- Better Products...
Max 30 X1 + 50X2 + 20X3
st.
.5x1 + 2x2 + .75x3 <= 40
x1 + x2 + .5 x3 <= 40
2x1 + 5x2 + 2x3 <= 100
x1 <= .5*(x1 + x2 + x3) simplifies to ( .5x1 - .5x2 -.5x3 <= 0)
x3 >= .2*(x1 + x2 +x3) simplifies to (-.2x1 - .2x2 +.8x3 >= 0)
x1, x2, x3 >= 0
Solver
- The Westchester Chamber .....
Max 100,000 XT + 18,000XR + 40,000 XN
st
2,000 XT + 300 XR + 600 XN <= 18,200
XT <= 10
XR <= 20
XN <= 10
XT >= .1*( XT + XR + XN) simplifies to (.9XT -.1XR -.1XN >=0)
XR <= .5*(XT + XR + XN) simplifies to (-.5XT +.5XR -.5XN <=0)
XT, XR, XN >= 0
Solver
- Shiploading
Xij is the tons of cargo i = (A, B, C) taken in hold j=(f, c, a)
MAx 25 *(XAf + XAc + XAa) + 30 *(XBf + XBc + XBa) + 20 *(XCf + XCc + XCa)
st
(Hold capacities)
XAf + XBf + XCf <= 2,000
XAc + XBc + XCc <= 3,200
XAa + XBa + XCa <= 1,900
(Hold volume Capacities in cf)
60 XAf + 50 XBf + 25 XCf <= 100,000
60XAc + 50 XBc +25 XCc <= 140,000
60XAa +50 XBa +25 XCa <= 95,000
(Available Cargo)
XAf + XAc + XAa <= 7,000
XBf + XBc + XBa <= 6,500
XCf + XCc + XCa <= 4,000
(Trim Constraints )
(XAf + XBf + XCf )/2,000 =( XAc + XBc + XCc)/ 3,200
(XAc + XBc + XCc)/ 3,200 = (XAa + XBa + XCa )/ 1,900
All xij >= 0
Solver
- Bootlegger..
Let xij be the fifth of ith whiskey (A,B,C) blended in the jth blend
(G,U,O). Profit margin of each of the variable is equal to selling price
of the blend less the cost of the whiskey.. For instance for x BG is
$6.00 - $ 5.5 = $ .5 per fifth. etc.
Max -.5XAU - 1.5 XAO+.5XBG -XBO + 3.1XCG+2.6XCU +1.6XCO
st:
Availability of Whiskeys A,B,C
XAG + XAU + XAO <= 2,800
XBG + XBU + XBO <= 2,000
XCG + XCU + XCO <= 2,000
Blend Specs...Gasp-for-air
XAG >= .6*(XAG + XBG + XCG) or -.4XAG + .6XBG + . 6CG <= 0
XBU <= .3*(XAG + XBG + XCG) or -.3XAG + .7XBG - .3XCG <= 0
XBG + XCG <=.4*(XAG + XBG + XCG) or .6XBG + .6XCG - .4XAG <= 0
Ulcer Maker:
XCU <=.5*(XAU + XBU + XCU) or -.5XAU - .5XBU + .5XCU <= 0
XAU >=.2*(XAU + XBU + XCU) or -.8 XAU +. 2XBU + .2XBU <= 0
Old Black:
XCO <=.7*(XAO + XBO + xCO) or -.7XAO - .7XBO + .3XCO <=0
all variables >=0
Solver
- Doug Star
Let A, B and C be the quantities of the grains.
Min 0.45 A + 0.38 B + 0.27 C
s.t 0.62 A + 0.55 B + 0.36 C >= 8/16
0.65 A + 0.10 B + 0.20 C >= 1/16
0.03 A + 0.02 B + 0.01 C <= 1/32
A, B, C >= 0
Solver
- Edwards manufacturing co.
Let xij be the number of ith component (1,2) purchased from the jth
(1,2,3) supplier.
Min 12 x11 + 13x12 + 14 x13 + 10 x21 + 11x22 + 10x23
st
Supplier capacities:
x11 + x21 <= 600
x12 + x22 <= 1000
x13 + x23 <= 800
Requirement Constraints:
x11 + x12 + x13 = 1000
x21 + x22 = x23 = 800
all variables >= 0
Solver
- Tim Burr company..
Let L be thousands of board feet of lumber produced and P be thousands
of square feet of plywood produced.
Max 45L + 60P
St
L + 2P <= 32 (Availability of spruce0
4L + 4P <= 72 (Availability of Douglas Fir)
L >= 5 (Market committment for Lumber)
P >= 12 (Market committmant for Plywood)
L, P >= 0
Solver
- Waiter Scheduling..
There are 12 scheduling periods running from 11-Noon to 9pm - 10 pm.
Let the number of part time employees starting their four hour shifts at
the start of period i be denoted by Xi. For instance, X1
is the number of waiters starting their shift at 11 (period 1) and will
be on hand for four consequtive periods (in the table X1 appears in
perids 1 through 4), likewise for other periods. The minimum
number of part-time employees needed for each period is computed net of
the full time employees available as: total number needed minus 1 if the
first full time employee is on minus 1 if the second full-time employee
is on for each period. Notice that the last group comes in at 6 pm to
work until closing.
Number starting their shift at
|
X1 |
X2 |
X3 |
X4 |
X5 |
X6 |
X7 |
X8 |
| Period |
Need |
11am |
12n |
1pm |
2pm |
3pm |
4pm |
5pm |
6pm |
| 11 - 12N |
(9-1-0)= 8 |
yes |
|
|
|
|
|
|
|
| 12N - 1 |
(9-1-0)= 8 |
yes |
yes |
|
|
|
|
|
|
| 1 - 2 |
(9-1-1)= 7 |
yes |
yes |
yes |
|
|
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|
|
| 2 - 3 |
(3-1-1)= 1 |
yes |
yes |
yes |
yes |
|
|
|
|
| 3 - 4 |
(3-0-1)= 2 |
|
yes |
yes |
yes |
yes |
|
|
|
| 4 - 5 |
(3-1-1)= 1 |
|
|
yes |
yes |
yes |
yes |
|
|
| 5 - 6 |
(6-1-0)= 5 |
|
|
|
yes |
yes |
yes |
yes |
|
| 6 - 7 |
(12-1-1)=10 |
|
|
|
|
yes |
yes |
yes |
yes |
| 7 - 8 |
(12-1-1)=10 |
|
|
|
|
|
yes |
yes |
yes |
| 8 - 9 |
(7-0-1)= 6 |
|
|
|
|
|
|
yes |
yes |
| 9 -10 |
(7-0-1)= 6 |
|
|
|
|
|
|
|
yes |
From the table notice that the number of waiters that will be on
hand during, say, perid 3pm -4pm will be X2 + X3 + X4 + X5
which needs to be at least 2 (the number of emplyees needed for that
period).. Likewise for all periods.
Therefore the problem is to:
Min X1 + X2 + X3 X4 +X5 + X6 + X7 + X8
st
X1 >= 8
X1 + X2 >= 8
X1 + X2 + X3 >= 7
X1 + X2 + X3 + X4 >= 1
+ X2 + X3 + X4 + X5 >= 2
+ X3 + X4 + X5 + X6 >= 1
+ X4 + X5 + X6 + X7 >= 5
+ X5 + X6 + X7 + X8 >= 10
+ X6 + X7 + X8 >= 10
+ X7 + X8 >= 6
+ X8 >= 6
X1, X2, X3, X4, X5,X6,X7,X8 >=0
Solver