x3 + .........+ a2nxn (≥, ≤, =) b2

...............................................................................

....am1x1 + am2x2 + am3x3 + .........+ amnxn (≥, ≤,

=) bm

& (non-negative restrictions) x1, x2,....., xn ≥ 0

Where all c's, a's, b's are constants and x's are decision variables. The expression (≥, ≤, =) means that each constraint may take only one of the three possible forms.

The expression xn≥ 0 means that the xn's must be non-negative.

Highway Materials (model formulation)The specification requires that the highway materials meet the limitations:

9000 cubic yard material is required. Four sources of materials are in the vicinity of the construction site. The composition and the unit cost of material delivered to the site are:

Determine the minimum cost solution that meets the specification requirement.

Percent Volumes

Lower Limit

Upper Limit

GravelSandSilt

5650

208515

Sources

Gravel (% )

Sand (%)

Silt (%)

Cost (Rs./ cu. yd)

River Gr.

15 75 10 60

Sea Gr. 30 70 00 80

Sand 00 100 00 50

On Site 05 30 65 00

Solution

Decision Variables: Let the material from River gravel = x1

Let the material from Sea gravel = x2

Let the material from Sand = x3

Let the material on Site = x4

Objective Function: Min. Z = 60 * x1 + 80 * x2 + 50 * x3 + 0 * x4

Constraints: x1 + x2 + x3 + x4 ≥ 9000 cu.yd (0.15x1 + 0.3x2 + 0.0x3 + 0.05x4)/(x1 + x2 + x3 + x4)*100 ≥ 5 (0.15x1 + 0.3x2 + 0.0x3 + 0.05x4)/(x1 + x2 + x3 + x4)*100 ≤ 20 (0.75x1 + 0.7x2 + 1.0x3 + 0.3x4)/(x1 + x2 + x3 + x4)*100 ≥ 65 (0.75x1 + 0.7x2 + 1.0x3 + 0.3x4)/(x1 + x2 + x3 + x4)*100 ≤ 85 (0.1x1 + 0.0x2 + 0.0x3 + 0.65x4)/(x1 + x2 + x3 + x4)*100 ≥ 00 (0.1x1 + 0.0x2 + 0.0x3 + 0.65x4)/(x1 + x2 + x3 + x4)*100 ≤ 15 x1≥ 0, x2≥ 0, x3≥ 0, x4≥ 0

Traffic Engineering (model formulation) City Nazim is studying the feasibility of introducing a mass transit bus system that will reduce the pollution by reducing in city driving. The initial study seeks the determination of the minimum number of buses that can handle the transportation needs

After collecting the information, the traffic engineer noticed that the minimum number of buses needed to meet demand vary with the time of day. He further discovered that the required number of buses can be assumed constant over successive interval of 4 hours each. (as shown in fig.)

It was decided that, to carry out the required daily maintenance, each bus could operate only 8 successive hours a day. Chalk out the policy to minimize buses.

12:00 AM

4:00 AM

8:00 AM

12:00 Noon

4:00 PM

8:00 PM

12:00 AM

Solution

Min. Z = x1 + x2 + x3 + x4 + x5 + x6

x1≥ 0, x2≥ 0, x3≥ 0, x4≥ 0,

x5≥ 0, x6≥ 0

x1 x3 x5

x6 x2 x4 x6

12:00 AM

4:00 AM

8:00 AM

12:00 Noon

4:00 PM

8:00 PM

12:00 AM

Constraints: x1 + x6 ≥ 4 (12:00 – 4:00) x1 + x2 ≥ 8 (4:00 – 8:00) x2 + x3 ≥ 10 (8:00 – 12:00) x3 + x4 ≥ 7 (12:00 – 4:00) x4 + x5 ≥ 12 (4:00 – 8:00) x5 + x6 ≥ 4 (8:00 – 12:00)

Inspection Problem (model formulation) A company has two grades of inspectors, 1 and 2 to undertake quality control inspection.

At least 1500 pieces must be inspected in an 8 hour day.

Grade 1 inspector can check 20 pieces in an hour with an accuracy of 96%, Grade 2 inspector checks 14 pieces an hour with an accuracy of 92%

The daily wages of Grade 1 inspectors are Rs. 200/hour while those of Grade 2 inspectors are Rs. 150/hour. Any error made by an inspector costs Rs. 100 to the company. If, there are, in all, 10 Grade 1 and 15 Grade2 inspectors in the company, find the optimal assignment of inspectors that minimizes the daily inspection cost.

Solution Let x1, x2 be the nos. of Grade 1 and Grade 2 inspectors.

Objective function is to minimize the daily cost of inspection which is equal to wages paid to the inspectors + the cost of their inspection error.

Cost of Grade 1 inspector/hr = 200 +(100*0.04*20) = Rs. 280

Cost of Grade 2 inspector/hr = 150+(100*0.08*14) =Rs. 262

Therefore, Z to minimize = 8(280x1+262x2) =2240 x1 +

2096x2

Constraints are x1 ≤ 10 and x2 ≤ 15(Grade 1) (Grade 2)

on the number of pieces to be inspected daily20 * 8x1 + 14 * 8x2 ≥ 1500

or 160x1 + 112x2 ≥ 1500

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4. Trim-Loss or Cutting Stock problem

• Three special orders for rolls of paper have been placed at a paper mill. The orders are to be cut from standard rolls of 10¢ and 20¢ widths.

Order Width Length1 5 10,000¢2 7‘ 30,000¢3 9 20,000¢

• Assumption: Lengthwise strips can be taped together

• Goal: Throw away as little as possible

13

Problem: What is trim-loss?

Decision variables: xj = length of roll cut using pattern, j = 1, 2, … ?

7

10

9'5'

5

20

5000'

14

Patterns possible

10 roll 20 roll

x1

5 2 0 0 4 2 2 1 0 07 0 1 0 0 1 0 2 1 09 0 0 1 0 0 1 0 1 2

Trim loss 0 3 1 0 3 1 1 4 2

min z = 10(x1+x2+x3) + 20(x4+x5+x6+x7+x8+x9)

s.t. 2x1 + 4x4 + 2x5 + 2x6 + x7 10,000

x2 +x5 + 2x7 + x8 30,000

x3 + x6 + x8 + 2x9 20,000

x2 x3 x4 x5 x6 x7 x8 x9

xj0, j = 1, 2,…,9

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Alternative Formulation

min z = 3x2 + x3 + 3x5 + x6 + x7 + 4x8 + 5y1 + 7y2 + 9y3

s.t. 2x1 + 4x4 + 2x5 + 2x6 + x7 – y1 = 10,000

x2 + x5 + 2x7 + x8 – y2 = 30,000

x3 + x6 + x8 + 2x9 – y3 = 20,000

xj0, j = 1,…,9; yi 0, i = 1, 2, 3

where yi is overproduction of width i

+ 2x9