The document describes a linear programming problem faced by a consumer products company. The company produces two sanitary napkin products, Product A and Product B, and must decide how many of each to produce to maximize profit. The objective is to maximize total profit subject to constraints on minimum production requirements, machine hours, and packaging hours. Solving the linear programming formulation reveals the optimal solution is to produce 300,000 units of Product A and 214,285 units of Product B for a maximum profit of $5,297,143.

1. 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

2. 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 (Bandaid, 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

3. 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)
Machine time per pack (in
minutes) – refer to working
note 1
200,000
100,000
300,000
0.1
0.7
180,000
Packaging time per pack (in
minutes)
0.5
(per month)
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)

4. 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.
Maximize
Minimum Requirement X1
Minimum Requirement X2
Machine Hours Constraint
Packaging Hours Constraint
Constraint 5
Constraint 6
Solution->
X1
8.8
1
0
0.1
0.5
1
0
300000
X2
12.4
0
1
0.7
0.7
0
1
214285.7
RHS
>=
>=
<=
<=
>=
>=
Dual
200000
0
100000
0
180000
0.1428564
300000
17.57143
0
0
0
0
$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.

5. 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

6. Iteration 8
0
0
0
Basic
8.8 12.4
surplus
surplus
0
0 artfcl surplus
Cj
Variables X1
X2
0 artfcl 1 1
0 artfcl 2 2
0 slack 3 0 slack 4 0 artfcl 5 surplus 5 6
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
(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

7. 2. Iterations for simplex method
0
0
0
Basic
8.8 12.4
surplus
surplus
0
0 artfcl surplus
Cj
Variables X1
X2
0 artfcl 1 1
0 artfcl 2 2
0 slack 3 0 slack 4 0 artfcl 5 surplus 5 6
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