


1
Winston Chapter 3.8, Page 91, Number 6 (Linear Programming)
Problem Statement: Bullco blends silicon and nitrogen to produce two types of fertilizers.
Fertilizer 1 must be at least 40% nitrogen and sells for $70/lb. Fertilizer 2 must be at
least 70% silicon and sells for $40/lb. Bullco can purchase up to 80 lbs. of nitrogen at
$15/lb and up to 100 lbs. of silicon at $10/lb. Assuming that all fertilizer produced can
be sold, formulate an LP to help Bullco maximize profits.
A. Summarize the problem in a table format.
B. Formulate the problem. Explain your decision variables and details of your
formulation.
C. Create the first simplex tableau for this problem. Do not solve manually.
D. Use Quant software to solve this problem. Print the input data as free format. Print
the solution as summary reports for the variables and the constraints.
E. Write a report of your solution for a hypothetical manager. Use a language
understandable to most people. Do not use mathematical abbreviations.
A. Problem Summarized in Table Format:
Fertilizer (lbs.) Price Per Pound N, Nitrogen (lbs.) Si, Silicon (lbs.)
F1
$70
Si1
N1 ( 40%)
F2
$40
N2
Si2 ( 70%)
Max 80 lbs.
Max 100 lbs.
Available 
$15
$10
Cost Per lb. 
Demand
No Limit
No Limit
B. Problem Formulation:
1. Decision Variables:
 F1 = Fertilizer 1, Pounds, Positive
 F2 = Fertilizer 2, Pounds, Positive
 N1 = Nitrogen used in Fertilizer 1, Pounds, Positive
 N2 = Nitrogen used in Fertilizer 2, Pounds, Positive
 Si1 = Silicon used in Fertilizer 1, Pounds, Positive
 Si2 = Silicon used in Fertilizer 2, Pounds, Positive
These decision variables are the unknowns that are needed to complete a formulation to
help maximize Bullco's profits.
2. Objective Function: Since profit is being maximized, the objective function is created
in the form of profit. Knowing that profit can be defined as the difference between the
cost of the ingredients and the selling price of the fertilizer, the objective function is
formulated as:


2

Max Z (Profit) = (Sum of Selling Prices) – (Sum of Ingredient Costs)
Max Z = (70F1 + 40F2) – (15N1 + 15N2 +10Si1 + 10Si2)
The objective function to maximize Bullco’s profits is:
Max Z = 70F1 + 40F2 – 15N1 – 15N2 – 10Si1 – 10Si2.
3. Constraints:







0.6N1 - 0.4Si1  0, F1 must contain at least 40% nitrogen.
0.3N2 - 0.7Si2  0, F2 must contain at least 70% silicon.
N1 + N2  80, A maximum of 80 lbs. nitrogen can be purchased.
Si1 + Si2  100, A maximum of 100 lbs. silicon can be purchased.
N1 + Si1 - F1 = 0, Fertilizer 1 is made of exactly N1 + Si1.
N2 + Si2 - F2 = 0, Fertilizer 2 is made of exactly N2 + Si2.
(Sii, Ni, Fi  0)
C. Initial Simplex Tableau:
Row
0
1
2
3
4
5
6
Z
1
0
0
0
0
0
0
N1
15
0.6
0
1
0
1
0
N2
15
0
0.3
1
0
0
1
Si1 Si2
10
10
-0.4
0
0
-0.7
0
0
1
1
1
0
0
1
F1
-70
0
0
0
0
-1
0
F2
-40
0
0
0
0
0
-1
Quant Input and Output:
Quant Input:
Free Format Model for p91 n6
Max -15N1-15N2-10Si1-10Si2+ 70F1+ 40F2
Subject to
(1) .6N1-.4Si1 >= 0
(2) .3N2-.7Si2 >= 0
(3) 1N1+ 1N2 <= 80
(4) 1Si1+ 1Si2 <= 100
(5) 1N1+ 1Si1-1F1 = 0
(6) 1N2+ 1Si2-1F2 = 0
Slack1 Slack2 Art1
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
Art2
0
0
1
0
0
0
0
Art3
0
0
0
0
0
1
0
Art4
0
0
0
0
0
0
1
Ex1
0
-1
0
0
0
0
0
Ex2 RHS
0
0
0
0
-1
0
0
80
0
100
0
0
0
0



3
Quant Output:
|------------------------------------------------------------------------------|
|
Summarized Report for p91 n6
Page : 1
|
|------------------------------------------------------------------------------|
|
|
|
|Opportunity| Objective | Minimum | Maximum |
|Number | Variable | Solution |
Cost
|Coefficient|Obj. Coeff.|Obj. Coeff.|
|-------+----------+-----------+-----------+-----------+-----------+-----------|
| 1
| N1
| +80.000000| 0
| -15.000000| -15.000000| + Infinity|
| 2
| N2
| 0
| 0
| -15.000000| - Infinity| -15.000000|
| 3
| Si1
| +100.00000| 0
| -10.000000| -70.000000| -10.000000|
| 4
| Si2
| 0
| 0
| -10.000000| -10.000000| +89.999992|
| 5
| F1
| +180.00000| 0
| +70.000000| +40.000000| + Infinity|
| 6
| F2
| 0
| +30.000000| +40.000000| - Infinity| +70.000000|
|------------------------------------------------------------------------------|
|
Maximized OBJ = 10400 Iteration = 6 Elapsed CPU second = 5.078125E-02
|
|------------------------------------------------------------------------------|
|------------------------------------------------------------------------------|
|
Summarized Report for p91 n6
Page : 2
|
|------------------------------------------------------------------------------|
|
|
|
|
Shadow | Slack or | Minimum | Maximum |
|Constr.| Status |
RHS
|
Price
| Surplus |
RHS
|
RHS
|
|-------+---------+------------+-----------+-----------+-----------+-----------|
| 1
| Loose | >0
| 0
| +8.0000010| - Infinity| +8.0000010|
| 2
| Tight | >0
| 0
| 0
| 0
| 0
|
| 3
| Tight | <+80.000000| +55.000000| 0
| +66.666664| + Infinity|
| 4
| Tight | <+100.00000| +60.000000| 0
| 0
| +120.00000|
| 5
| Tight | =0
| -70.000000| 0
| - Infinity| +180.00000|
| 6
| Tight | =0
| -70.000000| 0
| 0
| +26.666668|
|------------------------------------------------------------------------------|
|
Maximized OBJ = 10400 Iteration = 6 Elapsed CPU second = 5.078125E-02
|
|------------------------------------------------------------------------------|
D. Report for a Manager:
The maximum amount of profit that can be made in this scenario is $10,400 as seen by
Quant’s maximized objective value. This maximum profit can be reached when the
optimum amounts of nitrogen and silicon are used in the manufacture of Fertilizer 1 and
2.
Quant shows that 80 pounds of nitrogen should be purchased for use in making Fertilizer
1. The cost for each pound is of nitrogen is $15. Therefore, eighty pounds of nitrogen
would cost $1,200. As well, Quant shows that 100 pounds of silicon should be purchased
for use in making Fertilizer 1. The cost for each pound of silicon is $10. Similarly, one
hundred pounds of silicon would cost $1,000. The total cost for materials would be
$2,200. Fertilizer 1, when mixed, will weigh 180 pounds.
It is not recommended to create any Fertilizer 2, as its profit margin is unfavorable
compared to Fertilizer 1.
Since 180 pounds of Fertilizer 1 should be produced and will be sold at $70 per pound,
cash receipts for this sale will equal $12,600. The profit for this scenario is $12,600 $2,200 = $10,400, the maximum amount of profit possible.