Page 92, Problem 6
By: Eric Malone
a)Summarize the problem in a table format.
Sale $ / lb. / Nitrogen / SiliconFertilizer 1 (F1) / $70 / N1 40% / S1
Fertilizer 2 (F2) / $40 / N2 / S2 70%
Max. Available / 80 lb. / 100 lb.
Cost for Resource / $15 / lb. / $10 / lb.
b)Formulate the problem. Explain your decision variables and details of your formulation.
Decision Variables: In this problem, the goal is to produce an optimal number of pounds of each type of fertilizer in order to maximize profits. Based on the statement of the problem, there are only two types of fertilizer to choose from: F1 and F2. Each fertilizer has a unique minimum requirement with regard to the amounts of nitrogen and silicon required for each. Therefore the decision is to choose an appropriate combination of these types of fertilizers and their respective chemicals in order to achieve the desired goal. The decision variables are defined as:
F1: lbs. of fertilizer #1 to produce
F2: lbs. of fertilizer #2 to produce
N1: lbs. of nitrogen to use in F1
N2: lbs. of nitrogen to use in F2
S1: lbs. of silicon to use in F1
S2: lbs. of silicon n to use in F2
Objective Function: As stated above, the overall goal of this situation is to maximize profits, while satisfying the composition requirements for each type of fertilizer. With the costs for each type of resource (nitrogen or silicon) and the sale price for each type of fertilizer given above, it is easy to develop a function that describes this objective mathematically:
Objective: maximize profit (revenue – purchase cost)
max z = 70F1 + 40F2 – 15N1 – 15N2 – 10S1 – 10S2
Constraints: From the problem statement, we saw that each type of fertilizer requires a minimum proportion of each resource. Also, we know that there limit on the amount of each type of resource that can be purchased. Using this information, we formulate the constraints as follows:
1) Max. Nitrogen Availability:N1 + N2 80
2) Max. Silicon Availability:S1 + S2 100
3) Fertilizer 1 composed only of N1 & S1:N1 + S1 – F1 = 0
4) Fertilizer 2 composed only of N2 & S2:N2 + S2 – F2 = 0
5) Fertilizer 1 40% Nitrogen:
6) Fertilizer 2 70% Silicon:
Sign Constraint: It would not make sense to purchase a negative amount of either type of resource in this problem, as it would to produce a negative amount of product so we apply the following sign constraint:
F1, F2, N1, N2, S1, S2 0
c)Create the initial simplex tableau for this problem. Do not solve manually.
In order to set up the simplex tableau, we must first rewrite the problem formulation in standard LP format
- 70F1 - 40F2 + 15N1 + 15N2 + 10S1 + 10S2 = 0
N1 + N2 + s1 = 80
S1 + S2 + s2 = 100
N1 + S1 – F1 + a3 = 0
N2 + S2 – F2 + a4 = 0
INITIAL SIMPLEX TABLEAU, P. 92 #6
z / F1 / F2 / N1 / N2 / S1 / S2 / s1 / s2 / a3 / a4 / a5 / a6 / e5 / e6 / RHS / BV1 / -70 / -40 / 15 / 15 / 10 / 10 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / z
0 / 0 / 0 / 1 / 1 / 0 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / 0 / 0 / 80 / s1
0 / 0 / 0 / 0 / 0 / 1 / 1 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / 0 / 100 / s2
0 / -1 / 0 / 1 / 0 / 1 / 0 / 0 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / 0 / a3
0 / 0 / -1 / 0 / 1 / 0 / 1 / 0 / 0 / 0 / 1 / 0 / 0 / 0 / 0 / 0 / a4
0 / 0 / 0 / 0.6 / 0 / -0.4 / 0 / 0 / 0 / 0 / 0 / 1 / 0 / -1 / 0 / 0 / a5
0 / 0 / 0 / 0 / -0.7 / 0 / 0.3 / 0 / 0 / 0 / 0 / 0 / 1 / 0 / -1 / 0 / a6
d)Use WinQSB software to solve this problem. Print the input data as spreadsheet format. Print the solution as summary reports for the variables and the constraints.
The following data tables represent the input and output obtained through the use of WinQSB.
LP Input for P.91, #6Variable --> / F1 / F2 / N1 / N2 / S1 / S2 / Direction / R. H. S.
Maximize / 70 / 40 / -15 / -15 / -10 / -10
C1 / 0 / 0 / 1 / 1 / 0 / 0 / <= / 80
C2 / 0 / 0 / 0 / 0 / 1 / 1 / <= / 100
C3 / -1 / 0 / 1 / 0 / 1 / 0 / = / 0
C4 / 0 / -1 / 0 / 1 / 0 / 1 / = / 0
C5 / 0 / 0 / 0.6 / 0 / -0.4 / 0 / >= / 0
C6 / 0 / 0 / 0 / -0.7 / 0 / 0.3 / >= / 0
LowerBound / 0 / 0 / 0 / 0 / 0 / 0
UpperBound / M / M / M / M / M / M
VariableType / Continuous / Continuous / Continuous / Continuous / Continuous / Continuous
Combined Report for P.91, #6
Decision / Solution / Unit Cost or / Total / Reduced / Basis / Allowable / Allowable
Variable / Value / Profit c(j) / Contribution / Cost / Status / Min. c(j) / Max. c(j)
1 / F1 / 180 / 70 / 12,600.00 / 0 / basic / 40 / M
2 / F2 / 0 / 40 / 0 / -30 / at bound / -M / 70
3 / N1 / 80 / -15 / -1,200.00 / 0 / basic / -45 / -15
4 / N2 / 0 / -15 / 0 / 0 / basic / -15 / 15
5 / S1 / 100 / -10 / -1,000.00 / 0 / basic / -10 / M
6 / S2 / 0 / -10 / 0 / 0 / at bound / -M / -10
Objective / Function / (Max.) = / 10,400.00 / (Note: / Alternate / Solution / Exists!!)
Left Hand / Right Hand / Slack / Shadow / Allowable / Allowable
Constraint / Side / Direction / Side / or Surplus / Price / Min. RHS / Max. RHS
1 / C1 / 80 / <= / 80 / 0 / 55 / 66.6667 / M
2 / C2 / 100 / <= / 100 / 0 / 60 / 0 / 120
3 / C3 / 0 / = / 0 / 0 / -70 / -M / 180
4 / C4 / 0 / = / 0 / 0 / -70 / 0 / 0
5 / C5 / 8 / >= / 0 / 8 / 0 / -M / 8
6 / C6 / 0 / >= / 0 / 0 / 0 / -M / 0
e)Write a report of your solution for a hypothetical manager. Use a language understandable to most people. Do not use mathematical abbreviations.
According to the table above, it is evident that the company should choose to produce 180 pounds of Fertilizer 1. This will require the purchase of 80 pounds of nitrogen and 100 pounds of silicon. It is not recommended that the company produce any amount of Fertilizer 2. This is because for every unit of Fertilizer 2 produced, the company’s profit will be reduced by $30. By adhering to these recommendations, the company’s total profits will be $10,400.
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