Case Study: Frontier Concrete Company

Case Study: Frontier Concrete Company
25th of November 2015
Microeconomic Theory
Mr. Larry Stockman
Oswaldo Noriega

Table of Contents

1. Executive Summary……………………………………2
2. Plant Analysis & Recommendations ………………..3-4
3. Appendix……………………………………………….5-8

3.1. Cost Analysis Data………………………………5-7
3.2. Break Even Analysis………………………………7
3.3. Regression Analysis…….…………………………8

1. Executive Summary

As we know, concrete is a product that is used to create things such as buildings, streets, walls, and so on. Since it is an everyday-product, it is constantly replaced over periods of time because new things are coming and the old ones have to be removed. Opening a new plant in another city is the main decision that we have to take as a company depending on our sales.
The information that has been gathered using the twelve plant models and the internal production, cost and external demand data, will lead us to the conclusion if it will be profitable to open a new plant or not.
There were twelve simulated plants in total and production outputs of 30, 45 & 60 yards per hour, each output having four plants with different combinations of Capital (K) and Labor (L) to produce said output. The cost of one unit of labor is $7 (in the future by $8.50), and for one unit of Capital is $18000 a year (a working year of 2080 hours or 260 days of 8 hours per day, meaning $8.65 an hour). Other Variable Costs for a cubic yard of concrete include:
Gravel $3.34
Cement $7.50
Average Delivery Costs $4.00

2. Plant Analysis & Recommendations
The most cost efficient plants (appendix one) are plant two (Q=30), plant six (Q=45) and plant ten (Q=60). These outcomes were indifferent between the original wage rate of $7 and the potentially higher wage rate of $8.50. The average cost for all the foregone plants is $16.17 (under a $7.00 wage rate) and $16.27 (under $8.50 wage rate). Marginal cost is the same for all 3 plants.
Using regression analysis (appendix two) we are able to determine the Demand function and therefore Marginal Revenue. Upon this we are able to find the profit maximizing quantity for Frontier. The demand and marginal revenue functions are as follows (see appendix three):
Demand Function= Q= 245.83 -9.41P
Revenue= P*Q= Q^2/9.41 +26.1243Q
Marginal Revenue= -0.21254Q + 26.1243
The marginal cost for Frontier is the same for all three of the most efficient plants, which is equal to the average cost of $16.17 (when L=$7.00) and $16.28 (when L=$8.50). When setting Marginal Revenue equal to marginal cost we get a profit maximizing quantity of 46 Units (yards per hour). By plugging this quantity into the price equation we get a total price of $21.34. Since Plant Six (Q=45) this would be the best option to maximize profits. Assuming Frontier is in operation 8 hours a day, 5-days a week for 52 weeks of the year we can work out total revenue and therefore the revenue with association of our total costs.
Total Revenue = $1,997,424
Total Costs= $1,513,704
Profit= $483,720
A simulation of Plant Ten was also requested by management who forecast fixed costs to $96,000 of the total capital cost. By working out variable costs, we are able to produce a break-even analysis for the certain prices levels requested by management ($19.00, $20.00 and $21.00). Simply, we can work out the required units that Frontier needs to sell in order to cover its costs (FC/(P-VC))….. see appendix one. After performing the break-even analysis we have concluded the following:

• At $19.00, 26,713 units (L=$7)
• At $20.00, 20,898 units (L=$7)
• At $21.00, 17,162 units (L=$7)

However when we use these price levels in the demand function we see some issues. According to the derived demand curve of Q= 245.83 -9.41P we see that the market is insufficient for prices of $20 and $21, where quantity demanded is 58 Units and 48 Units respectively. This value of quantity demanded means the Frontier will be operating at less than capacity.
After this analysis, including analysis of production costs, the demand function and break-even analysis, I have come to the conclusion that it is best to progress with the opening of plant Six. This is because at this production level of 45 Units and a price level of $21.34 (profit maximization), there will be enough market demand to cover the production level and not have any unused capacity, like seen with Plant Ten. Frontier would also be making $483,720 in profit which provides enough incentive to open plant 6, considering this will be a 24.2% return on sales.

3-Appendix
3.1: Appendix One: Cost Analysis Data
1) Cost of Capital (K)= 18000/ (8*260) = $8.65
Other Costs= 3.34 + 7.50 +4= $14.84
The following cost information is under the assumption that labor will cost $7 per unit and Capital $8.65 per cubic yard of concrete. As well as other costs including: Gravel ($3.34), Cement ($7.50) and Average Daily Costs ($4.00), all per cubic yard of concrete also.

2) Cost of Capital (K)= 18000/ (8*260) = $8.65
Other Costs= 3.34 + 7.50 +4= $14.84
The following cost information is under the assumption that labor will cost $8.50 per unit and Capital $8.65 per cubic yard of concrete. As well as other costs including: Gravel ($3.34), Cement ($7.50) and Average Daily Costs ($4.00), all per cubic yard of concrete also

3.2: Appendix Two: Break-Even Analysis
Break-Even = Fixed Costs/ (Selling Price – Variable Cost)
Fixed Costs =$96,000
Variable Costs= Other costs + Variable capital + Labor ($7.00)
Variable Costs= 14.84 + $0.47 +$0.20
Variable Costs= $15.41

3.3: Appendix Three: Regression Analysis