Question

A power company has a maximum daily capacity of 5 million kilowatt-hours (kWh) of electric power available. The daily demand for power from customers is the total of the peak hours demand and off-peak hours demand. The following functions estimate the high-demand (peak) hours and low-demand (off-peak) hours (unit in millions of kWh):

High demand : 10 – 0.08ph + 0.007pl

Low demand : 9 – 0.18pl + 0.004ph , where variable pl represents the price per kilowatt-hour during low-demand hours, and ph is the price per kilowatt-hour during high-demand (peak) hours. Formulate and solve a nonlinear programming model to determine the price structure (per kWh) that will maximize revenue.

(I have to make one word file which clearly states the objective function, decision variables, constraints, and solution of all questions 2. one workbook which contains the spreadsheet file with formula and answer report for all questions (using separate worksheets for different questions) here is photo of question.

20Assignment 2019-20B%20(2).pdf A power company has a maximum daily capacity of 5 million kilowatt-hours (kWh) of electric power available. The daily demand for power from customers is the total of the peak hours demand and off-peak hours demand. The following functions estimate the high-demand (peak) hours and low-demand (off-peak) hours (unit in millions of kWh): High demand : 10 - 0.08p. + 0.007p, Low demand :9-0.18p, + 0.004pn , where variable p, represents the price per kilowatt-hour during low-demand hours, and pn is the price per kilowatt-hour during high-demand (peak) hours. Formulate and solve a nonlinear programming model to determine the price structure (per kWh) that will maximize revenue. e Date: 5pm, 15 March 2020 (Sunday) ading is based on the total score of 100. Late penalty: deduct 10% of the score.

Answer 1

First, we input the functions, total demand and total revenue as below :

5,000,000 Maximum daily capacity 2 PL 3PH 4 High Demand Hours 5 Low Demand Hours 6 Total Demand Hours 7 Total Revenue 10,000,000 9,000,000 19,000,000

5000000 1 Maximum daily capacity 2 PL 3 PH 4 High Demand Hours 5 Low Demand Hours 6 Total Demand Hours 7 Total Revenue =(10-(0.08*B3)+(0.007*B2)) 1000000 =(9-(0.18*B2)+(0.004*B3))*1000000 =B4+B5 =(B2*B5)+(B3*B4)

The functions estimate the demand in millions of kWh. Hence, we multiply the given functions with 1000000.

Next, we enter the inputs into Solver as below :

с D E F G H 1 Maximum daily capacity 2 PL B 5,000,000 Solver Parameters Set Objective: SB571 3 PH 4 High Demand Hours 5 Low Demand Hours 6 Total Demand Hours 7 Total Revenue 10,000,000 9,000,000 19,000,000 TO: O Max Min value of: 0 By Changing Variable Cells: $B$2:$B$3 Subject to the Constraints: $B$6 <= $B$1 Add Change Delete Reset All Load/Save Make Unconstrained Variables Non-Negative Select a Solving GRG Nonlinear Method: Options

Solving the model, we get the following output. This is the price structure to maximize revenue.

5,000,000 44.67 1 Maximum daily capacity 2 PL 3 PH 4 High Demand Hours 5 Low Demand Hours 6 Total Demand Hours 7 Total Revenue 82.52 3,710,843 1,289,157 5,000,000 363,819,877 $

в 1 Maximum daily capacity 5000000 2 P 44.6718684772355 3 PH 82.5232467557665 4 High Demand Hours = (10-10.08*B3)+(0.007*B2))*1000000 5 Low Demand Hours =(9-(0.18*B2)+(0.004*B3))*1000000 6 Total Demand Hours =B4+B5 7 Total Revenue = (B2*B5)+(B3*B4)