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School
York University
Department
Administrative Studies
Course
ADMS 3330
Professor
hassanq
Semester
Summer

Description
ZAD MASRI Assignment #1 20961521 Question 1 A) B=Number of ounces of Beef tips G=Number of ounces of gravy P=Number of ounces of Peas C=Number of ounces of carrots D=Number of ounces of Dinner Roll B) 1) 54B+20G+15P+8C+40D>300 2) 54B+20G+15P+8C+40D<350 3) 2.8B+9G-4.5P-2.4C-2D<0 4) 15P+90C>500 5) 1G+3P+1C>15 6) 8B+1P+1C+1D>30 7) 1B>3 8) 1C<5 9) -0.5B+1G>0 LINEAR PROGRAMMING PROBLEM MIN 0.4B+0.375G+0.25P+0.2C+0.38D S.T. 1) 54B+20G+15P+8C+40D>300 2) 54B+20G+15P+8C+40D<350 3) 2.8B+9G-4.5P-2.4C-2D<0 4) 15P+90C>500 5) 1G+3P+1C>15 6) 8B+1P+1C+1D>30 7) 1B>3 8) 1C<5 9) -0.5B+1G>0 OPTIMAL SOLUTION Objective Function Value = 3.761 Page 1 of 11 ZAD MASRI Assignment #1 20961521 Variable Value Reduced Costs --------------------------------------------- B 3.281 0.000 G 1.641 0.000 P 3.333 0.000 C 5.000 0.000 D 0.000 0.013 Constraint Slack/Surplus Dual Prices --------------------------------------------- 1 0.000 -0.009 2 50.000 0.000 3 3.047 0.000 4 0.000 -0.007 5 1.641 0.000 6 4.583 0.000 7 0.281 0.000 8 0.000 0.547 9 0.000 -0.191 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit ------------ -------------------------- --------------- B -0.188 0.400 0.420 G 0.148 0.375 0.416 P 0.159 0.250 No Upper Limit C No Lower Limit 0.200 0.747 D 0.367 0.380 No Upper Limit RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit ------------ -------------------------- --------------- 1 282.000 300.000 326.712 2 300.000 350.000 No Upper Limit 3 -3.047 0.000 No Upper Limit 4 492.642 500.000 518.000 5 No Lower Limit 15.000 16.641 6 No Lower Limit 30.000 34.583 7 No Lower Limit 3.000 3.281 8 4.780 5.000 5.090 9 -1.944 0.000 0.453 Page 2 of 11 ZAD MASRI Assignment #1 20961521 C) The dinner should contain 3.281 ounces of beef tip, 1.641 ounces of gravy, 3.333 ounces of peas, 5 ounces of carrots, and no dinner roll The optimal cost of the dinner 3.761 D) NUMBER 2, 3 and 8 will involve slack variables: 1)54B+20G+15P+8C+40D>300 2) 54B+20G+15P+8C+40D<350 3) 2.8B+9G-4.5P-2.4C-2D<0 4) 15P+90C>500 5) 1G+3P+1C>15 6) 8B+1P+1C+1D>30 7) 1B>3 8) 1C<5 9) -0.5B+1G>0 E)binding constraint has 0 slack/surplus at optimal solution. Constraints 1,4,8 and 9. Constraint Slack/Surplus Dual Prices --------------------------------------------- 1 0.000 -0.009 2 50.000 0.000 3 3.047 0.000 4 0.000 -0.007 5 1.641 0.000 6 4.583 0.000 7 0.281 0.000 8 0.000 0.547 9 0.000 -0.191 F) If costs of peas go down 5 cents, the new cost will b0.25-0.05=0.20 is within the optimality range for peas the optimal solution will still be optimal. OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit ------------ -------------------------- --------------- B -0.188 0.400 0.420 G 0.148 0.375 0.416 P 0.159 0.250 No Upper Limit C No Lower Limit 0.200 0.747 D 0.367 0.380 No Upper Limit Page 3 of 11 ZAD MASRI Assignment #1 20961521 g) Since 515 is within the feasibility ranges for constraint 4 (vitamin A), dual price will still be applicable. Since the dual price is negative the cost will increase by 0.007(515-500) =0.105. The dinner cost will increase by $0.105 RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit ------------ -------------------------- --------------- 1 282.000 300.000 326.712 2 300.000 350.000 No Upper Limit 3 -3.047 0.000 No Upper Limit 4 492.642 500.000 518.000 Constraint Slack/Surplus Dual Prices ---------------------------------------------- 1 0.000 -0.009 2 50.000 0.000 3 3.047 0.000 4 0.000 -0.007 h) The current cost/ounce of dinner roll should change. It should be reduced by 0.103 the value of its reduced cost for NY-11 to possibly include some dinner Variable Value Reduced Costs ---------------------------------------------- B 3.281 0.000 G 1.641 0.000 P 3.333 0.000 C 5.000 0.000 D 0.000 0.013 Question 2) A) Variables Let XAR=Number of pounds of component A in Regular. XBR=Number of pounds of component B in Regular. XAS=Number of pounds of component A in Super. XBS=Number of pounds of component B in Super. XAU=Number of pounds of component A in Ultra. XBU=Number of pounds of component B in Ultra. B) LP MODEL MIN 0.3 (XAR+XAS+XAU) + 0.5 (XBR+XBS+XBU) s.t 0.14XAR+0.2XBR > 0.1 (XAR+XBR) 0.14XAS+0.2XBS > 0.12 (XAS+XBS) Page 4 of 11 ZAD MASRI Assignment #1 20961521 0.14XAU+0.2XBU > 0.15 (XAU+XBU) 0.18XAR+0.24XBR > 0.16(XAR+XBR) 0.18XAS+0.24XBS > 0.2(XAS+XBS) 0.18XAU+0.24XBU > 0.18(XAU+XBU) XAR+XBR=1000 XAS+XBS=2000 XAU+XBU=2000 XAR, XBR, XAS, XBS, XAU, XBU >0 Question 3) A) 1200 RS 3 5 1 WF 2 SA 2500 1500 SS 4 5 4 2 5 2 6 NS 9 MV 2000 2000 OF 3 B) Let XXYY number of shipment from XX to location YY RS means riverside SS means Sunny Slope OF means old farm WF means Waterford NS means Northside SA means Sanderson MV means Millville MIN 1RSSS+5RSWF+3RSSA+4SSWF+5SSNS+6OFWF+3OFNS+2WFNS+2WFSA+4WFMV+2NSWF+5N SSA+9NSM V+2SAMV S.T. Page 5 of 11 ZAD MASRI Assignment #1 20961521 1) 1RSSS+1RSWF+1RSSA<1200 2) 1SSWF+1SSNS<1500 3) 1OFWF+1OFNS<2000 4) 1RSWF+1SSWF+1OFWF-1WFNS-1WFSA-1WFMV+1NSWF=0 5) 1OFNS+1WFNS-1NSWF-1NSSA-1NSMV=0 6) 1RSSA+1WFSA+1NSSA-1SAMV+1MVSA=2500 7) 1WFMV+1NSMV+1SAMV-1MVSA=2000 C) Ship 1200 From Riverside to Sanderson to (Continue to all ) The optimal cost of shipping is $29,200 C) We add a DU as the supple is only 4700, and the demand is 5500. DU is 800 and is a Dummy 1200 RS 3 5 1 2 SA 2500 1500 SS 4 WF 4 5 2 5 2 6 9 MV 3000 3 NS 2000 OF DU 8
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