LINEAR PROGRAMMING

• WHAT IS LINEAR PROGRAMMING?

• CONDITIONS UNDER WHICH LINEAR PROGRAMMING CAN BE APPLIED

• ASSUMPTIONS

STATEMENT OF LPP IN MATRIX FORM

MAXIMISATION PROBLEM MINIMISATION PROBLEMMaximise Z = cx Minimise Z = cxSubject to: ax ≤ b Subject to: ax ≥ b x ≥ 0 x ≥ 0

Where, c is row matrix containing coefficients in the objective functionx is column matrix containing decision variablesa is matrix containing the coefficients of constraintsb is column matrix containing right hand side of constraints

GENERAL STATEMENT OF MAXIMISING LPP

Maximise Z = c1x1 + c2x2+…+cnxn Subject to,a11x1 + a12x2+…+a1nxn ≤ b1

a21x1 + a22x2+…+a2nxn ≤ b2 .. .am1x1+ am2x2+...+amnxn ≤ bm

x1, x2,…xn ≥ 0

cj,aij, bi (i=1,2,…m; j=1,2,…n)

xj cj

aij biGeneralised formZ = cjxj (For j=1,2,…n)Subject to,aij xj ≤ bi (For i=1,2,…m) xj ≥0 (For j=1,2,…n)

CONSTRAINTS

OBJECTIVE FUNCTION

NON NEGATIVITY RESTRICTION

ARE KNOWN AS CONSTANT

ARE DECISION VARIABLESARE TERMED AS PROFIT COEFFICIENTS

ARE RESOURCE VALUES ARE CALLED TECHNOLOGICAL COEFFICIENTS

GENERAL STATEMENT OF MINIIMISING LPP

Minimise Z = c1x1 + c2x2+…+cnxn Subject to,a11x1 + a12x2+…+a1nxn ≥ b1

a21x1 + a22x2+…+a2nxn ≥ b2 .. .am1x1+ am2x2+...+amnxn ≥ bm

x1, x2,…xn ≥ 0

cj,aij, bi (i=1,2,…m; j=1,2,…n)

xj cj

aij biGeneralised formZ = cjxj (For j=1,2,…n)Subject to,aij xj ≥ bi (For i=1,2,…m) xj ≥0 (For j=1,2,…n)

CONSTRAINTS

OBJECTIVE FUNCTION

NON NEGATIVITY RESTRICTION

ARE KNOWN AS CONSTANTS

ARE DECISION VARIABLESARE TERMED AS PROFIT COEFFICIENTS

ARE RESOURCE VALUES ARE CALLED TECHNOLOGICAL COEFFICIENTS

GRAPHICAL SOLUTION OF LPP

STEPS FOR GRAPHICAL SOLUTION OF LPP

• PREPARE AN INFORMATION SUMMARY TABLE

• STATE THE LPP IN A MATHEMATICAL FORM

• PLOT THE LINES REPRESENTED BY CONSTRAINT EQUATIONS .

• IDENTIFY THE FEASIBLE REGION

• USE EXTREME POINTS METHOD TO FIND OPTIMAL SOLUTION

• USE ISO PROFIT OR ISO COST METHOD TO FIND OPTIMAL SOLUTION

NUMERICAL EXAMPLE

Q. 1 (MAXIMISATION BY GRAPHICAL METHOD) A PAINT MANUFACTURING COMPANY PRODUCES TWO TYPES OF PAINTS – SYNTHETIC ENAMEL (SYN) AND SPECIAL ENAMEL (SPL) WHICH ARE SOLD IN FIVE LITRE BUCKET PACKING AT A PROFIT OF Rs 40 AND Rs 35 PER BUCKET RESPECTIVELY. RAW MATERIALS R1 AND R2 REQUIRED TO PRODUCE ONE BUCKET (FIVE LITRES) OF SYN ARE 2 LITRES AND 4 Kgs RESPECTIVELY AND TO PRODUCE ONE BUCKET OF SPL ARE 3 LITRES AND 3 Kgs RESPECTIVELY. IF THE COMPANY FINDS THAT FOR THE NEXT WEEK, AVAILABILITY OF R1 IS ONLY 60 LITRES AND AVAILABILITY OF R2 IS ONLY 96 Kgs, DETERMINE HOW MANY BUCKETS OF SYN AND SPL SHOULD THE COMPANY PRODUCE TO MAXIMISE PROFIT.

NUMERICAL EXAMPLE

Q. 2 ( MINIMIZATION BY GRAPHICAL METHOD) AN ALLOY MANUFACTURING UNIT HAS RECEIVED AN ORDER TO SUPPLY 180 KGS OF AN ALLOY CONTAINING COPPER AND TIN IN THE RATIO OF 2:1. THE ALLOY MANUFACTURER BUYS EMPTY ARTILLERY SHELLS DISPOSED OFF BY ARMY. EACH SHELL WEIGHS 10 KGS. THERE ARE TWO TYPES OF SHELLS – A AND B. TYPE A SHELLS CONTAIN 80 PERCENT COPPER AND 20 PERCENT TIN. TYPE B SHELLS CONTAIN 50 PERCENT COPPER AND 50 PERCENT TIN. THE PRICE OF TYPE A SHELL IS RS 2400 EACH AND TYPE B SHELL IS RS 3000 EACH. WHAT IS THE NUMBER OF SHELLS OF TYPE A AND B THAT THE ALLOY MANUFACTURER MUST PURCHASE TO PRODUCE 180 KGS OF ORDERED ALLOY SO THAT THE COST IS MINIMIZED.

SIMPLEX METHOD - STEPS

1. ADD SLACK, SURPLUS OR ARTIFICIAL VARIABLES AND REWRITE THE LPP

2. PREPARE INITIAL FEASIBLE SOLUTION TABLE SHOWING CONSTRAINT COEFFICIENTS AND OBJECTIVE FUNCTION COEFFICIENTS (Cj)

3. DETERMINE ELEMENTS OF SOLUTION ROW

4. DETERMINE ELEMENTS OF Zj ROW

5. DETERMINE ELEMENTS OF j ROW

6. DETERMINE INCOMING VARIABLE

7. DETERMINE OUTGOING VARIABLE

8. PREPARE NEXT ITERATION

9. REPEAT STEPS 1 TO 8 TILL OPTIMAL SOLUTION IS OBTAINED

SIMPLEX METHOD – FLOW CHART.

FORMULATE BUSINESS PROBLEM AS A LPP – WRITE OBJECTIVE FUNCTION AND CONSTRAINT EQUATIONS

ADD SLACK, SURPLUS AND ARTIFICIAL VARIABES AS REQUIRED

PREPARE INITIAL FEASIBLE SOLUTION TABLE AND WRITE VALUES WRT ALL THE VARIABLES IN c j, zj, SOLUTION AND j ROWS

IS THE LPP MAXIMISATION OR MINIMISATION

IS j VALUE OF ONE OR MORE VARIABLES

-IVE

IS j VALUE OFONE OR MORE VARIABLES

+IVE

MAXIMISATION MINIMISATION

OPTIMAL SOLUTION NONO

YES YESIDENTIFY INCOMING VARIABLE HAVING LARGEST -IVE j VALUE. CALCULATE REPLACEMENT RATIOS

IDENTIFY INCOMING VARIABLE HAVING LARGEST +IVE j VALUE. CALCULATE REPLACEMENT RATIOS

+IVE VALUE OF ALL REPL RATIOS

NO

IDENTIFY PIVOT ROW AND PIVOT ELEMENT

USE SIMPLEX ALGORITHM TO DETERMINE REPL ROW AND OTHER ROWS FOR THE NEXT SIMPLEX TABLE

YES

IF ALL REPL RATIOS ARE ZERO OR NEGATIVE, THENSOLUTION IS UNBOUNDED.

IF ONE OR MORE REPL RATIOS HAVE ZERO VALUE,

SOLN IS DEGENERATE

SIMPLEX METHOD

COEFFICIENTS & SIGNS OF ADDL VARIABLES IN OBJECTIVE FUNCTION

CONSTRAINT EQUATION SIGN

ADDITIONALVARIABLE REQUIRED

COEFFS. & SIGNS OF ADDL VARIABLES IN THEOBJECTIVE FUNCTION MAX LPP MIN LPP

WILL ADDL VARIABLE BE A BASIC VARIABLE IN THE FIRST SIMPLEX TABLE

≤ SLACK VARIABLE

0 0 YES

≥ i) SURPLUS VARIABLE

ii) ARTIFICIAL VARIABLE

0 0

-M +M

NO (Because they have –ive sign)

YES

= ARTIFICIAL VARIABLE

-M +M

YES

NUMERICAL 3

A DYES MANUFACTURING COMPANY PRODUCES THREE TYPES OF DYES – D1,D2 AND D3 USING 3,5 AND 2 Kgs RESPECTIVELY OF RAW MATERIAL R1, 4,4 AND 4 Kgs RESPECTIVELY OF RAW MATERIAL R2 AND 2,4 AND 5 Kgs RESPECTIVELY OF RAW MATERIAL R3 FOR PRODUCING EACH DRUM OF DYES D1,D2 AND D3. THE CONTRIBUTION FOR DYES D1, D2 AND D3 IS Rs 50, 100 AND 80 PER DRUM RESPECTIVELY. IF THE AVAILABILITY OF RAW MATERIALS R1, R2 AND R3 IS 60,72 AND 100 Kgs RESPECTIVELY, HOW MANY DRUMS OF DYES D1,D2 AND D3 SHOULD THE COMPANY PRODUCE SO AS TO MAXIMISE PROFITS?

SOLUTION OF NUMERICAL 3INFORMATION SUMMARY

R1 R2 R3

D1 3 4 2 50 x1

D2 5 4 4 100 x2

D3 2 4 5 80 x3

DYE RAW MATLREQD/DRUM

CONTRIBUTION

PRODUCT MIX

MAXIMUM AVAILABLE

60 72 100

ADD SLACK VARIABLES TO RECONSTITUTE THE LPP AS FOLLOWSMaximise Z = 50x1 +100x2 +80x3 + 0s1 + 0s2 + 0s3 Subject to 3x1 +5x2 +2x3 + s1 + 0s2 + 0s3 ≤ 60

4x1 +4x2 +4x3 + 0s1 + s2 + 0s3 ≤ 722x1 +4x2 +5x3 + 0s1 + 0s2 + s3 ≤ 100

x1, x2, x3 ≥ 0

NUMERICAL4

(R4 LPP WITH MIXED CONSTRAINTS)THE MANAGER OF AN OIL REFINERY HAS TO DECIDE UPON THE OPTIMAL MIX OF TWO BLENDING PROCESSES. THE INPUTS AND OUTPUTS PER PRODUCTION RUN FOR THESE PROCESSES ARE GIVEN BELOW.PROCESS INPUT- CRUDE OIL O U T P U TTYPE C1 C2 PETROL DIESEL

1 5 3 5 3 2 4 5 4 4

THE MAXIMUM AMOUNTS OF CRUDE OIL TYPE C1 ND C2 THAT ARE AVAILABLE ARE 190 UNITS AND 140 UNITS RESPECTIVELY. MARKET REQUIREMENTS ARE THAT AT LEAST 100 UNITS OF PETROL AND EXACTLY 84 UNITS OF DIESEL MUST BE PRODUCED. PROFITS PER PRODUCTION RUN OF PROCESS TYPE 1 AND TYPE 2 ARE Rs 40 AND Rs 50 LAKHS RESPECTIVELY. FORMULATE THIS AS A LPP FOR SOLVING USING THE SIMPLEX ALGORITHM. DETERMINE THE OPTIMUM NUMBER OF PRODUCTION RUNS OF EACH TYPE OF PROCESS IN ORDER TO MAXIMISE PROFITS.

SOLUTION OF NUMERICAL 4

INFORMATION SUMMARY

C1 C2 PETROL DIESEL

P1 5 3 5 3 40 x1

P2 4 5 4 4 50 x2

PROCESS INPUTTYPE OF CRUDE OIL

(UNITS)

OUTPUT(UNITS)

PROFIT/PROCESS RUN (Rs LAKHS)

NO. OF RUNS OF PROCESS

MAXIMUM AVAILABLE

190 140

Maximise Z = 40x1 +50x2

Subject to 5x1 +4x2 ≤ 1903x1 +5x2 ≤ 1405x1 +4x2 ≥ 1003x1 +4x2 = 84 x1, x2 ≥ 0

100 MINIMUM REQUIREMENT

84 EXACT REQUIREMENT

SOLUTION OF NUMERICAL 4

INFORMATION SUMMARY

C1 C2 PETROL DIESEL

P1 5 3 5 3 40 x1

P2 4 5 4 4 50 x2

PROCESS INPUTTYPE OF CRUDE OIL

(UNITS)

OUTPUT(UNITS)

PROFIT/PROCESS RUN (Rs LAKHS)

NO. OF RUNS OF PROCESS

MAXIMUM AVAILABLE

190 140

ADD SLACK AND SURPLUS VARIABLES TO RECONSTITUTE THE LPP AS FOLLOWSMaximise Z = 40x1 +50x2 + 0s1 + 0s2 - 0s3 - 0s4- MA1- MA2 Slack variables have positive sign. Surplus variables have negative sign. Since it is a maximisation LPP, therefore artificial variable A1 and A2 will have negative signs. Subject to 5x1 +4x2 + s1 + 0s2 - 0s3 - 0A1- 0A2 = 190

3x1 +5x2 + 0s1 + s2 - 0s3 - 0A1- 0A2 = 1405x1 +4x2 + 0s1 + 0s2 - s3 + A1- 0A2 = 1003x1 +4x2 + 0s1 + 0s2 - 0s3 - 0A1+ A2 = 84

x1, x2 ≥ 0

100 MINIMUM REQUIREMENT

84 EXACT REQUIREMENT

NUMERICAL 4 - SIMPLEX TABLE 4

.

BASIS x1 x2 s1 s2 s3 A1 A2 bi bi / aij

s1 0 0 -8/3 1 0 0 0 -5/3 50

s2 0 0 1 0 1 0 0 1 56

x1 40 1 4/3 0 0 0 0 1/3 28

s3 0 0 8/3 0 0 1 -1 5/3 40

cj 40 50 0 0 0 -M -M

SOLN 28 0 50 56 40 0 0

Zj40 160/3 0 0 0 0 40/3

j =cj -Zj0 -10/3 0 0 0 -M -M-40/3

SIMPLEX TABLE 4- Optimal Soln

LARGEST POSITIVE VALUE

SOLUTION OF NUMERICAL 4

.

6040302010

40

10

60

20

50

30

50

0

0

A

D

E

O

PQ

C

B

28

47.5

3828

2521

46.7

H

G

F

Maximise Z = 4x1 +5x2 Subject to 5x1 +4x2 ≤ 190 LINE AB

3x1 +5x2 ≤ 140 LINE CD5x1 +4x2 ≥ 100 LINE EF3x1 +4x2 = 84 LINE GH x1, x2 ≥ 0

VALUE OF Z AT Q(107); H(112)HENCE H GIVES THE ANSWERx1=28 x2=0

FEASIBLE REGIONIS ONLY THE LINE QHBECAUSE OF LINE GH

NUMERICAL 5

(LPP WITH MIXED CONSTRAINTS)

SOLVE THE FOLLOWING LPP

Maximise Z = 2x1 +4x2 Subject to 2x1 + x2 ≤ 18

3x1 +2x2 ≥ 30 x1 +2x2 = 27

x1, x2 ≥ 0

NUMERICAL 6 (INFEASIBILITY)SOLVE THE FOLLOWING LPP

Maximise Z = 8x1 +12x2 Subject to x1 + 3x2 ≤ 30

2x1 + x2 ≤ 40 x1 ≥ 40

x2 ≥ 0

NUMERICAL 7(1) UNBOUNDEDNESS

(UNBOUNDEDNESS)SOLVE THE FOLLOWING LPP

Maximise Z = 6x1 +20x2 Subject to 3x1 + 4x2 ≥ 36

x1 + 3x2 ≥ 18 x1 , x2 ≥ 0

NUMERICAL 7(2) (UNBOUNDEDNESS) (UNBOUNDEDNESS)SOLVE THE FOLLOWING LPP

Maximise Z = 12x1 + 30x2 Subject to x1 + 2x2 ≥ 9

x1 + 5x2 ≥ 15 x1, x2 ≥ 0

NUMERICAL 7(2) (UNBOUNDEDNESS)

Graph to be made

NUMERICAL 7(2) (UNBOUNDEDNESS) Simplex tables to be made

NUMERICAL 7(3) (UNBOUNDEDNESS) (UNBOUNDEDNESS)SOLVE THE FOLLOWING LPP

Maximise Z = 4x1 + 8x2 Subject to x1 - 2x2 ≤ 6

x1 ≤ 10 x2 ≥ 1

x1, x2 ≥ 0

NUMERICAL 8

(DEGENERACY)

Maximise Z = 25x1 +32x2 Subject to 2x1 + x2 ≤ 6

3x1 + x2 ≤ 8 4x1 +5x2 ≤ 30 x1, x2 ≥ 0

NUMERICAL 8

(DEGENERACY)

REFORMULATED LPP

Maximise Z = 25x1 +32x2 + 0s1 + 0s2 + 0s3 Subject to 2x1+ x2 + s1 + 0s2 + 0s3 = 6

3x1+ x2 + 0s1 + s2 + 0s3 = 8 x2 + 0s1 + 0s2+ s3 = 30

x1,x2 ≥ 0

COMPLICATIONS IN LPPs

3. DegeneracyGraph is to be made

COMPLICATIONS IN LPPs

4. Multiple optimality

Maximise Z = 10x1 +20x2 Subject to x1 + x2 ≤ 8

x2 ≤ 5 x1 + 2x2 ≤ 12 x1, x2 ≥ 0

DUAL OF PRIMAL OF A LPP

• STATEMENT OF DUAL OF A LPP IN MATRIX FORM

• STEPS FOR WRITING A DUAL

• EXAMPLES OF DUALS

STATEMENT OF DUAL OF A LPP IN MATRIX FORM

IF THE PRIMAL LPP IS: THEN CORRESPONDING DUAL IS

Maximise Z = cx Minimise G = b’ySubject to: ax ≤ b Subject to: a’y ≥ c’ x ≥ 0 y ≥ 0

DUAL OF PRIMAL OF LPP FOR NUMERICAL 1

PRIMALMAXIMISE Z = 40x1 +35x2 SUBJECT TO: 2x1 +3x2 ≤ 60

4x1 +3x2 ≤ 96 x1,x2 ≥ 0

DUAL OF PRIMAL LPP OF NUMERICAL 1

DUALMinimise G = 60y1 +96y2 [ G = b’y]Subject to: 2y1 +4y2 ≥ 40 [a’y ≥ c’ ]

3y1 +3y2 ≥ 35 x1,x2 ≥ 0

RELATIONSHIP BETWEEN PRIMAL AND DUAL

PRIMAL THEN DUAL

1. IS MAXIMISATION LPP IS MINIMISATION LPP2. HAS n VARIABLES HAS n CONSTRAINTS 3. HAS m CONSTRAINTS HAS m VARIABLES

4. HAS ≤ TYPE OF CONSTRAINTS HAS ≥ TYPE OF CONSTRAINTS

5. VARIABLE xj IS UNRESTRICTED CONSTRAINT j IS OF = TYPE

6. CONSTRAINT i IS = TYPE VARIABLE yi IS UNRESTRICTED

7. OBJECTIVE FUNCTION COEFF. OF jth VARIABLE BECOMES

RHS CONSTANT FOR jth CONSTRAINT

8. RHS CONSTANT FOR ith CONSTRAINT BECOMES

OBJECTIVE FUNCTION COEFFICIENT OF ith VARIABLE

9. COEFFICIENT aij OF jth VARIABLE IN ith CONSTRAINT BECOMES

COEFFICIENT OF ith VARIABLE IN jth CONSTRAINT

NUMERICAL 1

.

BASIS x1 x2 s1 s2 bi

s1 35 0 1 2/3 -1/3 8

s2 40 1 0 -1/2 1/2 18

cj 40 35 0 0

SOLN 18 8 0 0

Zj 40 35 10/3 25/3

Dj = cj -Zj0 0 (-) 10/3 (-) 25/3

OPTIMAL SOLN OF THE PRIMAL

BASIS y1 y2 s1 s2 A1 A2 bi

y2 96 0 1 -1/2 1/3 1/2 -1/3 25/3

y1 60 1 0 1/2 -2/3 -1/2 2/3 10/3

cj 60 96 0 0 M M

SOLN 10/3 25/3 0 0 0 0

Zj60 96 -18 -8 18 16

j =cj -Zj0 0 18 8 M-18 M-16

OPTIMAL SOLN OF THE DUAL

INTERPRETATION OF DUAL OF PRIMAL OF LPP OF NUMERICAL 1

ITEM CONTENT PER BUCKET

MIXING MACHINE TIME PER BUCKET

BAKING MACHINE TIME PER BUCKET

MINIMUMUMRENTAL VALUE PER BUCKET

SYNTHETIC ENAMEL

5 LITRES 2 MINUTES 4 MINUTES RS 40

SPECIAL ENAMEL

5 LITRES 3 MINUTES 3 MINUTES RS 35

RENTAL VALUE/MINUTE

y1 y2

60 MINUTES 96 MINUTESRENTABLE MACHINE TIMES/DAY

DUAL OF PRIMAL OF LPP FOR NUMERICAL2.

PRIMALMinimise Z = 2400x1 +3000x2 Subject to 8x1+ 5x2 ≥ 120

2x1+ 5x2 ≥ 60 x1, x2 ≥ 0

DUAL OF PRIMAL OF LPP FOR NUMERICAL2.

DUALMaximise G =120y1+ 60y2 Subject to: 8y1 + 2y2 ≤ 2400

5y1 + 5y2 ≤ 3000 x1,x2 ≥ 0

NUMERICAL 10-1 (DUAL OF LPP HAVING MIXED RESTRICTIONS)

.

PRIMALMaximise Z = 10x1 - 5x2 + 8x3 + 4x4 Subject to 4x1 + 3x2 + 6x3 + x4 ≤ 40

-x1+ 2x2 + 3x3 + x4 ≤ 59x1 - 5x2 + 7x3 - x4 ≥ 60

6x2 +2x3+4 x4 = 47 x1, x2 ,x3, x4 ≥ 0

NUMERICAL 10-1 (DUAL OF LPP HAVING MIXED RESTRICTIONS).

REFORMULATION OF PRIMAL WITH APPROPRIATE SIGNS

Maximise Z = 10x1 - 5x2 + 8x3 + 4x4 Subject to 4x1 + 3x2 + 6x3 + x4 ≤ 40

-x1+ 2x2 + 3x3 + x4 ≤ 5 -9x1+ 5x2 - 7x3 + x4 ≤ -60

0x1 - 6x2 - 2x3- 4 x4 ≤ -47 0x1+ 6x2 +2x3+4 x4 ≤ 47 x1, x2 ,x3, x4 ≥ 0

NUMERICAL 10-1 (DUAL OF LPP HAVING MIXED RESTRICTIONS)

.

DUALMinimise G =40y1+ 5y2 - 60y3 - 47y4+ 47y5 Subject to: 4y1 - y2 - 9y3 ≥ 10

3y1+2y2+ 5y3 - 6y4+ 6y5 ≥ -5 6y1+3y2 - 7y3 - 2y4+ 2y5 ≥ 8 y1+ y2+ y3 - 4y4 + 4y5 ≥ 4 y1, y2, y3, y4, y5 ≥ 0

NUMERICAL 10-1 (DUAL OF LPP HAVING MIXED RESTRICTIONS)

.

REFORMULATED DUAL Minimise G =40y1+ 5y2 - 60y3 - 47y6 Subject to: 4y1 - y2 - 9y3 ≥ 10

3y1+2y2+ 5y3 - 6y6 ≥ -5 6y1+3y2 - 7y3 - 2y6 ≥ 8 y1+ y2+ y3 - 4y6 ≥ 4 y1, y2, y3 ≥ 0,

0 ≥ y6 ≥ 0 (y6 has unrestricted sign)

NUMERICAL 10-2 (DUAL OF A LPP HAVING UNRESTRICTED VARIABLE)

.

PRIMAL (n=3, m=2)Maximise Z = 5x1 + 4x2 + 7x3 Subject to 4x1 + x2 - 2x3 ≤ 5 …I

2x1+ 3x2 + x3 ≥ 21 …II x1, x2 ≥ 0; x3 has unrestricted

sign

NUMERICAL 10-2 (DUAL OF A LPP HAVING UNRESTRICTED VARIABLE)

.

We may rewrite the primal LPP asMaximise Z = 5x1 + 4x2 + 7 x4 - 7x5 Subject to 4x1 + x2 -2 x4 + 2x5 ≤ 5 …I

-2x1 - 3x2 - x4 + x5 ≤ - 21 …II x1, x2 , x4, x5 ≥ 0

NUMERICAL 10-2 (DUAL OF A LPP HAVING UNRESTRICTED VARIABLE)

.

DUAL (n=2; m=4)Minimise G =5y1+ 21y2 Subject to: 4y1 - 2y2 ≥ 10 … I

y1- 3y2 ≥ 4 … II -2y1- y2 ≥ 7 … III

2y1+ y2 ≥ -7 … IV y1, y2 ≥ 0

NUMERICAL 10-2 (DUAL OF A LPP HAVING UNRESTRICTED VARIABLE)

.

REFORMULATED DUAL (n=2, m=3)Minimise G =5y1+ 21y2 Subject to: 4y1 - 2y2 ≥ 10

y1- 3y2 ≥ 4 -2y1- y2 = 7

y1, y2 ≥ 0

NUMERICAL 11 (LPP HAVING VARIABLES WITH POSITIVE VALUES )

.

Maximise Z = 8x1 + 24x2 Subject to 2x1 + x2 ≤ 41 …I

3x1 + 4x2 ≤ 104 …II x1≥ 10 x2≥ 12

NUMERICAL 11(LPP HAVING VARIABLES WITH POSITIVE VALUES )

.

Put x1= (10 + x3 ) and x2= (12 + x4) where x3, x4 ≥ 0 We may rewrite the LPP as and solve it by Simplex Method

Maximise Z = 8 x3 + 24x4 + 368 Subject to 2x1 + x2 ≤ 9 …I

3x1 + 4x2 ≤ 26 …II x3, x4 ≥ 0

NUMERICAL 11 (LPP HAVING VARIABLES WITH POSITIVE VALUES ) Solve by Simplex method

NUMERICAL 12 (SHADOW PRICES IN MAX LPP)

A MANUFACTURING COMPANY FINDS THAT THE CONTRIBUTION OF PRODUCT ‘A’ IS Rs 7 PER PIECE AND IT REQUIRES 3 UNITS OF RAW MATERIAL AND 2 MAN-HOURS OF LABOUR TO PRODUCE. FOR ANOTHER PRODUCT ‘B’, THE CONTRIBUTION IS Rs 5 PER PIECE AND THE REQUIREMENT OF RAW MATERIAL IS ONE UNIT AND LABOUR IS ONE MAN-HOUR. AVAILABILITY OF RAW MATERIAL IS 48 UNITS AND OF LABOUR IS 40 MAN-HOURS. i) FORMULATE IT AS A LPPii) WRITE ITS DUALiii) SOLVE THE DUAL WITH SIMPLEX METHOD AND FIND THE OPTIMAL PRODUCT MIX AND SHADOW PRICES OF RAW MATERIAL AND LABOUR.

NUMERICAL 13

(INFEASIBLITY, DEGENERATION, MULTIPLE OPTIMALITY IN MAX LPP)

THE SIMPLEX TABLE OF AN LPP IS GIVEN BELOW. BASIS x1 x2 s1 s2 bi x2 5 1 1 1 0 10s2 0 1 0 -1 1 3Cj 4 5 0 0Zj 5 5 5 0Soln 0 10 0 3Δj = Cj-Zj -1 0 -5 0

ANSWER THE QUESTIONS GIVEN IN THE NEXT SLIDE GIVING REASONS FOR YOUR ANSWERS.

NUMERICAL 13 contd.a) IS THIS SOLUTION OPTIMAL?b) ARE THERE MORE THAN ONE OPTIMAL SOLUTION?c) IS THIS SOLUTION DEGENRATE?d) IS THIS SOLUTION FEASIBLE?e) IF s1 IS THE SLACK IN MACHINE ‘A’ (IN HRS PER WEEK) AND s2 IS THE SLACK IN MACHINE ‘B’ (IN HRS PER WEEK), WHICH OF THESE MACHINES IS BEING USED TO FULL CAPACITY WHEN PRODUCTION IS GOING ON ACCORDING TO THIS SOLUTION?f) A CUSTOMER IS WILLING TO HAVE ONE UNIT OF THE PRODUCT P1 FOR WHICH x1 UNITS ARE PRODUCED, AND IS WILLING TO PAY A PRICE IN EXCESS TO THE NORMAL PRICE IN ORDER TO GET IT. HOW MUCH SHOULD THE PRICE BE INCREASED IN ORDER THAT THERE IS NO REDUCTION IN PROFITS?g) HOW MANY UNITS OF P1 AND P2 SHOULD THE FIRM PRODUCE?h) MACHINE ‘A’ (ASSOCIATED WITH SLACK s1 HAS TO BE SHUT DOWN FOR REPAIRS FOR 2 HRS NEXT WEEK. WHAT WILL BE THE EFFECT OF SHUT DOWN ON PROFITS? i) HOW MUCH WOULD ONE BE WILLING TO PAY FOR ANOTHER HOUR PER WEEK ON MACHINE ‘A’ AND ON MACHINE ‘B’?

NUMERICAL 14 ((MAX LPP, SENSITIVITY ANALYSIS)A COMPANY HAS FACILITIES FOR PRODUCING FIVE PRODUCTS P1, P2, P3, P4, AND P5, WHICH REQUIRE SAME RAW MATERIAL AND THE SAME PRODUCTION FACILITIES AND SAME FINISHING AND PACKING FACILITIES, AS PER DETAILS GIVEN BELOW. PROD- CONTRI- RAW MATL. PRODN HRS. FINISH. & PACK. UCTS BUTION REQUIR(Kgs) REQUIREMENT HRS. REQUIREM P1 150 10 10 30P2 120 10 20 20P3 160 20 10 20P4 160 30 10 20P5 100 20 20 10MAX.AVAIL.(,000) 50 80 140 THE MANAGER OF THE COMPANY INSISTS THAT THE PRODUCTS P3 AND P4 SHOULD BE GIVEN TOP PRIORITY SINCE THEY YIELD THE MAXIMUM CONTRIBUTION. FORMULATE THIS AS AN LPP AND BTAIN THE OPTIMAL SOLUTION. STATE WHETHER YOU AGREE WITH THE MANAGER’S POINT OF VIEW. ALSO ATTEMPT THE FOLLOWING.a) WRITE THE DUAL OF THE PROBLEMb) DETERMINE MARGINAL PROFITABILITY OF THE FOLLOWING:

i) RAW MATERIALii) PRODUCTION HOURSiii) FINISHING AND PACKING HOURS

c) OVER WHAT RANGE OF VALUES OF THE RESPECTIVE CONSTRAINTS THE MARGINAL PROFITABILITIES DETERMINED BY YOU IN b) WILL BE VALID?d) OBTAIN THE OPTIMAL VALUES OF THE DUAL VARIABLESe) VERIFY THAT THE OBJECTIVE FUNCTION VALUES OF THE PRIMAL AND DUAL PROBLEMS ARE IDENTICALf) THE MARKETING MANAGER FEELS THAT THE PRICE OF THE PRODUCT P2 HAS TO BE REVISED DOWNWARDS AS A RESULT OF WHICH THE CONTRIBUTION WILL COME DOWN TO Rs 116. WHAT WOULD BE THE NEW PRODUCT MIX AND THE VALUE OF THE OBJECTIVE FUNCTION?

NUMERICAL 15 (MIN LPP BY SIMPLEX METHOD)

FOOD ‘A’ CONTAINS 20 UNITS OF VITAMIN V1 AND 40 UNITS OF VITAMIN V2 PER gm. FOOD ‘B’ CONTAINS 30 UNITS EACH OF VITAMINS V1 AND V2. THE DAILY MINIMUM HUMAN REQIREMENTS OF VITAMIN OF V1 AND V2 ARE 900 AND 1200 UNITS RESPECTIVELY. HOW MANY gms OF EACH TYPE OF FOOD SHOULD BE CONSUMED SO AS TO MINIMISE THE COST IF FOOD ‘A’ COSTS 60 PAISE PER gm AND FOOD ‘B’ COSTS 80 PAISE PER gm?

NUMERICAL 16 (MIN LPP BY SIMPLEX METHOD)

A FINISHED PRODUCT MUST WEIGH EXACTLY 150 gms. THE TWO RAW MATERIALS USED TO MANUFACTURE THE PRODUCT ARE ‘A’ AND ‘B’ WHICH COST Rs 2 AND Rs 8 PER UNIT RESPECTIVELY. AT LEAST 14 UNITS OF ‘B’ AND NOT MORE THAN 20 UNITS OF ‘A’ MUST BE USED. EACH UNIT OF ‘A’ AND ‘B’ WEIGHS 5 AND 10 gms RESPECTIVELY. HOW MUCH OF EACH TYPE OF RAW MATERIAL SHOULD BE USED FOR EACH UNIT OF THE FINAL PRODUCT IN ORDER TO MINIMISE COST?

NUMERICAL 17(MIN LPP BY SIMPLEX METHOD)

A DIETICIAN HAS PRESCRIBED VITAMIN ‘A’ AND ‘D’ TO A PATIENT WITH MINIMUYM DAILY REQUIREMENT OF 30 AND 40 UNITS RESPECTIVELY. THE DIETICIAN FINDS THAT THE PATIENT LIKES TO EAT FOOD ITEMS F1 AND F2 WHICH IN 100 gms HELPINGS WHICH RESPECTIVELY CONTAIN 1 AND 3 UNITS OF VITAMIN ‘A’ AND 2 EACH UNITS OF VITAMIN ‘D’. GIVEN THAT 100gms OF FOOD ITEM F1 AND F2 COST Rs 5 AND Rs 8 RESPECTIVELY, DETERMINE HOW MUCH QUANTITY OF FOOD ITEMS F1 AND F2 SHOULD THE PATIENT TAKE DAILY SO THAT IT COSTS THE MINIMUM. ASSUME THAT IF THE VITAMINS ARE TAKEN IN EXCESS OF PRESCRIBED REQUIREMENT, THERE ARE NO HARMFUL EFFECTS.

ANSWER FOR NUMERICAL 17(MIN LPP BY SIMPLEX METHOD)

Minimise Z = 5x1 +8x2 - 0s1 - 0s2 + MA1+ MA2

Subject to x1+ 3x2 - s1 - 0s2 + A1 + 0A2 = 30 2x1+ 2x2 - 0s1 - s2 + 0A1+ A2 = 40

x1,x2 ≥ 0

Minimise Z = 5x1 +8x2

Subject to x1+ 3x2 ≥ 30 2x1+ 2x2 ≥ 40

x1,x2 ≥ 0

ITEM VITAMIN AUNITS

VITAMIN BUNITS

COST PER 100 gms

NO. OF FOOD ITEMS TO BE TAKEN DAILY

FOOD F1 1 2 Rs 5 x1

FOOD F2 3 2 Rs 8 x2

30 40MIN REQUIREMENT

NUMERICAL 18 (FORMULATION OF MIN LPP)

A DAIRY HAS TWO MILK PROCESSING PLANTS P1 AND P2 AND EACH CAN WORK FOR 16 HRS PER DAY. IN THE PLANT P1 PROCESSING TIME FOR ONE TANKER CONTAINING 10 KILOLITRES OF FULL CREAM MILK (FCM) IS 3 HOURS AND THE COST IS Rs 30,000; PROCESSING TIME FOR ONE KILOLITRE OF DOUBLE TONED MILK (DTM) IS ONE HOUR AND THE COST IS Rs 5,000. IN THE PLANT P2, PROCESSING TIME FOR ONE TANKER (10 KL) OF FCM IS 2 HOURS AND THE COST IS Rs 36,000; PROCESSING TIME FOR ONE KILOLITRE OF DTM IS 1.5 HOURS AND THE COST IS Rs 4,000. THE DAIRY MUST PROCESS AT LEAST 10 TANKERS OF FCM AND 8 KILOLITRES OF DTM EVERY DAY. FORMULATE THIS PROBLEM AS AN LPP AND DETERMINE HOW MUCH FCM AND DTM SHOULD BE PROCESSED AT THE PLANTS P1 AND P2.

ANSWER FOR NUMERICAL 18 (FORMULATION OF MIN LPP)

Minimise Z = 5x1 +8x2 - 0s1 - 0s2 + MA1+ MA2

Subject to x1+ 3x2 - s1 - 0s2 + A1 + 0A2 = 30 2x1+ 2x2 - 0s1 - s2 + 0A1+ A2 = 40

x1,x2 ≥ 0

Minimise Z = 5x1 +8x2

Subject to x1+ 3x2 ≥ 30 2x1+ 2x2 ≥ 40

x1,x2 ≥ 0

PRODUCT PROCESSING TIME IN HOURS AT FACTORIES

VITAMIN BUNITS

COST PER 100 gms

NO. OF FOOD ITEMS TO BE TAKEN DAILY

F1 F2

FCM 3 2 Rs 5 x1

DTM 1 3/2 Rs 8 x2

MAX AVAILABLE

16 16

MIN REQUIREMENT

NUMERICAL 1

.

BASIS x1 x2 s1 s2 bi

s1 35 0 1 2/3 -1/3 8

s2 40 1 0 -1/2 1/2 18

cj 40 35 0 0

SOLN 8 18 0 0

Zj 40 35 10/3 25/3

Dj = cj -Zj0 0 (-) 10/3 (-) 25/3

OPTIMAL SOLN OF THE PRIMAL

BASIS y1 y2 s1 s2 A1 A2 bi

y2 96 0 1 -1/2 1/3 1/2 -1/3 25/3

y1 60 1 0 1/2 -2/3 -1/2 2/3 10/3

cj 60 96 0 0 M M

SOLN 10/3 25/3 0 0 0 0

Zj60 96 -18 -8 18 16

j =cj -Zj0 0 18 8 M-18 M-16

OPTIMAL SOLN OF THE DUAL