Op Scheduling Lesson 8

OPERATION SCHEDULING

ACTIVITIES INVOLVES IN OPERATION SCHEDULING

1. ASSIGNING JOB ORDERS TO DIFFERENT M/C (OR WORK CENTRES).

2. SEQUENCING ON PRIORITIZATION – DECIDING THE SEQENCE OF PROCESSING OF DIFFERENT MACHINES ON BASIS OF SOME PRIORITY RULE.

3. ROUTING – PLANING ROUTE OF MOVEMENT OF MATERIAL FROM ONE DEPARTMENT TO ANOTHER DURING PROCESSING.

4. DISPATCHING – ISSUING DISPATCH LIST TO VARIOUS WORK CENTRES (WC) CONTAINING INFORMATION ABOUT

WCs – A CUSTOMER ORDER SHOULD BE PROCESSED AT.

THE CUSTOMER ORDER TO BE PROCESSED FIRST.

AMOUNT OF TIME PROCESSING SHOULD TAKE.

5. EXPEDITING – TRACKING THE PROGRESS OF SCHEDULED JOBS AND IMPLEMENTATION OF SCHEDULES, REVISING SCHEDULES IN CASE OF DELAY AND EXPEDITING COMPLETION OF CERTAIN JOBS.

PROBLEMS DUE TO ABSENCE OF PROPER SCHEDULING

ABSENCE OF PROPER

SCHEDULING

DELAYS IN MEETING THE DUE DATES OF CUSTOMER ORDERS

HIGH WORK IN PROCESS

INVENTORY

HIGH AVERAGE COMPLETION TIME OF JOBS

LOW UTILISATION OF WORKERS AND MACHINES (HIGH IDLE TIME)

NO ACCURATE INFORMATION AVAILABLE OF THE CURRENT STATUS OF A JOB

HIGHER COST OF PRODUCTION / OPERATIONS

HIGHER SET-UP TIME (OVERALL) OF MACHINES

SEQUENCING OR PRIORITIZATION

SCHEDULING

SEQUENCING n JOBS

SEQUENCING TWO JOBS ON m MACHINES IN DIFFERENT ORDERS (AKERS METHOD)

ASSIGNINGn JOBS TO

m M/C MMMMMM

ON ONE MACHINE

ON TWO MACHINES

ON THREE MACHINES

ON m MACHINES

FIRST COME, FIRST

SERVED (FCFS)

METHOD

LAST COME, FIRST

SERVED (LCFS)

METHOD

DUE DATE METHOD

RANDOM METHOD

SHORTEST PROCESSING TIME (SPT) METHOD

JOHNSON’S METHOD

ASSIGNMENT MODEL

IN THE SAME JOB SEQUENCE

JOHNSONS METHOD

SEQUENCING n JOBS ON m MACHINES

JOHNSONS METHOD CAN ONLY BE APPLIED IN SEQUENCING n JOBS ON m MACHINES M1, M2, ………… Mm

IN THE ORDER

M1, M2, ………… Mm ONLY IF ANY ONE OR BOTH OF THESE CONDITIONS ARE FULLFILLED.

THESE CONDITIONS ARE:

1. MINIMUM PROCESSING TIME ON M/C M1 ≥ MAXIMUM PROCESSING TIME ON M/C M2 , M3 , ……. M(m-1).

2. MINIMUM PROCESSING TIME ON M/C Mm ≥ MAXIMUM PROCESSIN TIME ON M/C M2 , M3 , ……. M(m-1).

SEQUENCING n JOBS ON THREE MACHINES

JOHNSONS METHOD CAN BE APPLIED IN PROBLEMS OF SEQUENCING n JOBS ON THREE MACHINES ONLY WHEN CERTAIN CONDITIONS ARE FULLFILLED. THESE CONDITIONS ARE.

1. MINIMUM PROCESSING TIME ON M/C M1 ≥ MAXIMUM PROCESSING TIME ON M/C M2 .

2. MINIMUM PROCESSING TIME ON M/C M3 ≥ MAXIMUM PROCESSIN TIME ON M/C M2 .

IF ANY ONE OR BOTH OF THE ABOVE TWO CONDITIONS ARE SATISFIED, JOHNSON’S METHOD CAN BE APPLIED.

ASSUMPTIONS :

TWO HYPOTHETICAL MACHINES X AND Y ARE CONSIDERED.

PROCESSING TIME OF VARIOUS JOBS ON MACHINE X IS THE SUM OF PROCESSING TIME OF CORRESPONDING JOBS ON MACHINES M1 , M2 , …. M(m-

1).

SIMILARLY, PROCESSING TIME OF VARIOUS JOBS ON MACHINE Y IS THE SUM OF PROCESSING TIMES OF THE CORRESPONDING JOBS ON MACHINES M2 , M3 , ……. Mm.

PROBLEM :

THERE ARE TEN JOBS TO BE PROCESSED THROUGH TWO MACHINES M1 AND M2 IN THE ORDER M1, M2. THE PROCESSSING TIME REQUIRED BY EACH JOB (IN HOURS) IS GIVEN IN THE TABLE BELOW. FIND THE OPTIMAL SEQUENCE FOR THE PERFORMANCE OF THESE JOBS IN THE MINIMUM POSSIBLE TIME.

JOB M1 M2A 2 7B 5 3C 1 4D 6 7E 2 8F 3 9G 5 10H 1 2I 6 1

J 8 5

SOLUTION :

SEQUENCING USING JOHNSON’S METHOD :

TABLE – I

JOB M1 M2A 2 7B 5 3C 1 4D 6 7E 2 8F 3 9G 5 10H 1 2I 6 1J 8 5

TABLE – II


TABLE – III


TABLE – IV


TABLE – V


OPTIMAL SEQUENCE I

H C A E F G D J B I

OPTIMAL SEQUENCE II

C H E A F G D J B I

OPTIMAL SEQUENCE III

H C E A F G D J B I

OPTIMAL SEQUENCE IV

C H A E F G D J B I

TABLE VI

JOBMACHINE 1 MACHINE 2 WAITING TIME

OF JOBSIN OUT IN OUTH 0 1 1 3 0C 1 2 3 7 (3-2)=1A 2 4 7 14 (7-4)=3E 4 6 14 22 (14-6)=8F 6 9 22 31 (22-9)=13G 9 14 31 41 (31-14)=17D 14 20 41 48 (41-20)=21J 20 28 48 53 (48-28)=20B 28 33 53 56 (53-33)=20I 33 39 56 57 (56-39)=17

IDEAL TIME OF MACHINES

MACHINE 1 = (57 – 39) = 18MACHINE 2 = (1 – 0) = 1

PROBLEM :

THE FOLLOWING TABLE GIVES THE PROCESSING TIMES (IN HOURS) OF SEVEN JOBS TO BE PROCESSED ON THREE MACHINES M1, M2, M3. SEQUENCE THESE JOBS USING JOHNSON’S METHOD AND FIND THE OVERALL PROCESSING TIME. FIND ALSO THE WAITING TIMES OF THE JOBS AND THE IDLE TIMES OF THE THREE MACHINES.

JOB M1 M2 M3A 1 7 8B 3 3 10C 7 8 9D 9 2 11E 4 8 9F 5 6 14G 2 1 12

SEQUENCING USING JOHHSON’S METHOD (n JOBS ON 3 MACHINES)

TABLE - I

JOB M1 M2 M3A 1 7 8B 3 3 10C 7 8 9D 9 2 11E 4 8 9F 5 6 14G 2 1 12

MIN OF M1 = 1MAX OF M2 = 8MIN OF M3 = 8

THUS, MAX OF M2 = MIN OF M3. THEREFORE, THE METHOD CAN BE APPLIED.

TABLE – II

JOB X YA 8 15B 6 13C 15 17D 11 13E 12 17F 11 20G 3 13

TABLE – III

JOB X YA 8 15B 6 13C 15 17D 11 13E 12 17F 11 20G 3 13

TABLE - IVJOB X Y

A 8 15B 6 13C 15 17D 11 13E 12 17F 11 20G 3 13

TABLE - VJOB X Y

A 8 15B 6 13C 15 17D 11 13E 12 17F 11 20G 3 13

TABLE - VIJOB X Y

A 8 15B 6 13C 15 17D 11 13E 12 17F 11 20G 3 13

OPTIMAL SEQUENCE I

G B A D F E C

OPTIMAL SEQUENCE II

G B A F D E C

TABLE – VII

JOBMACHINE I MACHINE II MACHINE III WAITING TIME

OF JOBIN OUT IN OUT IN OUTG 0 2 2 3 3 15 0B 2 5 5 8 15 25 (15-8)=7A 5 6 8 15 25 33 (8-6)+(25-15)=12D 6 15 15 17 33 44 (33-17)=16F 15 20 20 26 44 58 (44-26)=18E 20 24 26 34 58 67 (26-24)+(58-34)=26C 24 31 34 42 67 76 (34-31)+(67-42)=28

IDEAL TIME OF MACHINES

MACHINE I = (76-31) = 45MACHINE II = (2-0)+(5-3)+(20-17)+(76-42) = 41MACHINE III = (3-0) = 3

PROBLEM :

THE FOLLOWING TABLE GIVES THE PROCESSING TIMES (IN HOURS) OF SEVEN JOBS TO BE PROCESSED ON FOUR MACHINES M1, M2, M3 AND M4 IN THE ORDER M1, M2, M3, M4. SEQUENCE THE GIVEN JOBS USING JOHNSON’S METHOD AND FIND THE OVERALL PROCESSING TIME.

JOB M1 M2 M3 M4A 3 1 4 12B 8 0 5 15C 11 3 8 10D 4 7 3 8E 5 5 1 10F 10 2 0 13G 2 5 6 9

SOLUTION :

SEQUENCING USING JOHHSON’S METHOD (n JOBS ON m MACHINES)

TABLE – IJOB M1 M2 M3 M4

A 3 1 4 12B 8 0 5 15C 11 3 8 10D 4 7 3 8E 5 5 1 10F 10 2 0 13

G 2 5 6 9

TABLE – II

JOB X YA 8 17B 13 20C 22 21D 14 18E 11 16F 12 15G 13 20

TABLE – III

JOB X YA 8 17B 13 20C 22 21D 14 18E 11 16F 12 15G 13 20

TABLE – IV

JOB X YA 8 17B 13 20C 22 21D 14 18E 11 16F 12 15

G 13 20

TABLE – V

JOB X YA 8 17B 13 20C 22 21D 14 18E 11 16F 12 15G 13 20

TABLE – VI

JOB X YA 8 17B 13 20C 22 21D 14 18E 11 16F 12 15G 13 20

MIN. OF M1 = 2MAX. OF M2 = 7MAX. OF M3 = 8MIN. OF M4 = 8

THUS,MIN. OF M4 = MAX. OF M3 AND MIN. OF M4 >

MAX. OF M2.

THEREFORE, THE METHOD CAN BE APPLIED.

OPTIMAL SEQUENCE I

A E F B G D C

OPTIMAL SEQUENCE II

A E F G B D C

TABLE – VII

JOBMACHINE I MACHINE II MACHINE III MACHINE IV

IN OUT IN OUT IN OUT IN OUTA 0 3 3 4 4 8 8 20E 3 8 8 13 13 14 20 30F 8 18 18 20 20 20 30 43B 18 26 26 26 26 31 43 58G 26 28 28 33 33 39 58 67D 28 32 33 40 40 43 67 75C 32 43 43 46 46 54 75 85

THE ASSIGNMENT MODEL FOR SCHEDULING

PROBLEM :

IN THE MACHINE SHOP OF XML MOTORS LTD, THERE ARE FIVE JOBS (J1 – J5) TO BE ASSIGNED TO FIVE LATHE MACHINES A, B, C, D AND E. THE PROCESSING TIMES (IN MINUTES) OF VARIOUS JOBS ON THE MACHINES, FROM PAST EXPERIENCE, ARE GIVEN IN THE TABLE GIVEN BELOW. JOB J5 CANNOT BE ASSIGNED TO MACHINE E, AS THIS MACHINE DOES NOT HAVE PROPER AUTOMATION TO HANDLE THIS JOB. ASSIGN THE JOBS TO THE MACHINES SUCH THAT THE TOTAL PROCESSING TIME IS MINIMUM. FIND ALSO THE MINIMUM TOTAL PROCESSING TIME OF THE JOBS.

TABLEA B C D E

J1 74 34 15 66 38J2 52 67 92 84 41J3 59 73 87 70 18J4 22 50 28 37 24J5 29 93 82 55 M

SOLUTION :

THE ASSIGNMENT HEURISTIC CAN BE APPLIED IN THE FOLLOWING STEPS.

MACHINES

JOBS

A B C D EJ1 74 34 15 66 38J2 52 67 92 84 41J3 59 73 87 70 18J4 22 50 28 37 24J5 29 93 82 55 M

STEP – 1 FIND THE SMALLEST VALUE IN EVERY ROW AND

SUBTRACT IT FROM EACH CELL VALUE IN THE CORRESPONDING ROW AS SHOWN IN THE FOLLOWING TABLE. FOR EXAMPLE, IN THE FIRST ROW THE SMALLEST VALUE IS 15; SUBTRACT 15 FROM EACH CELL VALUE OF THE FIRST ROW. THUS THE REVISED TABLE IS OBTAINED.

STEP – 2 FIND THE SMALLEST VALUE IN EVERY COLUMN

AND SUBTRACT IT FROM EACH CELL VALUE IN THE CORRESPONDING COLUMN. FOR EXAMPLE, IN THE FIRST COLUMN THE SMALLEST VALUE IS 0; SUBTRACT 0 FROM EACH VALUE OF THE FIRST COLUMN. THUS THE REVISED TABLE IS OBTAINED.

MACHINES

JOBS

A B C D EJ1 59 19 0 51 23J2 11 26 51 43 0J3 41 55 69 52 0J4 0 28 6 15 2J5 0 64 53 26 M

MACHINES

JOBS

A B C D EJ1 59 0 0 36 23J2 11 7 51 28 0J3 41 36 69 37 0J4 0 9 6 0 2J5 0 45 53 11 M

STEP – 3 MAKE HORIZONTAL AND VERTICAL LINES TO

COVER ALL THE ZEROS IN THE TABLE (SHOWN BY GREY SHADED RECTANGLES IN THE FOLLOWING TABLE). THE LINES DRAWN CAN BE ALL HORIZONTAL, ALL VERTICAL, OR A COMBINATION OF VERITICAL AND HORIZONTAL LINES. IF THE NUMBER OF LINES REQUIRED TO COVER ALL THE ZEROS IN THE TABLE IS EQUAL TO THE NUMBER OF MACHINES (AS THE NUMBER OF JOBS, AS BOTH EQUAL), THE OPTIMAL SOLUTION IS OBTAINED AND WE CAN SKIP STEP 4 AND DIRECTLY GO TO STEP 5. IN OUR EXAMPLE, THE NUMBER OF LINES REQUIRED TO COVER ALL ZEROS IS FOUR i.e., LESS THAN THE NUMBER OF MACHINES, THEREFORE, WE FOLLOW STEP 4.

STEP – 4 SELECT THE SMALLEST VALUE OUT OF THOSE

NOT COVERED BY ANY OF THE LINES. IN OUR EXAMPLE, IT IS 7. SUBTRACT THIS VALUE FROM ALL THE VALUES NOT COVERED BY ANY OF THE LINES (i.e., 7, 51, 28, 36, 69, 37, 45, 53 AND 11) AND ADD IT TO THOSE AT THE INTERSECTION OF ANY TWO LINES (i.e., 59, 23, 0 AND 2). AGAIN MAKE HORIZONTAL AND VERTICAL LINES TO COVER ALL THE ZEROS IN THE TABLE. IF THE NUMBER OF LINES REQUIRED TO COVER ALL THE ZEROS IN THE

MACHINES

JOBS

A B C D EJ1 59 0 0 36 23J2 11 7 51 28 0J3 41 36 69 37 0J4 0 9 6 0 2J5 0 45 53 11 M

TABLE IS EQUAL TO THE NUMBER OF MACHINES (OR THE NUMBER OF JOBS), THE OPTIMAL SOLUTION IS OBTAINED.

STEP – 5 A SINGLE ZERO IN ANY ROW OR COLUMN IS

ASSIGNED FIRST. FOR EXAMPLE, THE FIFTH ROW CONTAINS A SINGLE ZERO, WHICH IS ASSIGNED, i.e., JOB J5 HAS BEEN ASSIGNED TO MACHINE A. SIMILARLY J3 AND J4 ARE ASSIGNED TO E AND D RESPECTIVELY. THE FIRST ROW CONTAINS TWO ZEROS THEN WE WILL CHECK FOR COLUMNS WHICH CONTAINS SINGLE ZERO AND HENCE J1 IS ASSIGNED TO C. NOW SHADE THE OTHER ZERO IN THE FIRST ROW WITH DARK GREY, NOW EXCLUDING THIS ONE SINGLE ZERO IN SECOND COLUMN AND IT IS ASSIGNED AND HENCE J2 TO B. SHADE THE OTHER ZERO IN SECOND ROW DARK GREY. HENCE, ALL THE JOBS HAVE BEEN ASSIGNED TO THE GIVEN MACHINES, J1 TO C, J2 TO B, J3 TO E, J4 TO D AND J5 TO A. THUS, THE MINIMUM TOTAL PROCESSING TIME IS 15 + 67+ 18 + 37 + 29 = 166 MIN.

MACHINES

JOBS

A B C D EJ1 66 0 0 36 30J2 11 0 44 21 0J3 41 29 62 30 0J4 7 9 6 0 9J5 0 38 46 4 M

MACHINES

JOBS

A B C D EJ1 66 0 0 36 30J2 11 0 44 21 0J3 41 29 62 30 0J4 7 9 6 0 9J5 0 38 46 4 M

LINEAR PROGRAMMING I FORMULATION AND GRAPHIC SOLUTION

LINEAR PROGRAMMING

TECHNIQUE – CHOOSING THE BEST ALTERNATIVE FROM A SET OF FEASIBLE ALTERNATIVES.

CONSTRAINTS CAN BE EXPRESSED AS LINEAR MATHEMATICAL FUNCTIONS.

THE REQUIREMENTS FOR FORMULATION ARE :

OBJECTIVE SHOULD BE CLEARLY IDENTIFIABLE AND MEASURABLE IN QUANTITATIVE TERMS. FOR EX. MAXIMISATION OF SALES, OF PROFIT, MINIMISATION OF COST AND SO ON.

ACTIVITIES TO BE DISTINCTLY IDENTIFIABLE AND MEASURABLE IN QUANTITATIVE TERMS. FOR INSTANCE, THE PRODUCTS INCLUDED IN A PRODUCTION PLANNING PROBLEM.

GOAL SHOULD BE IDENTIFIABLE AND MEASURABLE QUANTITATIVELY.

RELATIONSHIPS REPRESENTING THE OBJECTIVE AS ALSO THE RESOURCE LIMITATION CONSIDERATIONS, REPRESENTED BY THE OBJECTIVE FUNCTION AND THE CONSTRAINT FUNCTION AND THE CONSTRAINT EQUATIONS OR INEQUALITIES, RESPECTIVELY MUST BE LINEAR IN NATURE.

THERE SHOULD BE SERIES OF ALTERNATIVE COURSES OF ACTION AVAILABLE TO THE DECISION MAKER.

WHEN THESE CONDITIONS ARE FULLFILLED, THE PROBLEM CAN BE EXPRESSED IN ALGEBRAIC FORM, CALLED THE LINEAR PROGRAMMING PROBLEM (LPP) AND THEN SOLVED FOR OPTIMAL DECISION.

FORMULATION OF LINEAR PROGRAMMING PROBLEMS

1. THE MAXIMISATION CASE

EXAMPLE : A FIRM IS ENGAGED IN PRODUCIN TWO PRODUCTS, A AND B. EACH UNIT OF PRODUCT A REQUIRES 2 KG OF RAW MATERIAL AND 4 LABOUR HOURS FOR PROCESSING, WHEREAS EACH UNIT OF PRODUCT B REQUIRES 3 KG OF RAW MATERIAL AND 3 HOURS OF LABOUR, OF THE SAME TYPE. EVERY WEEK, THE FIRM HAS AN AVAILABILITY OF 60 KG OF RAW MATERIAL AND 96 LABOUR HOURS. ONE UNIT OF PRODUCT A SOLD YIELDS Rs.40 AND ONE UNIT OF PRODUCT B SOLD GIVES Rs.35 AS PROFIT.

FORMULATE THIS PROBLEM AS A LINEAR PROGRAMMING PROBLEM ; ALSO DETERMINE AS TO HOW MANY UNITS OF EACH OF THE PRODUCTS SHOULD BE PRODUCED PER WEEK SO THAT THE FIRM CAN EARN MAXIMUM PROFIT. ASSUME THAT THERE IS NO MAEKETING CONSTRAINT SO THAT ALL THAT IS PRODUCED CAN BE SOLD.

THE OBJECTIVE FUNCITON :

IDENTIFY THE GOAL IN TERMS OF THE OBJECTIVE FUNCTION.

GOAL IS THE MAXIMISATION OF PROFIT, OBTAINED BY PRODUCING AND SELLING PRODUCTS A AND B.

x1 AND x2 REPRESENT THE NUMBER OF UNITS PRODUCED PER WEEK OF THE PRODUCTS OF A AND B RESPECTIVELY.

TOTAL PROFIT, Z = 40x1 + 35x2, AS THE UNIT PROFIT ON THE TWO PRODUCTS IS Rs.40 AND Rs.35 RESPECTIVELY.

THE CONSTRAINTS :

RESOURCES MUST BE IN LIMITED SUPPLY.

MATEMATICAL RELATIONSHIP USED TO EXPLAIN THIS LIMITATION IS INEQUALITY. THIS LIMITATION ITSELF IS KNOWN AS CONSTRAINT.

AS A REQUIRES 2 Kg AND B REQUIRES 3 Kg OF RAW MATERIALS, THE TOTAL CONSUMPTION WOULD BE 2x1 + 3x2 WHICH CANNOT EXCEED THE TOTAL AVAILABILITY OF 60 Kg EVERY WEEK. WE CAN EXPLAIN THIS CONSTRAINT AS 2x1 + 3x2 ≤ 60. SIMILARLY ONE UNIT OF A AND B REQUIRES 4 AND 3 LABOUR HOURS FOR PRODUCTION RESPECTIVELY. WITH THE AVAILABILITY OF 96 HOURS A WEEK, WE HAVE 4x1 + 3x2 ≤ 96 AS THE LABOUR HOURS CONSTRAINT.

NON – LINEARITY CONDITION :

QUITE OBVIOUSLY, x1 AND x2, BEING THE NUMBER OF UNITS PRODUCED, CANNOT HAVE NEGATIVE VALUES. THUS, BOTH OF THEM CAN ASSUME VALUES ONLY GREATER THAN OR EQUAL TO ZERO. THIS IS NON-NEGATIVITY CONDITION. THIS CAN SYMBOLICALLY EXPRESSED AS x1 ≥ 0 AND x2 ≥ 0.

NOW WE CAN WRITE THE PROBLEM AS FOLLOWS :

MAXIMISEZ = 40x1 + 35x2 PROFIT

SUBJECT TO2x1 + 3x2 ≤ 60 RAW MATERIAL CONSTRAINT4x1 + 3x2 ≤ 96 LABOUR HOURS CONSTRAINT x2, x2 ≥ 0 NON-NEGATIVITY RESTRICTION

2. THE MINIMISATION CASE :

EXAMPLE :

THE AGRICULURAL RESEARCH INSTITUTE SUGGESTED TO A FARMER TO SPREAD OUT AT LEAST 4800 Kg OF A SPECIAL PHOSPHATE FERTILISER AND NOT LESS THAN 7200 Kg OF A SPECIAL NITROGEN FERTILISER TO RAISE PRODUCTIVITY OF CROPS IN HIS FIELDS. THERE ARE TWO SOURCES FOR OBTAINING THESE – MIXTURES A AND B. BOTH OF THESE ARE AVAILABLE IN BAGS WEIGHING 100 Kg EACH AND THEY COST Rs.40 AND Rs.24 RESPECTIVELY. MIXTURE A CONTAINS PHOSPHATE AND NITROGEN EQUUIVALENT OF 20 Kg AND 80 Kg RESPECTIVELY, WHILE MIXTURE B CONTAINS THESE INGREDIENTS EQUIVALENT OF 50 Kg EACH.

WRITE THIS AS A LINEAR PROGRAMMING PROBLEM AND DETERMINE HOW MANY BAGS OF EACH TYPE SHOULD THE FARMER BUY IN ORDER TO OBTAIN THE REQUIRED FERTILISER AT MINIMUM COST.

THE OBJECTIVE FUNCTION :

IF x2 AND x2 ARE TAKEN TO REPRESENT THE NUMBER OF BAGS OF MIXTURES A AND B RESPECTIVELY, THE OBJECTIVE FUNCTION CAN BE EXPRESSED AS FOLLOWS :

MINIMISE Z = 40x1 + 24x2 COST

CONSTRAINTS :

IN THIS PROBLEM THERE ARE TWO CONSTRAINTS, NAMELY, A MINIMUM OF 4800 Kg OF PHOSPHATE AND 7200 Kg OF NITROGEN INGREDIENTS ARE REQUIRED. IT IS KNOWN THAT EACH BAG OF MIXTURE A CONTAINS 20 Kg

OF EACH BAG OF MIXTURE B CONTAINS 50 Kg OF PHOSPHATE. THE PHOSPHATE REQUIREMENT CAN BE EXPRESED AS 20x1 + 50x2 ≥ 4800. SIMILARLY THE NITROGEN REUIREMENT WOULD BE 80x1 + 50x2 ≥ 7200.

NON – NEGATIVE CONDITION :

AS BEFORE, IT LAYS THAT THE DECISION VARIABLES, REPRESENTING THE NUMBER OF BAGS OF MIXTURES A AND B, WOULD BE NON-NEGATIVE. THUS, x1 ≥ 0 AND x2 ≥ 0.

THE LINEAR PROGRAMMING PROBLEM CAN NOW BE EXPRESSED AS FOLLOWS :

MINIMISEZ = 40x1 + 24x2 COST

SUBJECT TO

20x1 + 50x2 ≥ 4800 PHOSPHATE REQUIREMENT80x1 + 50x2 ≥ 7200 NITROGEN REQUIREMENT x2, x2 ≥ 0 NON-NEGATIVITY

RESTRICTION

THE SOLUTION TO LINEAR PROGRAMMING PROBLEMS

GRAPHIC METHOD :TO USE THE GRAPHIC METHOD FOR

SOLVING LINEAR PROGRAMMING PROBLEMS, THE FOLLOWING STEPS ARE REQUIRED :

IDENTIFY THE PROBLEMS – THE DECISION VARIABLES, THE OBJECTIVE FUNCTION AND THE CONSTRAINT RESTRICTIONS.

DRAW A GRAPH THAT INCLUDES ALL THE CONSTRAINTS / RESTRICTIONS AND IDENTIFY THE FEASIBLE REGION.

OBTAIN THE POINT ON THE FEASIBLE REGION THAT OPTIMISES THE OBJECTIVE FUNCTION – THE OPTIMAL SOLUTION.

INTERPRET THE RESULTS.

THE MAXIMISATION CASE :

WE CONSIDER THE SAME EXAMPLE AGAIN. FOR THIS PROBLEM, THE DECISION VARIABLES ARE x2 AND x2, THE NUMBER OF UNITS OF THE PRODUCTS A AND B RESPECTIVELY. THE OBJECTIVE FUNCTION AND THE CONSTRAINTS ARE REPRODUCED AS

MAXIMISEZ = 40x1 + 35x2 PROFIT

SUBJECT TO2x1 + 3x2 ≤ 60 RAW MATERIAL CONSTRAINT4x1 + 3x2 ≤ 96 LABOUR HOURS CONSTRAINT x2, x2 ≥ 0

GRAPHIC PLOT OF CONSTRAINTS

x1

No 32 OF 30

UNIT 28

OF 26

PRO 24

DUCT 22

B 20

18 16 14

12

10

8

6

4

2

x2

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

No. OF UNITS OF PRODUCT A

LABOUR HOURS CONSTRAINT

RAW MATERIALS

CONSTRAINT

DETERMINATION OF OPTIMAL SOLUTION USING ISO PROFIT LINE

THE MINIMISATION CASE :

x1

No 32 OF 30

UNIT 28

OF 26

PRO 24

DUCT 22

B 20

18 16 14

12

10

8

6

4

2

x2

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

No. OF UNITS OF PRODUCT A

FEASIBLE REGION

ISO PROFIT LINEPROFIT = Rs.280

LABOUR HOURS CONSTRAINT

RAW MATERIALS

CONSTRAINT

ISO-PROFIT LINES

WE SHALL CONSIDER THE GRAPHICAL SOLUTION TO THEIR LINEAR PROGRAMMING PROBLEMS OF THE MINIMISATION NATURE. THE SAME EXAMPLE IS RECONSIDERED.

MINIMISEZ = 40x1 + 24x2 COST

SUBJECT TO20x1 + 50x2 ≥ 4800 PHOSPHATE REQUIREMENT80x1 + 50x2 ≥ 7200 NITROGEN REQUIREMENT x2, x2 ≥ 0

GRAPHIC SOLUTION TO THE LPP

x1

No OF

BAGS 168

OF 156

MIX 144

TURE 132

A 120

108 96 84

72

60

48

36

24

12

x2

20 40 60 80 100 120 140 160 180 200 220 240 260 280 300

No. OF BAGS OF MIXTURE A

RAW MATERIALS

CONSTRAINT

PHOSPHATE REQUIREMENT

FEASIBLE REGION

NITROGEN REQUIREMENT

ISO-COST LINE

CASE STUDY :

AN ELECTRONIC GOODS COMPANY HAS DISTRIBUTORS WHO WILL ACCEPT SHIPMENTS OF TRANSISTORS, RADIOS OR ELECTRONIC CALCULATORS TO STOCK FOR DIWALI INVENTORY. WHEREAS THE RADIOS CONTRIBUTE Rs.10 PER UNIT AND CALCULATORS Rs.15 PER UNIT OF PROFIT. EACH RADIO REQUIRES FOUR DIODES AND FOUR RESITORS, WHILE EACH CALCULATOR REUIRES TEN DIODES AND TWO RESISTORS. THE RADIO TAKES 12 MINUTES AND CALCULATORS TAKE 9.6 MINUTES ON THE COMPANY’S ELECTRONIC TESTING MACHINE.

THE PRODUCTION MANAGER ESTIMATES THAT 160 HOURS OF THE TEST TIME ARE AVAILABLE. THE FIRM HAS 8000 DIODES AND 3000 RESISTORS IN THE INVENTORY. DETERMINE THROUGH GRAPHICAL SOLUTION OF LINEAR PROGRAMMING, THE PRODUCT MIX TO MAXIMISE PROFIT.

(A) FORMULATE THE OBJECTIVE AND CONSTRAINT FUNCTIONS AND NON-NEGATIVITY RELATIONSHIP.

(B) DRAW THE GRAPH AND INDICATE THE FEASIBLE REGION AND FEASIBLE POINT SOLUTION.

(C) DETERMINE THE OPTIMUM MIX AND PROFIT.

(D) SUMMARISE THE GRAPHICAL SOLUTION PROCEDURE FOR MAXIMISATION PROBLEMS AND EXPLAIN THE MEANING OF ISO-PROFIT LINES.

SOLUTION :

DECISION VARIABLES

x1 AND x2 REPRESENTS THE NUMBER OF UNITS CONTRIBUTED PER UNIT OF THE PRODUCTS A (RADIO) AND B (CALCULATOR).

THE OBJECTIVE FUNCTION

THE TOTAL PROFIT

Z = 10x1 + 15x2

GOAL

MAXIMISATION OF PROFIT

CONSTRAINTS

4x1 + 10x2 ≤ 8000 DIODE CONSTRAINT

4x1 + 2x2 ≤ 3000 RESISTOR CONSTRAINT

12x1 + 9.6x2 ≤ 9600TIME CONSTRAINT

x1, x2 ≥ 0 NON-NEGATIVE RESTRICTION

x2

250 500 750 1000 1250 1500 1750 2000 2250 2500 2750

No. OF UNITS OF RADIO

1600

1500

1400No. OF

1300UNITS

12001100

OF 1000

CALCU- 900 LATOR 800

700600

500400

300

200

100

0

FEASIBLE REGION

x1

x2

250 500 750 1000 1250 1500 1750 2000 2250 2500 2750

No. OF UNITS OF RADIO

1600

1500

1400No. OF

1300UNITS

12001100

OF 1000

CALCU- 900 LATOR 800

700600

500400

300

200

100

0

FEASIBLE REGION

Op Scheduling Lesson 8

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