PT-MBA 1st year, 3rd Trimester, NMIMS

Decision Analysis and Modeling Assignment LPP formulation, Graphical solution and Simplex Method

Submitted to :

Prof. Shailaja Rego Submitted by :

Neha Kumar – A-029 Nikita Thakkar – B-029 3/23/2013

Overview:

Johnson and Johnson Consumer Products Division is divided into 5 main product categories.

Baby Care, Women’s Health Franchise (sanitary hygiene products), Skin Care (Neutrogena,

Clean and Clear), Over-the-counter products (like Benadryl, Nicorette, etc), Wound care (Band-

aid, Savlon) and Oral Care (Listerine).

Women’s Health Franchise consists of sanitary protection products. Stayfree is a very important

range of product under Women’s Health Franchise. It’s one of the highest revenue generating

brands for consumer products division. Any decision pertaining to Stayfree is very crucial for the

company given the contribution it generates for the company.

Stayfree range consists of 7 variants depending on their features to cater to needs of women

across SECs. The company has 2 products – product A and product B, both of which are

manufactured on the same machine. One packet contains 8 pads each for both products.

However, there are 2 major differences between the two products:

1. First difference is the cover which forms the top most surface of the napkin. Product A

has a non-woven fabric cover which gives a soft feel. Product B has a dry net cover

which gives a dry feel.

2. Product A is directly packed in polybags. Product B is first folded and then each napkin

is packed in individually folded pouches before packing them in the polybag.

Both these products are very important since they contribute significantly towards the turnover

of the Stayfree portfolio. Both these products cater to two different consumer needs, namely,

soft feel and dry feel. Therefore, demand for both is exclusive of each other. If Product A is

discontinued then the consumers of Product A are not likely to shift to Product B and vice versa.

The company therefore needs to continue manufacturing and marketing both products.

Since there is only a cover difference between the two products, they are manufactured on the

same machine with a change in only one raw material i.e. the top cover of the napkin. There

is an additional process involved for Product B where it gets folded and packed in individual

pouches. This process is done with the help of operating the machine along with an

extension which increases the machine time in case of Product B. The volumes of both

products are quite high. The gross profit margin is same for both at 40%. Details of the same

are given in the following table.

Product A Product B

Selling Price (MRP per

pack)

22 31

Less: Variable Cost 13.2 18.6

Contribution 8.8 12.4

Decision making: The company has one manufacturing unit which has 5 identical machines.

Each machine can be operated for maximum 20 hours per day and gives the same amount of

output. The machine hours, packaging time for both products are mentioned in the table below.

The company is obliged to supply minimum 200,000 packs of Product A and 100,000 packs

of product B to ensure that it does not lose its current market share. The company needs to

maximize its profit by producing the optimal quantities of both products.

Product A Product B Total

Total Demand (packs per

month)

350,000 250,000 600,000

Minimum requirement

(packs per month)

200,000 100,000 300,000

Machine time per pack (in

minutes) – refer to working

note 1

0.1 0.7 180,000

(per month)

Packaging time per pack (in

minutes)

0.5 0.7 300,000

(per month)

Linear Programming Problem Formulation:

The company has to decide the optimal quantity to be produced of both products to maximize its

profit. The company faces constraints with regards to the minimum quantity required of both

products to maintain current market share, the machine hours and the packaging hours.

Given this information, we need to formulate this problem into a linear programming problem:

Let X1 be the number of packs of product A that need to be produced by the company.

Let X2 be the number of packs of product B that need to be produced by the company.

Objective function: Maximize Z (Profit Maximization) = 8.80 x1 + 12.40 x2

Subject to constraints:

X1 ≥ 200,000 (Minimum requirement constraint for X1)

X2 ≥ 100,000 (Minimum requirement constraint for X2)

0.10 X1 + 0.70 X2 ≤ 180000 (Machine hours constraint)

0.50 X1 + 0.70 X2 ≤ 300000 (Packaging hours constraint)

X1, X2 ≥ 0 (Non-negativity constraint)

To solve this LPP formulation, we need to convert all these inequalities to equalities. We can

enter objective (i.e. profit maximization in this case) and constraint values in QM for windows

software and click on solve. We get the following output.

X1 X2

RHS Dual

Maximize 8.8 12.4

Minimum Requirement X1 1 0 >= 200000 0

Minimum Requirement X2 0 1 >= 100000 0

Machine Hours Constraint 0.1 0.7 <= 180000 0.1428564

Packaging Hours Constraint 0.5 0.7 <= 300000 17.57143

Constraint 5 1 0 >= 0 0

Constraint 6 0 1 >= 0 0

Solution-> 300000 214285.7

$5,297,142.88

We also get the corner points and the area highlighted in pink is the feasible area.

But as we can see in the above graph, the optimal solution is the point which is circled. We need

to produce 300,000 packs of X1 (Product A) and 214,285 packs of X2 (Product B). The optimal

solution will give us a profit of Rs. 5,297,143/-. The Z values for all 4 points of the feasible

region are mentioned below.

Optimal solution - Range Table:

Variable Value Reduced

Cost

Original Val Lower

Bound

Upper

Bound

X1 300000 0 8.8 1.77 8.86

X2 214285.7 0 12.4 12.32 61.6

Constraint Dual Value Slack/Surplus Original Val Lower Bound Upper

Bound

Minimum

Requirement X1

0 100000 200000 -Infinity 300000

Minimum

Requirement X2

0 114285.7 100000 -Infinity 214285.7

Machine Hours

Constraint

0.1428564 0 180000 116000 220000

Packaging Hours

Constraint

17.57143 0 300000 260000 620000

Constraint 5 0 300000 0 -Infinity 300000

Constraint 6 0 214285.7 0 -Infinity 214285.7

Iteration 8

(Iterations 1 to 7 are included in the working note for reference)

The optimal product mix is X1 = 300,000 and X2 = 214285.7 and total maximum

profit contribution is Z = Rs. 5,297,143/-

The objective function coefficient of X1 is 8.8 and its allowable decrease is 1.77 and

allowable increase is 8.86.

The objective function coefficient of X2 is 12.4 and its allowable decrease is 12.32 and

allowable increase is 61.6.

If the value of the objective function coefficients goes below the allowable decrease or

above the allowable increase then the optimal solution will change.

The RHS range for machine hours is 116,000 to 220,000 and for packaging hours is

260,000 to 620,000.

The shadow price value for machine hours is Rs. 0.1429 and packaging hours is Rs.

17.5714. Subject to the cost of increasing one unit of these resources, we should choose

the resource which has a higher shadow price as it will give us a higher contribution per

unit.

Working Notes:

1. Calculation of Total available machine hours per month

Number of

Machines

Maximum

Number of

Operating Hours

Maximum

Number of

minutes

Number of

days per month

Total Machine

Hours per

month (in

minutes)

5 20 60 30 180000

Cj

Basic

Variables

8.8

X1

12.4

X2 0 artfcl 1

0

surplus

1 0 artfcl 2

0

surplus

2 0 slack 3 0 slack 4 0 artfcl 5

0

surplus 5

0 artfcl

6

0

surplus

6 Quantity

Iteration 8

0 surplus 5 0 0 0 0 0 0 -2.5 2.5 -1 1 0 0 300,000.00

0 surplus 6 0 0 0 0 0 0 1.7857 -0.3571 0 0 -1 1 214,285.72

0 surplus 2 0 0 0 0 -1 1 1.7857 -0.3571 0 0 0 0 114,285.72

0 surplus 1 0 0 -1 1 0 0 -2.5 2.5 0 0 0 0 100,000.00

8.8 X1 1 0 0 0 0 0 -2.5 2.5 0 0 0 0 300,000.00

12.4 X2 0 1 0 0 0 0 1.7857 -0.3571 0 0 0 0 214,285.72

zj 8.8 12.4 0 0 0 0 0.1429 17.5714 0 0 0 0 5,297,142.88

cj-zj 0 0 0 0 0 0 -0.1429 -17.5714 0 0 0 0

2. Iterations for simplex method

Cj

Basic

Variables

8.8

X1

12.4

X2 0 artfcl 1

0

surplus

1 0 artfcl 2

0

surplus

2 0 slack 3 0 slack 4 0 artfcl 5

0

surplus 5

0 artfcl

6

0

surplus

6 Quantity

Iteration 1

0 artfcl 1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000

0 artfcl 2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000

0 slack 3 0.1 0.7 0 0 0 0 1 0 0 0 0 0 180,000

0 slack 4 0.5 0.7 0 0 0 0 0 1 0 0 0 0 300,000

0 artfcl 5 1 0 0 0 0 0 0 0 1 -1 0 0 0

0 artfcl 6 0 1 0 0 0 0 0 0 0 0 1 -1 0

zj 6.8 10.4 0 1 0 1 0 0 0 1 0 1 300,000

cj-zj 2 2 0 -1 0 -1 0 0 0 -1 0 -1

Iteration 2

0 artfcl 1 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 artfcl 2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000

0 slack 3 0 0.7 0 0 0 0 1 0 -0.1 0.1 0 0 180,000

0 slack 4 0 0.7 0 0 0 0 0 1 -0.5 0.5 0 0 300,000

8.8 X1 1 0 0 0 0 0 0 0 1 -1 0 0 0

0 artfcl 6 0 1 0 0 0 0 0 0 0 0 1 -1 0

zj 8.8 10.4 0 1 0 1 0 0 2 -1 0 1 300,000

cj-zj 0 2 0 -1 0 -1 0 0 -2 1 0 -1

Iteration 3

0 artfcl 1 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 artfcl 2 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000

0 slack 3 0 0 0 0 0 0 1 0 -0.1 0.1 -0.7 0.7 180,000

0 slack 4 0 0 0 0 0 0 0 1 -0.5 0.5 -0.7 0.7 300,000

8.8 X1 1 0 0 0 0 0 0 0 1 -1 0 0 0

12.4 X2 0 1 0 0 0 0 0 0 0 0 1 -1 0

zj 8.8 12.4 0 1 0 1 0 0 2 -1 2 -1 300,000

cj-zj 0 0 0 -1 0 -1 0 0 -2 1 -2 1

Iteration 4

0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 artfcl 2 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000

0 slack 3 0 0 -0.1 0.1 0 0 1 0 0 0 -0.7 0.7 160,000.00

0 slack 4 0 0 -0.5 0.5 0 0 0 1 0 0 -0.7 0.7 200,000

8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000

12.4 X2 0 1 0 0 0 0 0 0 0 0 1 -1 0

zj 8.8 12.4 1 0 0 1 0 0 1 0 2 -1 100,000

cj-zj 0 0 -1 0 0 -1 0 0 -1 0 -2 1

Iteration 5

0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 surplus 6 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000

0 slack 3 0 0 -0.1 0.1 -0.7 0.7 1 0 0 0 0 0 90,000.00

0 slack 4 0 0 -0.5 0.5 -0.7 0.7 0 1 0 0 0 0 130,000.00

8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000

12.4 X2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000

zj 8.8 12.4 1 0 1 0 0 0 1 0 1 0 0

cj-zj 0 0 -1 0 -1 0 0 0 -1 0 -1 0

Iteration 6

0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 surplus 6 0 0 0 0 1 -1 0 0 0 0 -1 1 100,000

0 slack 3 0 0 -0.1 0.1 -0.7 0.7 1 0 0 0 0 0 90,000.00

0 slack 4 0 0 -0.5 0.5 -0.7 0.7 0 1 0 0 0 0 130,000.00

8.8 X1 1 0 1 -1 0 0 0 0 0 0 0 0 200,000

12.4 X2 0 1 0 0 1 -1 0 0 0 0 0 0 100,000

zj 8.8 12.4 8.8 -8.8 12.4 -12.4 0 0 0 0 0 0 3,000,000

cj-zj 0 0 -8.8 8.8 -12.4 12.4 0 0 0 0 0 0

Iteration 7

0 surplus 5 0 0 1 -1 0 0 0 0 -1 1 0 0 200,000

0 surplus 6 0 0 -0.1429 0.1429 0 0 1.4286 0 0 0 -1 1 228,571.43

0 surplus 2 0 0 -0.1429 0.1429 -1 1 1.4286 0 0 0 0 0 128,571.43