Linear Programming: Optimal Solutions & Constraints in Civil Systems

WESTERN MICHIGAN UNIVERSITY CCE-6100-100 - Civil Systems Analysis HW.3 Done by: Aws Tarawneh

3. Consider the following linear program: a. Use the graphical solution procedure to find the optimal solution Intersection 1 Intersection 2 At Intersection 1 Y=3 X=2 min=52 At Intersection 2 Y=2 X=3 min= 48 Min is at Intersection 2 Min=48

b. Assume that the objective function coefficient for X changes from 8 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. 1x+3y≥ 9 Slope = -1/3 2x+2y ≥ 10 slope =-1 -1≤ -c 1 /12 ≤ -1/3 4≤ c 1 ≤ 12 X can be between 4 and 12 without changing the optimal solution When changing X from 8 to 6 the optimal solution will not change but the min will change min= 42 c. Assume that the objective function coefficient for X remains 8, but the objective function coefficient for Y changes from 12 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. 1x+3y≥ 9 Slope = -1/3 2x+2y ≥ 10 slope =-1 -1≤ -8/C 2 ≤ -1/3 -3≤ -c 2 /8 ≤ -1 8≤ c 2 ≤ 24

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Intersection 1 Intersection 2 When changing Y from 12 to 6 the optimal solution will change According to the above figure the Min will be at Intersection 1 X=2 Y=3 min= 34 d. The computer solution for the linear program in part (a) provides the following objective coefficient range information:

How would this objective coefficient range information help you answer parts (b) and (c) prior to re-solving the problem? The objective coefficient range gives us the allowed range that either X or Y changes without any change in the optimal solution. 7. Investment Advisors, Inc., is a brokerage firm that manages stock portfolios for a number of clients. A particular portfolio consists of U shares of U.S. Oil and H shares of Huber Steel. The annual return for U.S. Oil is $3 per share and the annual return for Huber Steel is $5 per share. U.S. Oil sells for $25 per share and Huber Steel sells for $50 per share. The portfolio has $80,000 to be invested. The portfolio risk index (0.50 per share of U.S. Oil and 0.25 per share for Huber Steel) has a maximum of 700. In addition, the portfolio is limited to a maximum of 1000 shares of U.S. Oil. The linear programming formulation that will maximize the total annual return of the portfolio is as follows: a. What is the optimal solution, and what is the value of the total annual return? The optimal solution U= 800 H= 1200 Total annual return =3*800+5*1200=8400 b. Which constraints are binding? What is your interpretation of these constraints in terms of the problem? Since the constraints 1&2 have a slack/surplus equal zero then these constraints are binding. 25U+50H≤80,000 0.5U+0.25D≤700 Constraint 1 represents the funds available, and constraint 2 represents the maximum risk. These constraints are limiting factors in the portfolio optimization. c. What are the dual values for the constraints? Interpret each. The dual values for constraints 1 and 2 are 0.09333 and 1.33333, respectively. The interpretation is as follows: For constraint 1: For every additional dollar of funds available the total return will increase $0.09333 For constraint 2: For every unit increased in the maximum risk the total return will increase $1.33333

d. Would it be beneficial to increase the maximum amount invested in U.S. Oil? Why or why not? Increasing the maximum amount invested in U.S. Oil will not be beneficial because the dual price is 0 which mean this constraint is not binding and will not increase the total annual return. 8. Refer to the above, which shows the computer solution of Problem 7. a. How much would the return for U.S. Oil have to increase before it would be beneficial to increase the investment in this stock? intersection 1 intersection 2 At intersection 1 U=800 H=1200 At intersection 2 U=1000 H=800 According to the computer solution the allowable increase for the return of the U.S. Oil (U) is 7 so when the U.S. Oil (U) is more than 10 the maximum will become at intersection 2 instead of intersection 1. The binding constraints will become:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

0.5U + 0.25H ≤ 700 1U ≤ 1000 which means it will be beneficial to increase the investment in this stock. from intersection 1 & intersection 2 U.S. Oil (U) will have to be increased from 800 to 1000 to maintain the maximum profit b. How much would the return for Huber Steel have to decrease before it would be beneficial to reduce the investment in this stock? The allowable decrease for Huber Steel (H) is 3.5 when the return for Huber Steel (H) is less than 1.5 it will be beneficial to reduce the investment in this stock. from intersection 1 & intersection 2 Huber Steel (H) shares will have to be decreased from 1200 to 800 to maintain the maximum profit c. How much would the total annual return be reduced if the U.S. Oil maximum were reduced to 900 shares? According the computer solution the allowable decrease in the 3 rd constraint is 200 which means the maximum can decreased to 800 without affecting the total annual return 20. Adirondack Savings Bank (ASB) has $1 million in new funds that must be allocated to home loans, personal loans, and automobile loans. The annual rates of return for the three types of loans are 7% for home loans, 12% for personal loans, and 9% for automobile loans. The bank’s planning committee has decided that at least 40% of the new funds must be allocated to home loans. In addition, the planning committee has specified that the amount allocated to personal loans cannot exceed 60% of the amount allocated to automobile loans. a. Formulate a linear programming model that can be used to determine the amount of funds ASB should allocate to each type of loan in order to maximize the total annual return for the new funds. H= Home loans P= Personal loans A= Automobile loans Max 0.07H+0.12P+0.09A Total amount available: H+P+A=1,000,000 At least 40% allocated to home loans: H≥0.4*1,000,000 H≥400,000 personal loans cannot exceed 60% of automobile loans P≤0.6A P-0.6A ≤ 0 using Lingo

Max 0.07H+0.12P+0.09A s.t. H+P+A=1000000 H>=400000 P-0.6A <=0 A >=0 H>=0 P>=0 b. How much should be allocated to each type of loan? What is the total annual return? What is the annual percentage return? Global optimal solution found. Objective value: 88750.00 Infeasibilities: 0.000000 Total solver iterations: 0 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 12 Nonlinear nonzeros: 0 Variable Value Reduced Cost H 400000.0 0.000000 P 225000.0 0.000000 A 375000.0 0.000000 Row Slack or Surplus Dual Price 1 88750.00 1.000000 2 0.000000 0.1012500 3 0.000000 -0.3125000E-01 4 0.000000 0.1875000E-01 5 375000.0 0.000000 6 400000.0 0.000000 7 225000.0 0.000000 From Lingo program H= $400000 P= $225000

A= $375000 Total annual return = $88750 Annual percentage return = (88750.00/1000000) *100% = 8.875% c. If the interest rate on home loans increased to 9%, would the amount allocated to each type of loan change? Explain. Max 0.09H+0.12P+0.09A s.t. H+P+A=1000000 H>=400000 P-0.6A <=0 A >=0 H>=0 P>=0 Global optimal solution found. Objective value: 96750.00 Infeasibilities: 0.000000 Total solver iterations: 0 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 12 Nonlinear nonzeros: 0 Variable Value Reduced Cost H 400000.0 0.000000 P 225000.0 0.000000 A 375000.0 0.000000

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Row Slack or Surplus Dual Price 1 96750.00 1.000000 2 0.000000 0.1012500 3 0.000000 -0.1125000E-01 4 0.000000 0.1875000E-01 5 375000.0 0.000000 6 400000.0 0.000000 7 225000.0 0.000000 From Lingo program H= $400000 P= $225000 A= $375000 Total annual return = $96750 The amount allocated to each type of loan will not change. Because the optimal solution didn’t change. Ranges in which the basis is unchanged: Objective Coefficient Ranges: Current Allowable Allowable Variable Coefficient Increase Decrease H 0.7000000E-01 0.3125000E-01 INFINITY P 0.1200000 INFINITY 0.3000000E-01 A 0.9000000E-01 0.3000000E-01 0.5000000E-01 Righthand Side Ranges: Current Allowable Allowable Row RHS Increase Decrease 2 1000000. INFINITY 600000.0 3 400000.0 600000.0 400000.0 4 0.000000 600000.0 360000.0 5 0.000000 375000.0 INFINITY 6 0.000000 400000.0 INFINITY 7 0.000000 225000.0 INFINITY We can also check through the range of H from 0 to0.1125 the optimal solution will not change d.Suppose the total amount of new funds available was increased by $10,000. What effect would this have on the total annual return? Explain. Max 0.07H+0.12P+0.09A s.t. H+P+A=1010000 (new funds) H>=400000 P-0.6A <=0 A >=0 H>=0 P>=0

Lingo solution Global optimal solution found. Objective value: 89762.50 Infeasibilities: 0.000000 Total solver iterations: 0 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 12 Nonlinear nonzeros: 0 Variable Value Reduced Cost H 400000.0 0.000000 P 228750.0 0.000000 A 381250.0 0.000000 Row Slack or Surplus Dual Price 1 89762.50 1.000000 2 0.000000 0.1012500 3 0.000000 -0.3125000E-01 4 0.000000 0.1875000E-01 5 381250.0 0.000000 6 400000.0 0.000000 7 228750.0 0.000000 H= $400000 P= $228750 A= $381250 Total annual return = $89762 Total annual return increment = 89762-88750=$1,012 The total annual return because the available funds constrain is a binding constrain which mean that any change in the available funds will directly change the Total annual return Since the range of the first constrain (available funds) is from 400,000 to infinity. 2 1000000. INFINITY 600000.0 And the dual price is 0.1012500 we can also calculate the Total annual return increment by multiplying the dual price by the increment or decrement in the available funds Total annual return increment =10,000*0.1012500=$1,012.5 e. Assume that ASB has the original $1 million in new funds available and that the planning committee has agreed to relax the requirement that at least 40% of the new funds must be allocated to home loans by 1%. How much would the annual return change? How much would the annual percentage return change? Max 0.07H+0.12P+0.09A

s.t. H+P+A=1000000 H>=390,000 (39% instead of 40%) P-0.6A <=0 A >=0 H>=0 P>=0 Lingo solution Global optimal solution found. Objective value: 89062.50 Infeasibilities: 0.000000 Total solver iterations: 0 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 12 Nonlinear nonzeros: 0 Variable Value Reduced Cost H 390000.0 0.000000 P 228750.0 0.000000 A 381250.0 0.000000 Row Slack or Surplus Dual Price 1 89062.50 1.000000 2 0.000000 0.1012500 3 0.000000 -0.3125000E-01 4 0.000000 0.1875000E-01 5 381250.0 0.000000 6 390000.0 0.000000 7 228750.0 0.000000 H= $390000 P= $228750 A= $381250 Total annual return = $89062.5 Total annual return increment = 89062.5-88750=$312.5 Or using dual price home loan change=400,000-390,000=-10,000 Total annual return increment = -0.03125*-10,000=$312.5 Annual percentage return = (89062.50/1000000) *100% = 8.9062% Annual percentage change = 8.9062%-8.875%=0.0312%

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

30. The program manager for Channel 10 would like to determine the best way to allocate the time for the 11:00–11:30 evening news broadcast. Specifically, she would like to determine the number of minutes of broadcast time to devote to local news, national news, weather, and sports. Over the 30-minute broadcast, 10 minutes are set aside for advertising. The station’s broadcast policy states that at least 15% of the time available should be devoted to local news coverage; the time devoted to the combination of local news and national news must be at least 50% of the total broadcast time; the time devoted to the weather segment must be less than or equal to the time devoted to the sports segment; the time devoted to the sports segment should be no longer than the total time spent on the local and national news; and at least 20% of the time should be devoted to the weather segment. The production costs per minute are $300 for local news, $200 for national news, $100 for weather, and $100 for sports. a. Formulate and solve a linear program that can determine how the 20 available minutes should be used to minimize the total cost of producing the program. L= Local news N=National news W=Weather S=Sports Min 300L+200N+100W+100S S.t. L+N+W+S=20 (30-10 advertisement) L >=3 L+N>=10 W-S <=0 S-L-N <=0 (doesn’t affect the solution because L+N is already more than 50% of the show) W>=4 L,N,W,S>=0 Lingo solution Global optimal solution found. Objective value: 3300.000 Infeasibilities: 0.000000 Total solver iterations: 2 Model Class: LP Total variables: 4 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 17 Nonlinear nonzeros: 0

Variable Value Reduced Cost L 3.000000 0.000000 N 7.000000 0.000000 W 5.000000 0.000000 S 5.000000 0.000000 Row Slack or Surplus Dual Price 1 3300.000 -1.000000 2 0.000000 -100.0000 3 0.000000 -100.0000 4 0.000000 -100.0000 5 0.000000 0.000000 6 5.000000 0.000000 7 1.000000 0.000000 Min cost=3300 L=3 N=7 W=5 S=5 b. Interpret the dual value for the constraint corresponding to the available time. What advice would you give the station manager given this dual value? according to the Lingo solution Dual price adding 1 minute to available time will increase the cost $100 but decreasing 1 minute will decrease the cost $100 c. Interpret the dual value for the constraint corresponding to the requirement that at least 15%of the available time should be devoted to local coverage. What advice would you give the station manager given this dual value? According so the Lingo solution Dual price if we decrease the percentage to 10% The total cost will decrease $100 to $3200 because the minute in Local news cost $300 while the minute in National news costs $200 so when decrease the time of the local news to 2 minutes the national news will become 8 minutes instead of 7 and we will save $100 d. Interpret the dual value for the constraint corresponding to the requirement that the time devoted to the local and the national news must be at least 50% of the total broadcast time. What advice would you give the station manager given this dual value? According to the Lingo solution Dual price if we decrease the percentage to 45% The total cost will decrease $100 to $3200 According to the allowable range for local and national news : 4 10.00000 2.000000 3.333333 The cost will decrease $100 per decreased minute until it goes under 6.6667 minutes for both local and national news.

e. Interpret the dual value for the constraint corresponding to the requirement that the time devoted to the weather segment must be less than or equal to the time devoted to the sports segment. What advice would you give the station manager given this dual value? There will be no deference in changing the constraint corresponding to the requirement that the time devoted to the weather segment must be less than or equal to the time devoted to the sports segment because both sports and weather cost $100 per minute. 31. Gulf Coast Electronics is ready to award contracts to suppliers for providing reservoir capacitors for use in its electronic devices. For the past several years, Gulf Coast Electronics has relied on two suppliers for its reservoir capacitors: Able Controls and Lyshenko Industries. A new firm, Boston Components, has inquired into the possibility of providing a portion of the reservoir capacitor needed by Gulf Coast. The quality of products provided by Lysehnko Industries has been extremely high; in fact, only 0.5% of the capacitors provided by Lyshenko had to be discarded because of quality problems. Able Controls has also had a high quality level historically, producing an average of only 1 unacceptable capacitors. Because Gulf Coast Electronics has had no experience with Boston Components, it estimated Boston Components’ defective rate to be 10%. Gulf Coast would like to determine how many reservoir capacitors should be ordered from each firm to obtain 75,000 acceptable-quality capacitors to use in its electronic devices. To nensure that Boston Components will receive some of the contract, management specified that the volume of reservoir capacitors awarded to Boston Components must be at least 10% of the volume given to Able Controls. In addition, the total volume assigned to Boston Components, Able Controls, and Lyshenko Industries should not exceed 30,000, 50,000, and 50,000 capacitors, respectively. Because of Gulf Coast’s long-term relationship with Lyshenko Industries, management also specified that at least 30,000 capacitors should be ordered from Lyshenko. The cost per capacitor is $2.45 for Boston Components, $2.50 for Able Controls, and $2.75 for Lyshenko Industries. A. Formulate and solve a linear program for determining how many reservoir capacitors should be ordered from each supplier to minimize the total cost of obtaining 75,000 acceptable-quality reservoir capacitors .

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

L = Number of copies done by Lyshenko Industries. A = Number of copies done by Able Controls. B = Number of copies done by Boston Components Min 2.45B + 2.5A + 2.75L S.t 0.995L+0.99A+0.9B=75,000 B<=30,000 A<=50,000 L<=50,000 B-0.1A>=0 L>=30,000 Lingo solution Global optimal solution found. Objective value: 197256.2 Infeasibilities: 0.000000 Total solver iterations: 2 Model Class: LP Total variables: 3 Nonlinear variables: 0 Integer variables: 0 Total constraints: 7 Nonlinear constraints: 0 Total nonzeros: 12 Nonlinear nonzeros: 0 Variable Value Reduced Cost B 4180.556 0.000000 A 41805.56 0.000000 L 30000.00 0.000000 Row Slack or Surplus Dual Price 1 197256.2 -1.000000 2 0.000000 -2.541667 3 25819.44 0.000000 4 8194.444 0.000000 5 20000.00 0.000000 6 0.000000 -0.1625000 7 0.000000 -0.2210417 optimal solution Min cost 197256.2 B=4,180.556 A= 41,805.56 L=30,000

b. Suppose that the quality level for reservoir capacitors supplied by Boston Components is much better than estimated. What effect, if any, would this quality level have? If the quality level for reservoir capacitors supplied by Boston Components is less than 97% nothing will change to the optimal solution but if the quality level for reservoir capacitors supplied by Boston Components is 98% the optimal solution will become Min cost 195,772.7 B=30,000 A= 15,909.09 L=30,000 c. Suppose that management is willing to reconsider their requirement that at least 30,000 capacitors must be ordered from Lyshenko Industries. What effect, if any, would this consideration have? According to the dual price for every unit reduced from Lyshenko Industries requirement the total cost will decrease by $0.2210 Case Problem 3. Truck Leasing Strategy S ij =number of trucks obtained from a short-term lease signed in month i for a period of j months L i =number of trucks obtained from the long-term lease used in month i Monthly fuel costs are 20 ($100) = $2000. For short term leased trucks: S 11 ,S 21 ,S 31 ,S 41 cost = $4000+$2000=$6000 S 12 ,S 22 ,S 32 cost =2*3700+2000= $9400 S 13 ,S 23 cost = 3*($3225) + $2000 = $11,675 S 14 cost = 4*($3040) + $2000 = $14,160 the long-term lease are already paid and the employes cannot be laid off the only cost is $2000 fuel TOTAL USED TRUCK IN PROJECT USED FOR LONG TERM LEASED TRUCK USED FOR SHORT TERM LEASED TRUCK MONTH 1 10 1 9 MONTH 2 12 2 10 MONTH 3 14 3 11 MONTH 4 8 1 7

Min 6000 S11 + 9400 S12 + 11675 S13 + 14160 S14 + 6000 S21 + 9400 S22 + 11675 S23 + 6000 S31 + 9400 S32 + 6000 S41 + 2000 L1 + 2000 L2 + 2000 L3 + 2000 L4 S.T. S11 + S12 + S13 + S14 + L1 = 10 S21 + S22 + S23 + S14 + S13 + S12 + L2 = 12 S31 + S32 + S23 + S22 + S14 + S13 + L3 = 14 S41 + S32 + S23 + S14 + L4 = 8 L1 < 1 L2 < 2 L3 < 3 L4 < 1 1. The optimal leasing plan Global optimal solution found. Objective value: 151660.0 Infeasibilities: 0.000000 Total solver iterations: 4 Model Class: LP Total variables: 14 Nonlinear variables: 0 Integer variables: 0 Total constraints: 9 Nonlinear constraints: 0 Total nonzeros: 42 Nonlinear nonzeros: 0 Variable Value Reduced Cost S11 0.000000 3515.000 S12 0.000000 3725.000 S13 3.000000 0.000000 S14 6.000000 0.000000 S21 0.000000 2810.000 S22 0.000000 210.0000 S23 1.000000 0.000000 S31 1.000000 0.000000 S32 0.000000 915.0000 S41 0.000000 3515.000 L1 1.000000 0.000000 L2 2.000000 0.000000 L3 3.000000 0.000000 L4 1.000000 0.000000 Row Slack or Surplus Dual Price 1 151660.0 -1.000000 2 0.000000 -2485.000 3 0.000000 -3190.000 4 0.000000 -6000.000 5 0.000000 -2485.000 6 0.000000 485.0000 7 0.000000 1190.000 8 0.000000 4000.000 9 0.000000 485.0000

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

2. The costs associated with the optimal leasing plan cost = $151660 3. The cost for Reep Construction to maintain its current policy of no layoffs In short term leasing plan, the Penn State company gives the driver along with the trucks. But Bob Reep chose to use his own drivers and pay them $20 an hour. So, the cost for the long term lease = 2000+20*160=$5200 Min 6000 S11 + 9400 S12 + 11675 S13 + 14160 S14 + 6000 S21 + 9400 S22 + 11675 S23 + 6000 S31 + 9400 S32 + 6000 S41 + 5200 L1 + 5200 L2 + 5200 L3 + 5200 L4 S.T. S11 + S12 + S13 + S14 + L1 = 10 S21 + S22 + S23 + S14 + S13 + S12 + L2 = 12 S31 + S32 + S23 + S22 + S14 + S13 + L3 = 14 S41 + S32 + S23 + S14 + L4 = 8 L1 = 1 L2 = 2 L3 = 3 L4 = 1 With the no layoffs policy the cost will be $174,060 Global optimal solution found. Objective value: 174060.0 Infeasibilities: 0.000000 Total solver iterations: 4 Model Class: LP Total variables: 10 Nonlinear variables: 0 Integer variables: 0 Total constraints: 5 Nonlinear constraints: 0 Total nonzeros: 30 Nonlinear nonzeros: 0

Variable Value Reduced Cost S11 0.000000 3515.000 S12 0.000000 3725.000 S13 3.000000 0.000000 S14 6.000000 0.000000 S21 0.000000 2810.000 S22 0.000000 210.0000 S23 1.000000 0.000000 S31 1.000000 0.000000 S32 0.000000 915.0000 S41 0.000000 3515.000 L1 1.000000 0.000000 L2 2.000000 0.000000 L3 3.000000 0.000000 L4 1.000000 0.000000 Row Slack or Surplus Dual Price 1 174060.0 -1.000000 2 0.000000 -2485.000 3 0.000000 -3190.000 4 0.000000 -6000.000 5 0.000000 -2485.000 6 0.000000 -2715.000 7 0.000000 -2010.000 8 0.000000 800.0000 9 0.000000 -2715.000

CCE 6100 HW3

Related Documents