LP Modeling Most common type of LP involves allocating resources to activities Determining this allocation involves choosing the levels of the activities.

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LP Modeling Most common type of LP involves allocating resources to activities Determining this allocation involves choosing the levels of the activities that achieve the best possible value of the overall measure of the performance

Regional Planning Example Three communities (kibbutz) Limited usable land and water allocation Kibbutz Usable Land (Acres) Water Allocation (Acres Feet) 1 2 3 400 600 300 800 375

Regional Planning-Cont. Three types of crops It is agreed every community will plant the same proportion of its irrigable land Any combination of the crops can be planted How many acres to devote to each of the three crops at each of the three kibbutz Crops Maximum Quota (Acres) Water allocation (Acres/feet) Net Return ($/Acres) Sugar beets Cotton Sorghum 600 500 325 3 2 1 1000 750 250

Defining Decision Variables 1) xj=No. of acres to devote to each of the crops at each of the three communities (j=1,2,…,9),or x1 =no. of acres that community 1 plants sugar beets x2=no. of acres that community 2 plants sugar beets x3=no. of acres that community 3 plants sugar beets x4=no. of acres that community 1 plants cotton x5=no. of acres that community 2 plants cotton x6=no. of acres that community 3 plants cotton x7=no. of acres that community 1 plants sorghum x8=no. of acres that community 2 plants sorghum x9=no. of acres that community 3 plants sorghum 2) xij=No. of acres to devote crop i (i=1,2, and 3) at community j (j=1,2, and 3)

Formulating Objective Function and Constraints 1) Max z=1000(x1 +x2+x3)+750(x4+x5+x6)+250(x7+x8+x9) or 2) Max z =1000(x11+x12+x13)+750(x21+x22+x23)+250(x31+x32+x33) Constraints 1) Usable land for each community x1 +x4+x7 <=400 (community 1) x2+x5+x8 <=600 (community 2) x3+x6+x9 <=300 (community 3)

Constraints-Cont. 2) Water allocation for each community 3x1 +2x4+x7 <=600 (community 1) 3x2+2x5+x8 <=800 (community 2) 3x3+2x6+x9 <=375 (community 3) 3) Total acreage for each crop x1 +x2+x3 <=600 (Sugar beets) x4+x5+x6 <=500 (Cotton) x7+x8+x9 <=325 (Sorghum)

Constraints-Cont. 4) Equal proportion of land planted Optimal Solution [(x1 +x4+x7 )/400 ] =[(x2+x5+x8 )/600] [(x2+x5+x8 )/600] =[(x3+x6+x9 )/300] [(x3+x6+x9)/300 ] =[(x1 +x4+x7 )/400] Optimal Solution z=$633,333.33 x1 =133.33, x2=100, x3=25, x4=100, x5=250, x6=150, x7 =0, x8 =0, x9=0

A Product Mix Example A clothing company produces two types of T-shirt, one with silk screen printing on the front and one with both sides, and two sweatshirts of the same configuration Company has to complete all production within 72 hours (between the end of game and truck pickup) Capacity of truck is 1,200 standard-size boxes A standard-size box holds one dozen T-shirt, whereas a box of one dozen sweatshirts is three times the size of a standard box. Has budgeted $25,000 for production run They have also 500 dozen blank sweatshirts and T-shirts each in stock ready for production.

A Product Mix Example-Cont. The resource requirements, unit costs, and profit per dozen for each type of shirt are: The company wants to know how many dozen (boxes) of each type of shirt to produce in order to maximize profit Product Processing Time (hr) per dozen Cost ($) per dozen Profit ($) per dozen Sweatshirts-F Sweatshirts-B/F T-shirt-F T-shirt B/F 0.1 0.25 0.08 0.21 36 48 25 35 90 125 45 65

Decision Variables and Objective Function x1 =sweatshirts, front printing x2=sweatshirts, back and front printing x3=T-shirts, front printing x4=T-shirts, back and front printing Objective Function Max z=90x1 +125x2+45x3+65x4

Model Constraints x1 +x2 500 dozen sweatshirts Availability of processing time--available processing time between the end of game and truck pickup 0.1x1 +0.25x2+0.08x3+0.21x4 72 hr Availability of shipping capacity--Boxes of sweatshirts are three times the standard-size box. Thus, each box of sweatshirt is equivalent to three boxes of T-shirts 3x1 +3x2+x3+x4 1,200 boxes Availability of the total budget--$25,000 for production run 36x1 +48x2+25x3+35x4 $25,000 Availability of the blank sweatshirts and T-shirts-- The company has 500 dozen blank sweatshirts and T-shirts each in stock x1 +x2 500 dozen sweatshirts x3+x4 500 dozen T-shirts

Model Summary and Solution Max z=90x1 +125x2+45x3+65x4 s.t. 0.1x1 +0.25x2+0.08x3+0.21x4 72 hr 3x1 +3x2+x3+x4 1,200 boxes 36x1 +48x2+25x3+35x4 $25,000 x1 +x2 500 dozen sweatshirts x3+x4 500 dozen T-shirts and xj0 (j=1,2,3,4) Computer solution x1 =175.56 boxes of front-only sweatshirts, x2=57.78 boxes of front and back sweatshirts, x3=500 boxes of front-only T-shirts, and z=45,522.22

A Diet Example A health and fitness center offers a healthy breakfast The breakfast should be high in calories, calcium, protein, and fiber, but low in fat and cholesterol The breakfast must have at least 420 calories, 5 mg of iron, 400 mg of calcium, 20 g of protein, and 12 g of fiber The dietitian wants to limit fat to no more than 20 g and cholesterol to 30 mg. The selected food items with corresponding nutrient contribution and cost are as follows

Breakfast Cal. Fat (g) Chol. (mg) Iron Pro. Fib. Cost ($) Barn cereal(cup) Dry cereal (cup) Oatmeal (cup) Oat barn (cup) Egg Bacon (slice) Orange Milk (cup) Orange J. (cup) Toast (slice) 90 110 100 75 35 65 120 2 5 3 4 1 270 8 12 6 20 48 30 52 250 26 7 9 0.18 0.22 0.10 0.12 0.09 0.40 0.16 0.50 0.07

Decisions Variables and Objective Function Decision Variables x1 =cups of barn cereal, x2=cups of dry cereal, x3=cups of oatmeal, x4=cups of oat barn, x5=eggs, x6=slice of bacon, x7=orange, x8= cups of milk, x9=cups of orange juice, and x10=slices of toast Objective Function Min z=0.18x1 +0.22x2+0.10x3+0.12x4+ 0.10x5+0.09x6+0.4x7+0.16x8+0.5x9 +0.7x10

Model Constraints At least 420 calories No more than 20 g of fat 90x1 +110x2+100x3+90x4+75x5+35x6+65x7+100x8 +120x9+65x10420 No more than 20 g of fat 2x2+2x3+2x4+5x5+3x6+4x8+x1020 No more than 30 mg of cholesterol 270x5+8x6+12x8  30

Model Constraints-Cont. At least 5 g of iron 6x1 +4x2+2x3+3x4+x5+x7+x105 At least 400 mg of calcium 20x1 +48x2+12x3+8x4+30x5+52x7+250x8+3x9+26x10400 At least 20 g of protein 3x1 +4x2+5x3+6x4+7x5+2x6+x7+9x8+x9+3x1020 At least 12 g of fiber 5x1 +2x2+3x3+4x4+x7+3x1012

Model Summary and Model Summary Min z=0.18x1 +0.22x2+0.10x3+0.12x4 +0.10x5+0.09x6+0.4x7+0.16x8+0.5x9+0.7x10 s.t. 90x1 +110x2+100x3+90x4+75x5+35x6+65x7+100x8+120x9+65x10420 2x2+2x3+2x4+5x5+3x6+4x8+x1020 270x5+8x6+12x8  30 6x1 +4x2+2x3+3x4+x5+x7+x105 20x1 +48x2+12x3+8x4+30x5+52x7+250x8+3x9+26x10400 3x1 +4x2+5x3+6x4+7x5+2x6+x7+9x8+x9+3x1020 5x1 +2x2+3x3+4x4+x7+3x1012 x1,x2,x3, x4,x5, x6, x7x8,,x9, and x100 Computer Solution x3=10.25 cups of oatmeal, x8= 1.241cups of milk, x10=2.975 slices of toast, and z=$0.509

An Investment Example $70,000 is available to invest among several alternatives Municipal bond yields 8.5% annual return CD yields 5% annual return Treasury bill yields 6.5% annual return Stock fund yields 13% annual return Some guidelines: No more than 20% of the total should be invested in MBs The amount invested in CDs should not exceed the amount invested in the other three alternatives At least 30% of the investment should be in TBs and CDs The amount invested in CDs and TBs to the MBs and SFs must be at least by a ratio of 1.2 to 1

Decision Variables and Objective Function x1 =amount ($) invested in MBs x2= amount ($) invested in CDs x3= amount ($) invested in TBs x4= amount ($) invested in SFs Objective Function Max 0.085x1 +0.05x2+0.06x3+0.13x4

Model Constraints No more than 20% of the total should be invested in MBs x4 14,000 The amount invested in CDs should not exceed the amount invested in the other three alternatives x2  x1 +x3 +x4 ;or x2 -x1 -x3 -x4  0 At least 30% of the investment should be in TBs and CDs x2+x3  21,000 The amount invested in CDs and TBs to the MBs and SFs must be at least by a ratio of 1.2 to 1 (x2 +x3 )/(x1 +x4 )  1.2; or -1.2 x1+ x2 +x3 -1.2 x4 0 Total available fund x2 +x1 +x3 +x4 70,000

Model Summary and Computer Solution Max 0.085x1 +0.05x2+0.06x3+0.13x4 s.t. x4 14,000 x2 -x1 -x3 -x4  0 x2+x3  21,000 -1.2 x1 + x2+x3 -1.2 x4 0 x2 +x1 +x3 +x4 70,000 x2,x1 ,x3 ,x4 0 Computer Solution x3=38181.84, x4=31818.18, and z=6,818.18

Personnel Scheduling How to schedule the agents to provide satisfactory service with the smallest personnel cost Minimum number of agents required on duty Time period Shift 1 Shift 2 Shift 3 Shift 4 Shift 5 Min No. of Agents 6 am to 8 am 8 am to 10 am 10 am to noon Noon to 2 pm 2 pm to 4 pm 4 pm to 6 pm 6 pm to 8 pm 8 pm to 10 pm 10 pm to midnight Midnight to 6 am Daily cost/agent * $170 $160 $175 $180 $195 48 79 65 87 64 73 82 43 52 15

Personnel Scheduling Shift 1: 6:00 am to 2:pm Shift 2: 8:am to 4:pm Shift 3: Noon to 8:pm Shift 4: 4:pm to midnight Shift 5: 10:pm to 6:am

Decision Variables and Objective Function xj=no. of agents assigned to shift j (j=1,2, 3,4,5) Min z=170 x1 +160 x2+175 x3+180 x4+195 x5

Model Constraints x1 48 (6-8 am) x1 +x2 79 (8-10 am) x1 +x2 65 (10-noon) x1 +x2+x3 87 (noon-2pm) x2+x3 64 (2-4 pm) x3+x4 73 (4-6pm) x3+x4 82 (6-8 pm) x4 43 (8-10 pm) x4+x5 52 (10-midnight x5 15 (midnight-6 am) xj 0 (j=1,2,3,4,5)

Distribution Network 30 units 50 units W1 $900 F1 $400 $200 $200 DC $300 Max 10 $100 $300 Max 80 F2 W2 40 units 60 units

Corrections to Problem 3.6A-1 The last statement of the problem should be corrected to: One unit A uses 2 units of M1 and 3 units of M2, and 1 unit of B uses 2 units of M1 and 6 units of M2.