W6_IBRAHEEM_PRICING OPTIONS

1.0 Problem definition:  Pricing options

A gas company intends to vary price with respect to demand to boost revenue. It is expected that a reduction in price will improve customer patronage. To achieve this, equation was established and feasible alternatives developed.  Demand, price and break-even inter-relationship will be developed by varying parameters on a cost-revenue and volume (demand) equation to identify the preferred path.

2.0 Development of feasible alternative:

Two feasible alternatives was singled-out from numerous alternatives:

1. Demand as a function of price using relationship p= NGN97-0.01 (D)
2. Fixed selling Price as independent of Demand

Where –slope =-0.01, intercept=97 and D = demand on alternative 1

Intercept= 97 on alternative 2. P equals selling price on both instances

The two scenarios will be analysed and the best alternative selected based on profit margin within maximum and minimum limits. 

3.0 Develop the outcome for each alternative:

The two scenarios will be subjected to numerical analysis using solution by hand and spreadsheet method

Alternative 1

From standard relationship Profit= Total Revenue-Total Cost

TR= p*D, p=a-b*d, TC=CT + CF

From first derivatives Δprofit/Δdemand, D* = a-cv/2b-  Equation (1)

Substituting relevant values into equation

D*          = (N97- N3)/2*0.01

                =4700 Liters/day

For profit to occur, a-c_v>0, which equals 94 in this case.

Ensuring maximum profit, 2nd derivatives equals -0.02

From standard relation above

Profit     = (aD-bD²)-(CF+c_v*D)                                                     Equation (2)

                = [97*(4700) - 0.01*(4700²)] - [200000+3*(4700)]

                =N235, 000- 214100

                =N20, 900

To break-even, total revenue=total cost

Equating equation (2) to zero:

-0.01D²+ (97-3) D-200,000

D’ = [-94±SQRT (94²-4(0.01*200000)]/2*0.01

D’₁ = [-94+SQRT (94²-4(0.01*200000)]/2*0.01

D’₂ = [-94-SQRT (94²-4(0.01*200000)]/2*0.01

Solving equation above

D’₁ = 3, 254.5 liters/day

D’₂ = 6, 145 liters/day

Spreadsheet analysis produces the result below

Produced graph is shown below

Alternative 2

Total Revenue =total cost at breakeven

Using the rel. D’₁                     = CF/ (p-C_V)

 = 200000/ (97-3)

=2,128 Liters per day

Spreadsheet analysis for alternative 2 produces the result below

Produced graph is shown below

4.0 Selection of acceptable Criteria:

Acceptable criteria will be based on justification on which of the two alternatives brings appreciable return within the breakeven points and profit margin.

5.0 Analysis and Comparison of alternatives:

From spreadsheet and hand calculation above on alternative 1, it is evident that the sales volume must be within 3600 to 6400 litres on the high and lower limits. This confirms the optimum volume falls within this sales margin.

Alternative 2 indicates a break even from sales volume of 2400 litres  And the continual increment of profit from this point is appreciable.

6.0 Selection of preferred alternative:

The most preferred alternative is alternative 2.  Alternative 1 exhibits a narrow acceptable range (profit margin). This makes substantial profit difficult to predict since loss occurs at increased demand.

7.0 Performance Monitoring & Post Evaluation of result:

Continual monitoring measures to be put in place to evaluate effect of price-demand relationship according to fluctuation in variable cost.

References:

Sullivan, W. G., Wicks, E.M., & Koelling, C.P. (2012). Engineering Economy (15^th ed.) (pp 42-72) New Jersey, NJ. Pearson Higher Education, Inc.

 Purdue OWL APA style. (2011). APA formatting and style guide. Retrieved from http://owl.english.purdue.edu/owl/resource/560/19/

Break Even Tool (2007): Break-even Analysis Tool. Retrieved from http://hbsp.harvard.edu/multimedia/flashtools/breakeven/index.html