basic terminology and application of simplex method using algorithm

1. LINEAR PROGRAMMING
PROBLEMS
SIMPLEX METHOD

2. BASIC TERMINOLOGY
• SLACK VARIABLE: A VARIABLE WHICH IS ADDED TO THE LEFT HAND SIDE OF A LESS THAN OR EQUAL CONSTRAINT TO MAKE IT AN
EQUALITY CONSTRAINT IS CALLED A SLACK VARIABLE.
• FOR EXAMPLE:
• 2X+3Y≤80
• 2X+3Y+𝒔𝟏 = 𝟖𝟎
• THIS SLACK VARIABLES, MUST BE NON-NEGATIVE.
• INTERPRETATION OF SLACK VARIABLE: SLACK VARIABLE REPRESENTS THE UNUSED CAPACITY.
• SURPLUS VARIABLE: A VARIABLE SUBTRACTED FROM THE LEFT HAND SIDE OF A GREATER THAN OR EQUAL TO CONSTRAINT TO MAKE IT
AN EQUALITY CONSTRAINT IS CALLED A SURPLUS VARIABLE.
• FOR EXAMPLE:
• 2X+3Y≥80
• 2X+3Y-𝒔𝟏 = 𝟖𝟎
• THIS SLACK VARIABLE MUST BE NON-NEGATIVE.
• BASIC SOLUTION: FOR A SYSTEM OF M SIMULTANEOUS LINEAR EQUATIONS IN N VARIABLES (N > M) A SOLUTION OBTAINED BY SETTING
(N-M) VARIABLES EQUAL TO ZERO AND SOLVING FOR THE REMAINING M VARIABLES IS CALLED A BASIC SOLUTION. THE (N-M) VARIABLES
WHICH ARE SET EQUAL TO ZERO IN ANY SOLUTION ARE CALLED NON-BASIC VARIABLES. THE OTHER M VARIABLES WHOSE VALUES ARE
OBTAINED BY SOLVING THE REMAINING SYSTEM OF EQUATIONS ARE REFERRED TO AS BASIC VARIABLES.

3. BASIC TERMINOLOGY
• DEGENERATE SOLUTION: A BASIC SOLUTION TO THE SYSTEM IS CALLED DEGENERATE IF
ONE OR MORE OF THE BASIC VARIABLE IS ZERO.
• BASIC FEASIBLE SOLUTION: THE BASIC SOLUTION WHICH SATISFY THE NON-NEGATIVITY
RESTRICTION OF AN LPP IS CALLED A BASIC FEASIBLE SOLUTION. IN THEOREM: THE SET OF
CORNER POINTS OF THE FEASIBLE REGION CORRESPONDS TO THE SET OF BASIC FEASIBLE
SOLUTIONS .
• CORNER POINTS ARE BASIC FEASIBLE SOLUTIONS AND VICE-VERSA.
• FUNDAMENTAL EXISTENCE THEOREM: IT STATES THAT WHENEVER THERE EXISTS AN
OPTIMUM SOLUTION TO A LINEAR PROGRAMMING PROBLEM, THERE EXISTS ONE WHICH IS
ALSO BASIC FEASIBLE SOLUTION THE SIMPLEX METHOD OF SOLVING AN LPP IS BASED ON
THIS THEOREM.

4. ALGORITHM
• FOR THE SOLUTION OF ANY LPP BY SIMPLEX ALGORITHM, THE EXISTENCE OF AN INITIAL BASIC
FEASIBLE SOLUTION IS ALWAYS ASSUMED.
• STEP 1: CHECK WHETHER THE OBJECTIVE FUNCTION OF THE GIVEN LPP IS TO BE MAXIMIZED OR
• MINIMIZED. IF IT IS TO BE MINIMIZED THEN FIRST CONVERT IT IN TO A PROBLEM OF
MAXIMIZATION
• TYPE BY MULTIPLYING THE OBJECTIVE FUNCTION BY (-1) THAT IS MINIMUM Z = - MAXIMUM (-1)
• STEP 2: CHECK WHETHER THE R.H.S OF ALL THE CONSTRAINTS DENOTED BY B ARE NON-
NEGATIVE. IF ANY ONE OF THE B, IS NEGATIVE THEN MULTIPLY THE CORRESPONDING
INEQUATION OF THE CONSTRAINT BY (-1), SO AS TO SET ALL B, NON-NEGATIVE.
• STEP 3: CONVERT ALL THE INEQUATIONS OF THE CONSTRAINTS IN TO EQUATIONS BY
INTRODUCING
• SLACK AND/OR SURPLUS VARIABLES IN THE CONSTRAINTS. ASSIGN A COST OF ZERO TO ALL THE
SLACK

5. ALGORITHM
• OBTAIN AN INITIAL BASIC FEASIBLE SOLUTION TO THE PROBLEM
BY CONSIDERING THE SLACK VARIABLES AS THE BASIC
VARIABLES.
• STEP4: DETERMINE THEIR VALUES DIRECTLY, BECAUSE THE
REMAINING VARIABLES ARE NON-BASIC AND THEREFORE EACH
HAS THE VALUE ZERO.
• STEP 5: SET UP THE INITIAL SIMPLEX TABLEAU AS FOLLOWS:
CB
Co-efficient of the
current basic
variable in objective
function
Variables in
Basis
𝑺𝒏
𝒄𝒋 𝑪𝟏 𝑪𝟐 𝑪𝟑 …
…
𝑪𝒏 0 0 0 0 MIN
RATIO
SOLUTI
ON
VALUE
S b
B
Basic
variables
𝑿𝟏 𝑿𝟐 𝑿𝟑 …. 𝑿𝒏 𝑺𝟏 𝑺𝟐 ….. 𝑺𝒏
𝑪𝑩𝟏 𝑺𝟏=𝒃𝟏 𝑿𝟏 𝒂𝟏𝟏 𝒂𝟏𝟐 𝒂𝟏𝟑 𝒂𝟏𝒏 𝒃𝟏
𝑪𝑩𝟐 𝑺𝟐=𝒃𝟐 𝑿𝟐 𝒂𝟐𝟏 𝒂𝟐𝟐 𝒂𝟐𝟑 𝒂𝟐𝒏 𝒃𝟐
𝑪𝑩𝟑 𝑺𝟑=𝒃𝟑 𝑿𝟑 𝒂𝟑𝟏 𝒂𝟑𝟐 𝒂𝟑𝟑 𝒂𝟑𝒏 𝒃𝟑
…….. …….. ………. …
…
……
𝑪𝑩𝒎 𝑺𝑵=𝒃𝑵 𝑿𝒏 𝒂𝒎𝟏 𝒂𝒎𝟐 𝒂𝒎𝟑 𝒂𝒎𝒏 𝒃𝒎
𝒁𝒋 − 𝑪𝒋
Index row
𝒁𝟏
− 𝑪𝟏
𝒁𝟐
− 𝑪𝟐
𝒁𝟑
− 𝑪𝟑
𝒁𝒏
− 𝑪𝒏

6. • (I) THE FIRST ROW CONTAINS 𝐶𝑗 WHICH REPRESENTS THE COEFFICIENTS OF THE
VARIABLES IN THE OBJECTIVE FUNCTION. THESE VALUES REMAINS UNCHANGED IN THE
WHOLE SIMPLEX METHOD.
• (II) THE SECOND ROW SHOWS THE COLUMN HEADINGS. THESE HEADINGS REMAIN THE
SAME IN
SUBSEQUENT SIMPLEX TABLES.
• (III) THE FIRST COLUMN LABELLED 𝐶𝐵SHOWS THE COEFFICIENTS (IN THE OBJECTIVE
FUNCTION) OF THE BASIC VARIABLES ONLY.
• (IV) THE SECOND COLUMN LABELLED "BASIC VARIABLES" GIVES THE NAMES OF BASIC
VARIABLES
AT EACH ITERATION.
• (V) THE THIRD COLUMN LABELLED "SOLUTION REPRESENTS THE SOLUTION VALUES OF THE
BASIC VARIABLES.
• (VI) THE BODY MATRIX UNDER THE NON-BASIC VARIABLES IN THE INITIAL SIMPLEX
TABLEAU CONSISTS OF THE COEFFICIENTS OF THE DECISION VARIABLES IN THE
CONSTRAINTS SET.
• (VII) THE IDENTITY MATRIX IS FORMED UNDER THE COLUMNS OF BASIC VECTORS IN
EVERY SIMPLEX TABLE

7. • (VIII) TO GET AN ENTRY IN THE 𝑍𝑗 ROW UNDER A COLUMN, WE MULTIPLY THE ENTRIES IN
THAT COLUMN BY THE CORRESPONDING ENTRIES OF THE 𝐶𝑗 COLUMN AND ADD THE
PRODUCTS. THE 𝑍𝑗 ENTRY UNDER THE "SOLUTION" COLUMN GIVES THE CURRENT VALUE OF
THE OBJECTIVE FUNCTION. THE OTHER 𝑍𝑗 ENTRIES REPRESENT THE DECREASE IN THE
OBJECTIVE FUNCTION THAT WOULD RESULT IF ONE OF THE VARIABLE NOT INCLUDED IN THE
SOLUTION WERE BROUGHT IN THE SOLUTION.
• (IX) THE LAST ROW LABELLED 𝐶𝑗 - 𝑍𝑗 CALLED THE NET EVALUATION ROW IS USED TO CHECK
IF THE CURRENT SOLUTION IS OPTIMAL OR NOT. 𝐶𝑗 - 𝑍𝑗 CORRESPONDING TO BASIC
VARIABLE IS ALWAYS ZERO. BUT THIS VALUE IS SIGNIFICANT FOR NON-BASIC VARIABLES. 𝐶𝑗 -
𝑍𝑗 ROW REPRESENT THE NET CONTRIBUTION TO THE OBJECTIVE FUNCTION THAT RESULTS
BY INTRODUCING ONE UNIT OF EACH OF THE RESPECTIVE COLUMN VARIABLES. A PLUS
VALUE INDICATES THAT A GREATER CONTRIBUTION CAN BE MADE BY BRINGING THE
VARIABLE FOR THAT COLUMN IN TO THE BASIS. A NEGATIVE VALUE INDICATES THE AMOUNT
BY WHICH CONTRIBUTION WOULD DECREASE IF ONE UNIT OF THE VARIABLE FOR THAT
COLUMN WERE BROUGHT IN TO THE SOLUTION.
• SHADOW PRICE: THE SHADOW PRICE FOR EACH RESOURCE IS SHOWN IN THE 𝐶𝑗 - 𝑍𝑗 ROW OF
THE FINAL SIMPLEX TABLE UNDER THE CORRESPONDING SLACK VARIABLE. IT REPRESENTS
THE MAXIMUM PRICE THAT ONE WOULD LIKE TO PAY FOR AN ADDITIONAL UNIT OF A

8. • OPPORTUNITY COST: IT IS THE PENALTY INCURRED IF WE PASS THE OPPORTUNITY OF
LEAVING THE SOLUTION AS IS AND BRING IN A NEW VARIABLE. THE TERMS IN Z, ROW
INDICATE THE OPPORTUNITY COST. IF WE DO NOT UTILIZE ONE UNIT OF X, THE LOSS OF
PROFIT IS 4 UNITS AND NON- UTILIZATION OF ONE UNIT OF Y COSTS 10 UNITS.
• (B) SHADOW PRICES: IT IS SHOWN IN THE ROW OF THE FINAL SIMPLEX TABLE UNDER THE
CORRESPONDING SLACK VARIABLE. IN THE FINAL SIMPLEX TABLE THE VARIABLES
CORRESPONDING TO SECOND CONSTRAINT HAS 𝐶𝑗 - 𝑍𝑗 , IF 𝐶𝑗 - 𝑍𝑗 =-2 FOR 𝑆2 THAT
SHADOW PRICE FOR THE SECOND RESOURCE IS +2. THIS MEANS THAT WE COULD
INCREASE THE OBJECTIVE FUNCTION BY 2 IF WE HAD AN ADDITIONAL UNIT OF THAT
RESOURCE. THUS IF THE MANAGER WERE TO PAY BELOW 2 FOR THE ADDITIONAL UNIT OF
SECOND RESOURCE, THEN PROFITS COULD BE INCREASED AND IF THE MANAGER WERE TO
PAY ABOVE THIS FIGURE THEN THE PROFITS WOULD DECREASE. THUS THE MAXIMUM PRICE
THAT THE MANAGER SHOULD PAY FOR AN ADDITIONAL UNIT OF SECOND RESOURCE IS 2.
AND IF THE SHADOW PRICES OF FIRST AND THIRD RESOURCE IS S₁=0 AND S₂=0
RESPECTIVELY. AN ADDITIONAL UNIT OF FIRST AND THIRD RESOURCE WOULD NOT S3
HELP SINCE THESE RESOURCES ARE NOT FULLY UTILIZED.

9. A PHARMACEUTICAL COMPANY HAS 100KG OF A,180KG OF B AND 120KG OF C INGREDIENTS AVAILABLE PER MONTH .COMPANY CAN USE THESE
MATERIALS TO MAKE THREE BASIC PHARMACEUTICAL PRODUCTS NAMELY 5-10-5,5-5-10,AND 20-5-10 WHERE THE NUMBERS IN EACH CASE REPRESENT
THE PERCENTAGE OF WEIGHT OF A,B AND C RESPECTIVELY, IN EACH OF THE PRODUCTS . THE COST OF THESE RAW MATERIAL IS AS FOLLOWS :
ingredient Cost per kg (Rs)
A 80
B 20
C 50
Inert ingredient 20
The selling prices of these products are Rs 40.5,Rs 43 and Rs 45
respectively. There is a company restriction of the company for
product 5-10-5 , because of which the company cant produce more
than 30kg per month determine how much of each of the product
the company should produce in order to maximize its monthly
profit
Solution :
Product A B C Inert (REMAIN UNUTILIZED)
5-10-5 5% 10% 5% 100-(5+10+5)=80%
5-5-10 5% 5% 10% 100-(5+5+10)=80%
20-5-10 20% 5% 10% 100-(20+5+10)=65%
Cost per kg 80 20 50 20%
COMPANY HAS 100kg (A) 180kg (B) 120kg (C)
Cost of product (5-10-5)= 5% of 80+10%of 20+5%50+80% of
20=4+2+2.50+16=Rs 24.5 per kg
Cost of product (5-5-10)= 5% of 80+5%of 20+10%50+80% of
20=4+1+5+16=Rs 26 per kg
Cost of product (20-5-10)=20% of 80+5%of 20+10%50+65% of
20=16+1+5+13=Rs 35 per kg
Selling price for (5-10-5)= Rs 40.5 (given )
Selling price for (5-5-10)=Rs 43 (given)
Selling price for (20-5-10)=Rs 45 (given)
Profit = selling price –cost price
Now profit on (5-10-5)=Rs40.5 -
Rs24.5=Rs 16
Profit on (5-5-10)=Rs43 –
Rs26=Rs 17
profit on (20-5-10)=Rs 45 – Rs
35=Rs 10

10. NOW OBJECTIVE AND CONSTRAINTS
ARE
• OBJECTIVE : TO MAXIMIZE PROFIT
• LET 𝑥1𝑢𝑛𝑖𝑡 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 5 − 5 − 10 , 𝑥2 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 5 − 10 − 5 𝑎𝑛𝑑 𝑥3 𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 20 −

11. OBJECTIVE FUNCTION:16𝑥1 + 17𝑥2 + 10𝑥3+0𝑠1+0𝑠2+0𝑠3 + 0𝑠3
• CONSTRAINTS :
• 𝒙𝟏 + 𝒙𝟐 + 𝟒𝒙𝟑+𝟏𝒔𝟏+0𝒔𝟐+0𝒔𝟑 + 𝟎𝒔𝟑=2000
• 𝟐𝒙𝟏 + 𝒙𝟐 + 𝒙𝟑 +0𝒔𝟏+1𝒔𝟐+0𝒔𝟑 + 𝟎𝒔𝟑=3600
• 𝒙𝟏 + 𝟐𝒙𝟐 + 𝟐𝒙𝟑 +0𝒔𝟏+0𝒔𝟐+1𝒔𝟑 + 𝟎𝒔𝟑=2400
• 𝒙𝟏 + 𝟎𝒙𝟐 + 𝟎𝒙𝟑 +0𝒔𝟏+0𝒔𝟐+0𝒔𝟑 + 𝟏𝒔𝟑=30
• 𝑥1, 𝑥2, 𝑥3 ≥ 0 𝑪𝑱 16 17 10 0 0 0 0 MIN RATIO
𝑐𝐵 B 𝑋𝐵 𝑋1 𝑋2 𝑋3 𝑆1 𝑆2 𝑆3 𝑆4 𝑋𝐵
𝑋2
0 𝑆1 2000 1 1 4 1 0 0 0 2000/1=2000
0 𝑆2 3600 2 1 1 0 1 0 0 3600/2=1800
0 𝑆3 2400 1 2 2 0 0 1 0 2400/1=2400
0 𝑆4 30 1 0 0 0 0 0 1 30/1=30
𝑍𝐽 0 0 0 0 0 0 0
𝐶𝐽-𝑍𝐽 16 17 10 0 0 0 0

12. Thank you