This document provides solutions for BUSN7051 Assignment at Australian National University. It includes the formulation of the problem, objective function, constraints, and graphical representation. It also covers a second problem with its optimal solution and sensitivity analysis. References are provided at the end.
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BUSN7051 ASSIGNMENT1 Australian National University BUSN7051 ASSIGNMENT Student Name
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BUSN7051 ASSIGNMENT2 2.Formulation of the problem that will aid in deciding the quantity of each product to make in the present week. (a)Determining variables that are not known. Quantity of X units produced in the present week week=x Quantity of Y units produced in the current week=y (b)Formulating the objective function for the program The organizations aim is to maximize a combine sum of units’ production for product X and those of Y. The objective function Z is thus: To maximize the number of units left in stock at the end of the week (x+30-75)+(y+90-95)=x+y-50 (c)The constraint of the Linear programming problem are: 50x + 24y≤2400 30x + 33y≤2100 x≥(75 – 30) x≥45 The amount of X produced≥demand (75) - initial stock quantity is 30, which ensures we meet demand y≥95 - 90 y≥5, The amount units of Y produced≥demand (95) – initial value of stock is (90), This will ensure the company demand is met. (d)We utilize the Iso-profit profit line method to solve the linear program developed above.To maximize our objective function, we draw the Iso profit line which will enable us find the maximum value of the objective function. Variables x45 y6.25 Objective Maximize1.25 Constraintsinequality 12400<=2400 21556.25<=2100 345>=45 46.25>=5 The point that maximizes x+y-50 is (45,6.25), The solution to the linear program is 45+6.25-50=1.25.
BUSN7051 ASSIGNMENT3 The graphical representation from desmos graphs is as below 3.(a). Minimize P=A+2B+3C+4D Subject to B+2C+2D≥100 2A+D≥30 A,B,C,D≥0
BUSN7051 ASSIGNMENT4 Variables A15 B0 C50 D0 Objective Minimize165 Constraints 1100>=100 230>=30 315>=0 40>=0 550>=0 60>=0 (b)The optimal solution of the Linear programming problem is A=15,B=0, C=50, D=0. (c).The minimum cost is 165. The Sensitivity analysis report Microsoft Excel 12.0 Sensitivity Report Worksheet: [LPP.xlsx]Sheet3 Report Created: 4/1/2019 10:09:01 AM Adjustable Cells FinalReduced CellNameValueGradient $B$2A150 $B$3B00 $B$4C500 $B$5D00 Constraints FinalLagrange CellNameValueMultiplier $B$111001.5 $B$12300.5 $B$13150 $B$1400.500000328 $B$15500 $B$1600.5 (d). If the value of the minimal requirement of the first constraint increased from 100 to 250 and the problem is run in solver, the output is shown below.
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BUSN7051 ASSIGNMENT5 Microsoft Excel 12.0 Answer Report Worksheet: [LPP.xlsx]Sheet5 Report Created: 4/1/2019 10:14:37 AM Target Cell (Min) CellNameOriginal ValueFinal Value $B$8Minimize390390 Adjustable Cells CellNameOriginal ValueFinal Value $B$2A1515 $B$3B00 $B$4C125125 $B$5D00 Constraints CellNameCell ValueFormulaStatusSlack $B$11250$B$11>=$D$11Binding0 $B$1230$B$12>=$D$12Binding0 $B$1315$B$13>=$D$13Not Binding15 $B$140$B$14>=$D$14Binding0 $B$15125$B$15>=$D$15Not Binding125 $B$160$B$16>=$D$16Binding0 The new cost obtained is 390, the solutions for the problem are A=15,B=0,D=0 and C=125. The sensitivity report is shown below. Microsoft Excel 12.0 Sensitivity Report Worksheet: [LPP.xlsx]Sheet5 Report Created: 4/1/2019 10:12:34 AM Adjustable Cells FinalReduced CellNameValueGradient $B$2A150 $B$3B00 $B$4C1250 $B$5D00 Constraints FinalLagrange CellNameValueMultiplier $B$112501.5 $B$12300.5 $B$13150 $B$1400.49999997 $B$151250 $B$1600.49999994
BUSN7051 ASSIGNMENT6 References Heizer, J. and Render, B. (2013).Operations Management, Global Edition. Edinburgh: Pearson Education. Hill, A. and Hill, T. (2012).Operations management. Houndmills, Basingstoke, Hampshire: Palgrave Macmillan. Stevenson, W. (2012).Operations management. New York, N.Y.: McGraw-Hill/Irwin.