1

2

3

Special Variables:

1 -Slack Variable (s1): represents the amount by which the available amount of the resources exceeds its usage by the activities.

To find the slack for a ≤ constraint algebraically, we add a slack variable to the constraint and convert it into an equality.

Ex: In Reddy Mikks model

The constraint 6x1+ 4x2 24 associated with the usage of raw material M1 is equivalent to 6x1+ 4x2 + s1 = 24 provided that s1 ≥ 0

The slack variable s1 =24 - 6x1 - 4x2 represents the unused amount of raw material M1

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2 -Surplus Variable (S1) : represents the excess of the left hand side over the minimum requirement.

To find the surplus for a ≥ constraint, we subtract a surplus variable from the left-hand side to make it an equality.

Ex: In Diet Model

The constraint representing minimum feed requirement, x1+ x2 ≥ 800 is equivalent to x1+ x2 - S1 = 800 provided that S1 ≥ 0

A positive value of S1 signifies that a surplus amount of feed ( over the minium requirement of 800Ib) will be produced.

3 -Unrestricted Variable : In both Reddy Mikks and Diet model, the nature of variables requires them to assume nonnegative values. But there are other situations where variable can assume any real value.

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Example:

McBurger fast – food restaurant sells quarter-pounder and cheeseburger. A quarter-pounder uses a quarter of pound of meat, and cheeseburger uses only 0.2 Ib. The restaurant starts the day with 200 Ib of meat but may order more at an additional cost of 25 cents per pound to cover the delivery cost. Any surplus meat at the end of the day is donated at HotSoup Charity. McBurger’s profits are 20 cents from a quarter pounder and 15 cents for a cheeseburger. All in all, McBurger doesn’t expect to sell more than 900 sandwich in any one day. How many of each sandwich should McBurger

make?

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Solution:

Assume x1 The daily number of quarter-pounders made

x2 The daily number of cheeseburger made

The constraints:0.25x1+ 0.2x2 200 In first case (Slack)

0.25x1+ 0.2x2 ≥ 200 In second case (Surplus)

We consider x3 as unrestricted that allows the variable to play the roles of both a slack and a surplus as desired

0.25x1+ 0.2x2 + x3 = 200

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The objective function: Maximize the total profit, less any additional cost that may be incurred as a result of ordering special delivery of additional pounds of meat

The additional cost is incurred only if x3 < 0 (role of surplus)

Dealing with an unrestricted variable is cumbersome. So we use standard substitution that converts the unrestricted variable into two nonnegative variables

X3 = x3+ - x3

- x3+, x3

- ≥ 0

If x3+ >0 and x3

- =0 x3+ plays a role of slack

If x3- >0 and x3

+ =0 x3- plays a role of surplus

The constraint can be expressed as:0.25x1+ 0.2x2 + x3

+ - x3- = 200

The objective function is expressed as:Maximize z =0.2 x1 + 0.15 x2 - 0.25 x3

-

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Sensitivty Analysis: Studying the effect of making changes in the model parameters on a given optimum LP solution.

1 )Changes in the objective function coefficient

The objective function (Two variable- maximize or minimize )can be expressed as:

z = c1x1 + c2x2

•The change in the coefficients c1 and c2 will change the slope of z which lead to changing the optimum solution to a different corner point of the solution

space. So we need to determine the range of optimality for the ratio or that will keep the optimum solution unchanged.

cc

2

1 cc

1

2

Corner point: A point that lies at the intersection of two (or possibly more) constraint lines on the boundary of the feasible region.

No interior points in the feasible region need be considered because at least one corner point is better than any interior point

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Example:

In Reddy Mikks problem.

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The objective function was Z = 5x1 + 4x2

If we change the objective function to Z = c1 x1 + c2 x2

The solution at C point will remain optimum . As the slope of z lies between the slopes of two line intersecting C--- namely

6x1+ 4x2 24 raw material M1 x1+ 2x2 6 raw material M2

We can express this relationship algebraically:

1

2

6

4

1

2 cc

or

4

6

2

1

2

1 cc

01c

02c

The line cannot to be horizontal

The line cannot to be vertical

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If c2 remain equal to 4, then

621 c

Similarly, If we specify c1 equal to 5, then

103

102 c

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2 )Unit Worth of a Resource ( Changes in in the right hand – side of the constraints)

Quantifies the rate of change in the optimum value in the objective function as a result of making changes in the availability of the resource.

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Example:

In Reddy Mikks model:The first constraints represent the limitations on the usage of raw material, M1 and M2, respectively. Determine the worth per unit for each resource.

Solution :The worth per unit for raw material M1

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End points D=(2.2) and G=(6.0) delineate the feasibility range for M1The amount of M1 associated with point D is computed as 6 x1 + 4 x2 =20 tons. Similarly, The amount of M1 associated with point G is computed as 6 x1 + 4 x2 =36 tons .Thus the range of feasibility for M1 is 36120 M

Let y1 represent the worth per unit of raw material M1

GtoDfromMinchange

GtoDfromzinchangey11

Given D= (2.2) and G= (6.0) then,z at D = 5*2 +4*2 = 18 (thousand dollars)z at G = 5*6 +4*0 = 30 (thousand dollars)

)1(4

3

2036

18301

Moftonperdollarsthousandy

As a result, the 1-ton change in M1 in range 20 M1 36 will change optimum value of z by 750$

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The worth per unit for raw material M2

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End points B=(4,0) and H=(8/3, 2) delineate the feasibility range for M1The amount of M2 associated with point B is computed as x1 + 2 x2 =4 tons. Similarly, The amount of M2 associated with point H is computed as x1 + 2 x2 =20/3 tons .Thus the range of feasibility for M1 is

3

2014 M

Let y2 represent the worth per unit of raw material M2

GtoDfromMinchange

GtoDfromzinchangey11

Given B= (4,0) and H= (8/3, 2) then,z at B = 5*4 +4*0 = 20(thousand dollars)z at H = 5*8/3 +4*2 = 64/3 (thousand dollars)

)2(2

1

4320

203

64

2Moftonperdollarsthousandy

As a result, the 1-ton change in M1 in range 20 M1 36 will change optimum value of z by 500$