1. 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

2. 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

3. 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

4. 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)

5. 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)

6. 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)

7. 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

8. 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.

9. 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

10. 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

11. Model Constraints x1 +x2 500 dozen sweatshirts
Availability of processing time--available processing time between the end of game and truck pickup
0.1x x2+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

12. Model Summary and Solution
Max z=90x1 +125x2+45x3+65x4
s.t.
0.1x x2+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 = 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

13. 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

14. 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

15. 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=0.18x x2+0.10x3+0.12x4+
0.10x5+0.09x6+0.4x7+0.16x8+0.5x9
+0.7x10

16. 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

17. 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

18. Model Summary and Model Summary
Min z=0.18x x2+0.10x3+0.12x x5+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

19. 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

20. 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 0.085x x2+0.06x3+0.13x4

21. 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 +x x4 0
Total available fund
x2 +x1 +x3 +x4 70,000

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

23. 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

24. 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

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

26. Model Constraints x1 48 (6-8 am) x1 +x2 79 (8-10 am)
x1 +x 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)

27. 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

28. 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.