LINEAR PROGRAMMING CASE STUDIES AND SOLUTIONS
Vassilis Kostoglou
E-mail:
[email protected] [email protected] URL: www.it.teithe.gr/~vkostogl www.it.teithe.gr/~vkostogl
Case study 1 Portfolio selection
Select a portfolio package from a set of alternative investments
Maximization Maximization of the expected return or minimization of the risk
Available capital
Company’s policy
Duration of investments’ economic life, potential growth rate, danger, liquidity
Case study 1 Portfolio selection
Select a portfolio package from a set of alternative investments
Maximization Maximization of the expected return or minimization of the risk
Available capital
Company’s policy
Duration of investments’ economic life, potential growth rate, danger, liquidity
Return data Expected annual return of investments Investment
Expected annual return rate (%)
Share A – manufacturing sector
15.4
Share B - manufacturing sector
19.2
Share C - food and beverage sector
18.7
Share D – food and beverage sector
13.5
Mutual fund E
17.8
Mutual fund Z
16.3
Requirements
Total amount = € 90000
Amount in shares of a sector no larger than 50% of total available
Amount in shares with the larger return of a sector less or equal to 80% of sector’s total amount Amount in manufacturing company Β less or equal to 10% of the whole share amount Amount in mutual funds less or equal to 25% of the amount in manufacturing shares
Solution Decision variables x1 = invested amount in share A of the manufacturing sector x2 = invested amount in share B of the manufacturing sector x3 = invested amount in share C of the food and beverage sector x4 = invested amount in share D of the food and beverage sector x5 = invested amount in mutual fund E x6 = invested amount in mutual fund Z
Summary of the model Max z = 0.154x 1 + 0.192x 2 + 0.187x3 + 0.135 x4 + 0.178x5 + 0.163x6 with constraints: 1. 2. 3. 4. 5. 6. 7.
x1 x1 -0.8 x1 -0.1 x1 -0.25 x1
+ + + + -
and xi >= 0, i = 1, 2,…, 6
x2 x2
+
x3
+
x4
x3
+
x4
0.2 x3 0.1 x3
-
0.8 x4 0.1 x4
+
x5
+
x6 <=
+
<= <= <= <= <= <=
0.2 x2 0.9 x2 0.25 x2
-
+
x5
x6
90.000 45.000 45.000 0 0 0 0
Summary of results Results of optimal solution Investment
Amount invested
(1)
(2)
Annual return rate expected (%)
Total expected return of the investment
(3)
(4) = (2) x (3)
Share A
27900
15.4
4296.6
Share B
8100
19.2
1555.2
Share C
36000
18.7
6732
Share D
9000
13.5
1215
Mutual fund E
9000
Mutual fund Z
0
Total
90000
17.8 16.3 17.11
1602 0 15400.8
Solution with QSB
QSB+ |---QSB+-------------- Solution Summary for case study 1–--------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | X1 | 27900 | 0 | -.070777 | .154 | .1764444| | 2 | X2 | 8100 | 0 | .154 | .192 | .394| | 3 | X3 | 36000 | 0 | .1638 | .187 | M| | 4 | X4 | 9000 | 0 | 0.042199 | .135 | .187| | 5 | X5 | 9000 | 0 | .163 | .178 | .2708001| | 6 | X6 | 0 | .015 | -M | .163 | .178| |----------------------------------------------------------------------------| | Maximized OBJ = 15400.8 Iteration = 7 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |----------------------------------------------------------------------------| |---------------- Constraint Summary for Ενότητα 3.2 ------------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (<)| .16184 | 0 | 53035.71 | 90000 | 101250 | | 2 |Loose (<)| 0 | 9000 | 36000 | 45000 | M | | 3 |Tight (<)| 0.018560 | 0 | 33750 | 45000 | 76363.63 | | 4 |Loose (<)| 0 | 20700 | -20700 | 0 | M | | 5 |Tight (<)| .052 | 0 | -36000 | 0 | 9000 | | 6 |Tight (<)| .038 | 0 | -8100 | 0 | 20700 | | 7 |Tight (<)| .01616 | 0 | -11250 | 0 | 36964.29 | |----------------------------------------------------------------------------|
Solution with LINDO
LINDO results (1) LP OPTIMUM FOUND AT STEP 6 OBJECTIVE FUNCTION VALUE 1) 15400.80 VARIABLE VALUE REDUCED COST X1 27900.000000 0.000000 X2 8100.000000 0.000000 X3 36000.000000 0.000000 X4 9000.000000 0.000000 X5 9000.000000 0.000000 X6 0.000000 0.015000 ROW SLACK OR SURPLUS DUAL PRICES 1) 0.000000 0.161840 2) 9000.000000 0.000000 3) 0.000000 0.018560 4) 20700.000000 0.000000 5) 0.000000 0.052000 6) 0.000000 0.038000 7) 0.000000 0.016160 NO. ITERATIONS=
6
LINDO results (2) RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 0.154000 0.022444 0.224778 X2 0.192000 0.202000 0.038000 X3 0.187000 INFINITY 0.023200 X4 0.135000 0.052000 0.092800 X5 0.178000 0.092800 0.015000 X6 0.163000 0.015000 INFINITY
ROW 1 2 3 4 5 6 7
CURRENT RHS 90000.000000 45000.000000 45000.000000 0.000000 0.000000 0.000000 0.000000
RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 11250.000000 36964.285156 INFINITY 9000.000000 31363.634766 11250.000000 INFINITY 20700.000000 9000.000000 36000.000000 20700.000000 8100.000000 36964.285156 11250.000000
Solution with Excel
Solution window
Answer report
Sensitivity analysis
Case Study 2 Financial programming problem
Initial amount: € 80000
Timeframe of investments’ decisions: 4 months
Two-month government bonds: return 3%
Three-month government bills: return 6.5%
Bank deposits: interest 1%
At the beginning of the 5 month at least € 40000 are needed
Maximum amount in two-month or three-month bonds: € 32000
th
Decision variables Β j = amount to be invested in government bonds at the beginning of the month j C j = amount to be invested in government bills at the beginning of the month j D j= amount to be invested in bank deposits at the beginning of the month j
Objective function (total return) Max z
=
0.03B 1
+
0.03B2
+
0.03B3
+
0.03B4
+
0.065C1
+
0.065C2
+
0.065C 3
+
0.065C4
+
0.01D1
+
0.01C2
+
0.01C3
+
0.01D4
Basic rule of capital flow: Invested amount (start t) + Cash available (start t) = Available amount (end t-1) + Cash available (end t-1)
Mathematical model 1.
B1 +
C1 +
D1
2.
B2 +
C2 -
1.01D1 +
3. - 1.03B1 +
B3 +
C3
-
1.01D2 +
4. - 1.03B2 +
B4 - 1.065C1 +
C4 -
5.
1.03B3 + 1.065C1 +
1.01D4
D2 D3 1.01D3 +
=
80000
=
0
=
0
D4 =
0
>= 40000
6.
B1
<= 32000
7.
B1 +
B2
<= 32000
8.
B2 +
B3
<= 32000
9.
B3 +
B4
<= 32000
10.
C1 +
11.
C1 +
C2
12.
C1 +
C2 +
C3
<= 32000
13.
C2 +
C3 +
C4
<= 32000
<= 32000
Bi, Ci, Di >= 0, for i = 1, 2, 3, 4
<= 32000
An alternative objective function Maximization of final value: (the intermediate returns are included after their reinvestment) Max Z = 1.03B3+ 1.03B4 + 1.065C2 + 1.065C 3 + 1.065C4 + 1.01D4
Summarized results Results of optimal solution Investment Amount invested (1) (2) B1 32000.000 B3 19557.842 B4 12442.157 C1 32000.000 C4 32000.000 D1 16000.000 D2 16160.000 D3 29723.750 D4 19658.834
Investment return Total return (3) (4) = (2) x (3) 0.03 960.000 0.03 586.735 0.03 373.265 0.065 2080.000 0.065 2080.000 0.01 160.000 0.01 161.600 0.01 297.237 0.01 196.588 Total 6895.425
Solution with QSB (1st)
Solution with QSB (2nd)
Solution with QSB (3rd alternative)
QSB+ (1-1) |------------------ Solution Summary for case study 2------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | Β1 | 32000 | 0 | 0.029899 | .03 | M| | 2 | Β2 | 0 | 0 | 0.019901 | .03 |0.03010097| | 3 | Β3 | 19557.84 | 0 | 0.019903 | .03 | .0401| | 4 | Β4 | 12442.16 | 0 | .0199 | .03 |0.07543910| | 5 | Γ1 | 32000 | 0 | 0.029606 | .065 | M| | 6 | Γ2 | 0 | 0.045348 | -M | .065 | .1103485| | 7 | Γ3 | 0 | 0.010200 | -M | .065 |0.07520001| | 8 | Γ4 | 32000 | 0 | 0.054799 | .065 | M| | 9 | Δ1 | 16000 | 0 | -M | .01 |0.01010097| | 10 | Δ2 | 16160 | 0 | -9.90E19 | .01 |0.02009801| | 11 | Δ3 | 29723.76 | 0 | 0.009900 | .01 |0.02009603| | 12 | Δ4 | 19658.84 | 0 | -.999996 | .01 | .0199| |----------------------------------------------------------------------------| | Maximized OBJ = 6895.425 Iteration = 13 Elapsed CPU seconds = .0546875 | |----------------------------------------------------------------------------|
QSB+ (1-2) |---------------- Constraint Summary for case study 2 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (=)| 0.050909 | 0 | 67805.39 | 80000 | 99168.72 | | 2 |Tight (=)| .040504 | 0 | -12316.5 | 0 | 19360.4 | | 3 |Tight (=)| 0.030201 | 0 | -12439.7 | 0 | 19554.01 | | 4 |Tight (=)| 0.020001 | 0 | -12564.1 | 0 | 19749.54 | | 5 |Tight (>)| -0.00990 | 0 | 20052.96 | 40000 | 52689.76 | | 6 |Tight (<)| 0.000100 | 0 | 19559.06 | 32000 | 32000 | | 7 |Tight (<)| 0.010098 | 0 | 32000 | 32000 | 44565.35 | | 8 |Loose (<)| 0 | 12442.16 | 19557.84 | 32000 | M | | 9 |Tight (<)| 0.009998 | 0 | 12250.46 | 32000 | 44564.12 | | 10 |Tight (<)| 0 | 0 | 32000 | 32000 | M | | 11 |Tight (<)| 0.035393 | 0 | 0 | 32000 | 32000 | | 12 |Tight (<)| 0 | 0 | 32000 | 32000 | M | | 13 |Tight (<)| 0.044998 | 0 | 12250.46 | 32000 | 44564.12 | |----------------------------------------------------------------------------| | Maximized OBJ = 6895.425 Iteration = 13 Elapsed CPU seconds = .0546875 | |----------------------------------------------------------------------------|
QSB+ (2-1) |--------------------- Solution Summary for case study 2---------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | Β1 | 32000 | 0 | -0.00010 | 0 | M| | 2 | Β2 | 0 | 0 | -0.01009 | 0 |0.00010097| | 3 | Β3 | 19557.84 | 0 | 1.019904 | 1.03 | 1.0401| | 4 | Β4 | 12442.16 | 0 | 1.0199 | 1.03 | 1.075439| | 5 | Γ1 | 32000 | 0 | -0.03539 | 0 | M| | 6 | Γ2 | 0 | 0.045348 | -M | 1.065 | 1.110349| | 7 | Γ3 | 0 | 0.010200 | -M | 1.065 | 1.0752| | 8 | Γ4 | 32000 | 0 | 1.0548 | 1.065 | M| | 9 | Δ1 | 16000 | 0 | -M | 0 |0.00010097| | 10 | Δ2 | 16160 | 0 | -9.90E19 | 0 |0.01009800| | 11 | Δ3 | 29723.76 | 0 | -0.00009 | 0 |0.01009602| | 12 | Δ4 | 19658.84 | 0 | 0.000001 | 1.01 | 1.0199| |----------------------------------------------------------------------------| | Maximized OBJ = 86895.42 Iteration = 13 Elapsed CPU seconds = 0.058593 | |----------------------------------------------------------------------------| |----------------------------------------------------------------------------|
Alternative objective function QSB+ (2-2) |-------------------- Constraint Summary for case study 2--------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (=)| 1.050909 | 0 | 67805.39 | 80000 | 99168.72 | | 2 |Tight (=)| 1.040504 | 0 | -12316.5 | 0 | 19360.4 | | 3 |Tight (=)| 1.030202 | 0 | -12439.7 | 0 | 19554.01 | | 4 |Tight (=)| 1.020002 | 0 | -12564.1 | 0 | 19749.54 | | 5 |Tight (>)| -0.00990 | 0 | 20052.96 | 40000 | 52689.76 | | 6 |Tight (<)| 0.000100 | 0 | 19559.06 | 32000 | 32000 | | 7 |Tight (<)| 0.010098 | 0 | 32000 | 32000 | 44565.35 | | 8 |Loose (<)| 0 | 12442.16 | 19557.84 | 32000 | M | | 9 |Tight (<)| 0.009998 | 0 | 12250.46 | 32000 | 44564.12 | | 10 |Tight (<)| 0 | 0 | 32000 | 32000 | M | | 11 |Tight (<)| 0.035393 | 0 | 0 | 32000 | 32000 | | 12 |Tight (<)| 0 | 0 | 32000 | 32000 | M | | 13 |Tight (<)| 0.044998 | 0 | 12250.46 | 32000 | 44564.12 | |----------------------------------------------------------------------------| | Maximized OBJ = 86895.42 Iteration = 13 Elapsed CPU seconds = 0.058593 | |----------------------------------------------------------------------------| |----------------------------------------------------------------------------|
Solution with Excel
Solution window
Answer report
Sensitivity analysis
Case Study 3 Investment choice problem in a limited capitals status
There are five independent investments
Maximization of the total present net value
Satisfaction of budget limitations
Cash inflows of the investments
Cash outflows of the investments
Each investment is divisible (investment rate)
Inflow data Cash inflows
Year
Investment 1
2
3
4
5
1
38
11
17
8
25
2
41
16
24
11
28
3
54
15
29
13
35
4
-
20
-
19
46
Outflow data Cash outflows Year
Investment
Amount
1
2
3
4
5
Available
0
34
10
16
9
31
55
1
13
5
8
4
10
28
2
14
6
10
6
13
30
3
17
6
11
7
12
37
4
-
8
-
5
16
30
Table of cash flows Cash flows of the problem Year
0 1 2 3 4 NPV
Investment
1 -34 25 27 37 38.14
2 -10 6 10 9 12 18.67
3 -16 9 14 18 17.26
10% discount rate of the interest
4 -9 4 5 6 14 12.83
5 -31 15 15 23 30 32.79
1.000 0.909 30.826 0.751 0.683
Net Present Value 1 (NPV1) = -34 *1 + 25 0.909 + 27 0.826 + 37 0.751 = 38.14
Decision variables – Objective function x j = investment rate j (j = 1, 2, 3, 4, 5) which is accepted in the optimum solution Max z = 38.14x 1 + 18.67x 2 + 17.26x3 +12.83x4 + 32.79x5
Constraints 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
34x1 13x1 14x1 17 x1
+ + + +
x1
+ -
x j >= 0, j = 1, 2, …5
10x2 5x2 6x2 6x 2 8x 2
+ + + +
16x3 8x 3 10x3 11x3
+ + + + +
31x4 4x 4 6x4 7x4 5x 4
x2
+ + + + +
31x5 10x5 13x5 12x5 16x5 x5
x3 x4 x5
<= <= <= <= <= <= <=
55 28 30 37 30 1 1 1 1 1
Summarized results Optimal solution results Variable (1) x1 x2 x3 x4 x5
Variable value (2) 1 1 0.125 1 1
NPV (*€ 1000) (3) 38.14 18.67 17.26 12.83 32.79 Total
Total NPV (*€ 1000) (4) = (2) x (3) 38.14 18.67 2.1575 12.83 0 71.7975
Solution with QSB
QSB+ |-QSB+------------ Solution Summary for case study 3 ------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | X1 | 1 | 0 | 36.6775 | 38.14 | M| | 2 | X2 | 1 | 0 | 10.7875 | 18.67 | M| | 3 | X3 | .125 | 0 | 16.92387 | 17.26 | 17.94823| | 4 | X4 | 1 | 0 | 9.70875 | 12.83 | M| | 5 | X5 | 0 | .6512495 | -M | 32.79 | 33.44125| |----------------------------------------------------------------------------| | Maximized OBJ = 71.7975 Iteration = 5 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |----------------- Constraint Summary for Ενότητα 3.4 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (<)| 1.07875 | 0 | 53 | 55 | 59.4 | | 2 |Loose (<)| 0 | 5 | 23 | 28 | M | | 3 |Loose (<)| 0 | 2.75 | 27.25 | 30 | M | | 4 |Loose (<)| 0 | 5.625 | 31.375 | 37 | M | | 5 |Loose (<)| 0 | 17 | 13 | 30 | M | | 6 |Tight (<)| 1.462499 | 0 | .6206896 | 1 | 1.058824 | | 7 |Tight (<)| 7.8825 | 0 | 0 | 1 | 1.2 | | 8 |Loose (<)| 0 | .875 | .125 | 1 | M | | 9 |Tight (<)| 3.12125 | 0 | 0 | 1 | 1.222222 | | 10 |Loose (<)| 0 | 1 | 0 | 1 | M |
Solution with Excel
Solution window
Answer window
Sensitivity analysis window
Case study 4 Advertising media selection Problem data Advertising media 1. 2. 3. 4. 5. 6.
Friday – day Saturday – day Sunday – day Friday – night Saturday – night Sunday - night
Cost of one view (in €) 400 450 450 500 550 550
Units of expected audience rate of one view 5000 5500 5700 7500 8200 8400
Other relevant data
Goal: Determination of views / records in order to maximize the total audience rate
Total available amount: € 45000
Maximum amount for Friday: € 11000
Maximum amount for Saturday: € 14400
Total daily view number: at least 20
Total nightly view number: at least 50% of the total
Maximum view number: each day 12, each night 18
Decision variables Χ1 = number of views on Friday (day) Χ2 = number of views on Saturday (day) Χ3 = number of views on Sunday (day) Χ4 = number of views on Friday (night) Χ5 = number of views on Saturday (night) Χ6 = number of views on Sunday (night)
Mathematical model Max z = 5000x 1 + 5500x2 + 5700x3 + 7500x4 + 8200x5 + 8400x 6 with constraints: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
400x1 400x1 x1 -0.5x1 x1
+
450x2 + 450x3 + +
+ -
450x2 x2 + 0.5x2 -
x j >= 0, j = 1, 2, …5
x3 0.5x3 +
500x4 500x4
0.5x4
+
550x5
+
+
550x5
+
+
0.5x5
+
550x6
0.5x6
x2 x3 x4 x5 x6
<= <= <= >= >= <= <= <= <= <= <=
45000 11000 14400 20 0 12 12 12 18 18 18
Results Variable (1) x1 x2 x3 x4 x5 x6
Value (number of views) (2) 5 10 12 18 18 18
Units of view Total audience rate audience rate of views (3) (4) = (2) x (3) 5000 25000 5500 55000 5700 68400 7500 135000 8200 147600 8400 151200 Total 582200
Solution with QSB
QSB+ |-QSB+------------ Solution Summary for case study 4 ------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | X1 | 5 | 0 | 0 | 5000 | 6000| | 2 | X2 | 10 | 0 | 0 | 5500 | 6709.091| | 3 | X3 | 12 | 0 | 0 | 5700 | M| | 4 | X4 | 18 | 0 | 6250 | 7500 | M| | 5 | X5 | 18 | 0 | 6722.222 | 8200 | M| | 6 | X6 | 18 | 0 | 0 | 8400 | M| |----------------------------------------------------------------------------| | Maximized OBJ = 582200 Iteration = 8 Elapsed CPU seconds = .0625 | |----------------------------------------------------------------------------| |----------------------------------------------------------------------------| |----------------- Constraint Summary for Ενότητα 3.5 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Loose (<)| 0 | 4300 | 40700 | 45000 | M | | 2 |Tight (<)| 12.5 | 0 | 9000 | 11000 | 13800 | | 3 |Tight (<)| 12.22222 | 0 | 11250 | 14400 | 15300 | | 4 |Loose (>)| 0 | 7 | -M | 20 | 27 | | 5 |Loose (>)| 0 | 13.5 | -M | 0 | 13.5 | | 6 |Loose (<)| 0 | 7 | 5 | 12 | M | | 7 |Loose (<)| 0 | 2 | 10 | 12 | M | | 8 |Tight (<)| 5700 | 0 | 5 | 12 | 21.55556 | | 9 |Tight (<)| 1250 | 0 | 12.4 | 18 | 22 | | 10 |Tight (<)| 1477.778 | 0 | 16.36364 | 18 | 23.72727 | | 11 |Tight (<)| 8400 | 0 | 0 | 18 | 25.81818 | |----------------------------------------------------------------------------|
Solution with LINDO
LINDO results (1) LP OPTIMUM FOUND AT STEP 7 OBJECTIVE FUNCTION VALUE 1) 582200.0 VARIABLE X1 X2 X3 X4 X5 X6
VALUE 5.000000 10.000000 12.000000 18.000000 18.000000 18.000000
ROW SLACK OR SURPLUS 1) 4300.000000 2) 0.000000 3) 0.000000 4) 7.000000 5) 13.500000 6) 7.000000 7) 2.000000 8) 0.000000 9) 0.000000 10) 0.000000 11) 0.000000 NO. ITERATIONS= 7
REDUCED COST 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 DUAL PRICES 0.000000 12.500000 12.222222 0.000000 0.000000 0.000000 0.000000 5700.000000 1250.000000 1477.777832 8400.000000
LINDO results (2) RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 5000.000000 1000.000000 5000.000000 X2 5500.000000 1209.090942 5500.000000 X3 5700.000000 INFINITY 5700.000000 X4 7500.000000 INFINITY 1250.000000 X5 8200.000000 INFINITY 1477.777832 X6 8400.000000 INFINITY 8400.000000 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 1 45000.000000 INFINITY 4300.000000 2 11000.000000 2800.000000 2000.000000 3 14400.000000 900.000000 3150.000000 4 20.000000 7.000000 INFINITY 5 0.000000 13.500000 INFINITY 6 12.000000 INFINITY 7.000000 7 12.000000 INFINITY 2.000000 8 12.000000 9.555555 7.000000 9 18.000000 4.000000 5.600000 10 18.000000 5.727273 1.636364 11 18.000000 7.818182 18.000000
Results with Excel
Solution window
Answer report
Sensitivity analysis
Case study 5 Marketing research
Personal interviews, Daily (D) and Nightly (N)
Households: with children, without children, of one person
Sample = 800 households
At least: 400 with children, 200 without children, 100 of one person
Respondents Night > Respondents Day
To be done during: At least 50% of interviews to households with children At least 60% of interviews to households without children At least 70% of interviews to households of one person
Cost elements Interview cost (in monetary units) Household category With children Without children One person
Day 1500 1300 1000
Night 1800 1600 1200
Decision variables x11 = Number of interviews to households with children carried out during the day x12 = Number of interviews to households with children carried out during the n ight x21= Number of interviews to households without children carried out during the day x22 = Number of interviews to households without children carried out during the night x31 = Number of interviews to households of one person carried out during the day x32 = Number of interviews to households of one person carried out during the night
Mathematical model Min z = 1.500x 11 + 1.800x 12 + 1.300x21 + 1.600x22 + 1.000x 31 + 1.200x 32 with constraints: 1. 2. 3. 4. 5. 6. 7. 8.
x11 + x11 +
-x11 + -0.5x11 +
x12 + x12
x21 +
x22 +
x21 +
x22 +
x12 x21 + 0.5x12 - 0.6x21 +
x22
-
x32
x31 + x31 +
x32 x32
0.4x22 -
and x
x31 +
>= 0, for i = 1, 2, 3 and j =1, 2
0.7x31 + 0.3 x32
= >= >= >= >= >= >= >=
800 400 200 100 0 0 0 0
Summarized results Variable (1) x11 x12 x21 x22 x31 x32 Total
Number of interviews’ (2) 200 200 80 120 60 140 800
Interview cost (3) 1500 1800 1300 1600 1000 1200
Total interview cost (4) = (2) x (3) 300000 360000 104000 192000 60000 168000 1184000
Solution with QSB
QSB+ |-------------------- Solution Summary for case study 5 ---------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | x11 | 200 | 0 | 480 | 1500 | 1800| | 2 | x12 | 200 | 0 | 1500 | 1800 | M| | 3 | x21 | 80 | 0 | 450 | 1300 | 1600| | 4 | x22 | 120 | 0 | 1300 | 1600 | M| | 5 | x31 | 60 | 0 | -M | 1000 | 1200| | 6 | x32 | 140 | 0 | 1000 | 1200 | 1685.714| |----------------------------------------------------------------------------| | Minimized OBJ = 1184000 Iteration = 9 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |----------------- Constraint Summary for Ενότητα 3.6 -----------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (=)| 1140 | 0 | 700 | 800 | M | | 2 |Tight (>)| 510 | 0 | 0 | 400 | 500 | | 3 |Tight (>)| 340 | 0 | 0.000015 | 200 | 300 | | 4 |Loose (>)| 0 | 100 | -M | 100 | 200 | | 5 |Loose (>)| 0 | 120 | -M | 0 | 120 | | 6 |Tight (>)| -300 | 0 | -60 | 0 | 200 | | 7 |Tight (>)| -300 | 0 | -60 | 0 | 80 | | 8 |Tight (>)| -200 | 0 | -60 | 0 | 60 |
Solution with LINDO
LINDO results (1) LP OPTIMUM FOUND AT STEP
8
OBJECTIVE FUNCTION VALUE 1)
1184000.
VARIABLE X11 X12 X21 X22 X31 X32
ROW 1) 2) 3) 4) 5) 6) 7) 8)
VALUE 200.000000 200.000000 80.000000 120.000000 60.000000 140.000000
SLACK OR SURPLUS 0.000000 0.000000 0.000000 100.000000 120.000000 0.000000 0.000000 0.000000
NO. ITERATIONS=
8
REDUCED COST 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
DUAL PRICES -1140.000000 -510.000000 -340.000000 0.000000 0.000000 -300.000000 -300.000000 -200.000000
LINDO results (2) RANGES IN WHICH THE BASIS IS UNCHANGED:
VARIABLE X11 X12 X21 X22 X31 X32
ROW 1 2 3 4 5 6 7 8
CURRENT COEF 1500.000000 1800.000000 1300.000000 1600.000000 1000.000000 1200.000000
OBJ COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 300.000000 1020.000000 INFINITY 300.000000 300.000000 850.000000 INFINITY 300.000000 200.000000 INFINITY 485.714294 200.000000
CURRENT RHS 800.000000 400.000000 200.000000 100.000000 0.000000 0.000000 0.000000 0.000000
RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE INFINITY 100.000000 100.000000 400.000000 100.000000 199.999985 100.000000 INFINITY 120.000000 INFINITY 200.000000 60.000000 80.000000 60.000000 60.000000 60.000000
Results with Excel
Solution window
Answer report
Sensitivity analysis report
Case study 6 Human resources management
Allocation of available human resources to different departments, work centers, shifts etc.
Recruitment of seasonal staff
Allocation of staff to shifts
Minimizing the number of employees who should work in various time periods during the day
Problem’s data Time period 07 a.m. – 09 a.m. 09 a.m. – 11 a.m. 11 a.m. – 13 p.m. 13 p.m. – 15 p.m. 15 p.m. – 17 p.m. 17 p.m. – 19 p.m. 19 p.m. – 21 p.m. 21 p.m. – 23 p.m. 23 p.m. – 07 a.m. Gross employee cost per day (in €)
1 + + + +
230
2 + + + +
220
Shift 3
+ + + +
225
4
5
+ + + + 240
+ 260
Minimum number of required employees 35 68 60 57 65 63 72 33 12
Mathematical model x j = number of employees starting in shift j Min z = 230x1 + 220x2 + 225x3 + 240x4 + 260x5 with constraints: 1. 2. 3. 4. 5. 6. 7. 8. 9.
x1 x1 x1 x1
+ + +
x2 x2 x2 x2
+ +
x3 x3 x3 x3
+ + +
x4 x4 x4 x4 x5
x j >= 0,
j = 1, 2,…,5
>= >= >= >= >= >= >= >= >=
35 68 60 57 65 63 72 33 12
(07 a.m. – 09 a.m.) (09 a.m. – 11 a.m.) (11 a.m. – 13 p.m.) (13 p.m. – 15 p.m.) (15 p.m. – 17 p.m.) (17 p.m. – 19 p.m.) (19 p.m. – 21 p.m.) (21 p.m. – 23 p.m.) (23 p.m. – 07 a.m.)
Model with less constraints Min z = 230x1 + 220x2 + 225x3 + 240x4 + 260x5 with constraints: 1. 2. 3. 4. 5. 6. 7.
x1 x1 x1
+ +
x2 x2 x2
+ +
x3 x3 x3
+ +
x4 x4 x4 x5
x j >= 0,
j = 1, 2,…,5
>= >= >= >= >= >= >=
35 68 57 65 72 33 12
(07 a.m. – 09 a.m.) (09 a.m. – 13 a.m.) (13 p.m. – 15 p.m.) (15 p.m. – 17 p.m.) (17 p.m. – 21 p.m.) (21 p.m. – 23 p.m.) (23 p.m. – 07 a.m.)
Summarized results Optimal solution results Variable
Value
(1) x1 x2 x3 x4 x5
(2) 35 33 39 33 12
Employee cost Total cost of employees per day per day (3) (4) = (2) x (3) 230 8050 220 7260 225 8770 240 7920 260 3120 Total 35125
Solution with QSB
QSB+ |----------------- Solution Summary for case study 6 ------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | X1 | 35 | 0 | 220 | 230 | M| | 2 | X2 | 33 | 0 | 0 | 220 | 230| | 3 | X3 | 39 | 0 | 0 | 225 | 240| | 4 | X4 | 33 | 0 | 225 | 240 | M| | 5 | X5 | 12 | 0 | 0 | 260 | M| |----------------------------------------------------------------------------| | Minimized OBJ = 35125 Iteration = 7 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |---------------- Constraint Summary for case study 6 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (>)| 10 | 0 | 0 | 35 | 68 | | 2 |Tight (>)| 220 | 0 | 35 | 68 | M | | 3 |Loose (>)| 0 | 50 | -M | 57 | 107 | | 4 |Loose (>)| 0 | 40 | -M | 65 | 105 | | 5 |Tight (>)| 225 | 0 | 33 | 72 | M | | 6 |Tight (>)| 15 | 0 | 0 | 33 | 72 | | 7 |Tight (>)| 260 | 0 | 0 | 12 | M | |----------------------------------------------------------------------------|
Solution with LINDO
LINDO results (1) LP OPTIMUM FOUND AT STEP 6 OBJECTIVE FUNCTION VALUE 1) 35125.00 VARIABLE VALUE REDUCED COST X1 35.000000 0.000000 X2 33.000000 0.000000 X3 39.000000 0.000000 X4 33.000000 0.000000 X5 12.000000 0.000000 ROW SLACK OR SURPLUS 1) 0.000000 2) 0.000000 3) 50.000000 4) 40.000000 5) 0.000000 6) 0.000000 7) 0.000000 NO. ITERATIONS= 6
DUAL PRICES -10.000000 -220.000000 0.000000 0.000000 -225.000000 -15.000000 -260.000000
LINDO results (2) RANGES IN WHICH THE BASIS IS UNCHANGED:
VARIABLE X1 X2 X3 X4 X5
ROW 1 2 3 4 5 6 7
CURRENT COEF 230.000000 220.000000 225.000000 240.000000 260.000000
CURRENT RHS 35.000000 68.000000 57.000000 65.000000 72.000000 33.000000 12.000000
OBJ COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE INFINITY 10.000000 10.000000 220.000000 15.000000 225.000000 INFINITY 15.000000 INFINITY 260.000000 RIGHTHAND SIDE RANGES ALLOWABLE INCREASE 33.000000 INFINITY 50.000000 40.000000 INFINITY 39.000000 INFINITY
ALLOWABLE DECREASE 35.000000 33.000000 INFINITY INFINITY 39.000000 33.000000 12.000000
Case study 7 Production planning
Planning horizon: A number of time periods Problem: Having a forecast for each period’s demand, determine the products’ quantities that can be produced with feasible methods in order to satisfy the total demand with the minimum cost. For two products Α and Β, there are demand forecasts for January, February and March.
Initial stock: 100 units of product A and 120 units of product B.
Minimum total required stock: 130 units of product Α and 110 of product Β
Unit production cost: Α = € 20 and Β = € 25
Maintenance cost per period and per unit: 2% on the unit production cost
Demand Bicycle demand Month January February March Total
Bicycle A 700 900 1000 2600
B 800 600 900 2.300
Capacity System’s capacity Month January February March
Machine capacity (machine hours) 3000 2800 3600
Available work (man-hours) 2500 2300 2400
Use of resources Required resources per product unit Bicycle A B
Machine hours 1.5 1.6
Man-hours 1.1 1.2
Decision variables xIJ = number of units of product I produced during month J, where I = A, B and J = 1, 2, 3 IIJ = number of units of product I maintained in stock at the end of the month J, where I = A, B and J = 1, 2, 3 J = 1 – January J = 2 – February J = 3 – March
Objective function Min z = 20X A1 + 20X A2 + 20X A3 + 25XB1 + 25XB2 + 25XB3 + 0.41I A1 + 0.41I A2 + 0.41I A3 + 0.5IB1 + 0.5IB2 + 0.5IB3 with the constraints: Stock (end t-1) + Production (t) = Demand (t) + Reserve (end t)
Mathematical model Min z = 20X A1 + 20X A2 + 20X A3 + 25XB1 + 25XB2 + 25XB3 + 0.41I A1 + 0.41I A2 + 0.41I A3 + 0.5IB1 + 0.5IB2 + 0.5IB3 with constraints: 1. X A1 - I A1 = 600 January demand 2. XB1 - IB1 = 680 January demand 3. X A2 + I A1 I A2 = 900 February demand 4. XB2 + IB1 IB2 = 600 February demand 5. X A3 + I A2 I A3 = 1000 March demand 6. XB3 + IB2 IB3 = 900 March demand 7. I A3 >= 130 Total stock in March 8. IB3 >= 110 Total stock in March 9. 1.5X A1 + 1.6XB1 <= 3000 Machine hours 10. 1.5X A2 + 1.6XB2 <= 2800 Machine hours 11. 1.5X A3 + 1.6XB3 <= 3600 Machine hours 12. 1.1X A1 + 1.2XB1 <= 2500 Man-hours 13. 1.1X A2 + 1.2XB2 <= 2300 Man-hours 14. 1.1X A3 + 1.2XB3 <= 2400 Man-hours Xij >= 0, i = A, B and j = 1,2,3 Iij >= 0, i = A, B and j = 1,2,3
Summary results Optimal solution results Variable
Value
Cost per product unit
Total product cost
X A1
600
20
12000
X A2
950
20
19000
X A3
1080
20
21600
XB1
680
25
17000
XB2
600
25
15000
XB3
1010
25
25250
I A1
0
0.4
0
I A2
50
0.4
20
I A3
130
0.4
52
IB1
0
0.5
0
IB2
0
0.5
0
IB3
110
0.5
55 Total
109997
Solution with QSB (1)
Solution with QSB (2)
QSB+ (1) |---------------- Solution Summary for case study 7 ------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | XA1 | 600 | 0 | 19.6 | 20 | M| | 2 | XA2 | 950.0001 | 0 | 19.6 | 20 | 20.05833| | 3 | XA3 | 1080 | 0 | 19.94167 | 20 | 20.4| | 4 | XB1 | 680 | 0 | 24.5 | 25 | M| | 5 | XB2 | 600 | 0 | 24.93636 | 25 | 25.5| | 6 | XB3 | 1010 | 0 | -.936363 | 25 | 25.06364| | 7 | IA1 | 0 | .4 | 0 | .4 | M| | 8 | IA2 | 50.00007 | 0 | -2.98E-8 | .4 | .4583333| | 9 | IA3 | 130 | 0 | -20.4 | .4 | M| | 10 | IB1 | 0 | .5 | 0 | .5 | M| | 11 | IB2 | 0 | 0.063636 | .4363636 | .5 | M| | 12 | IB3 | 110 | 0 | -25.4363 | .5 | M| |----------------------------------------------------------------------------| | Minimized OBJ = 109977 Iteration = 10 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |----------------------------------------------------------------------------|
QSB+ (2) |---------------- Constraint Summary for case study 7 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Tight (=)| 20 | 0 | 0 | 600 | 1274.667 | | 2 |Tight (=)| 25 | 0 | 0 | 680 | 1312.5 | | 3 |Tight (=)| 20 | 0 | -50.0000 | 900 | 1176.667 | | 4 |Tight (=)| 25 | 0 | 0 | 600 | 859.3749 | | 5 |Tight (=)| 20.4 | 0 | 949.9999 | 1000 | 1276.667 | | 6 |Tight (=)| 25.43636 | 0 | 854.1666 | 900 | 1153.611 | | 7 |Tight (>)| 20.8 | 0 | 79.99993 | 130 | 406.6666 | | 8 |Tight (>)| 25.93636 | 0 | 64.16661 | 110 | 363.611 | | 9 |Loose (<)| 0 | 1012 | 1988 | 3000 | M | | 10 |Loose (<)| 0 | 414.9999 | 2385 | 2800 | M | | 11 |Loose (<)| 0 | 364.0001 | 3236 | 3600 | M | | 12 |Loose (<)| 0 | 1024 | 1476 | 2500 | M | | 13 |Loose (<)| 0 | 534.9999 | 1765 | 2300 | M | | 14 |Loose (<)| 0 | 1024 | 2095.667 | 2400 | 2455 | |----------------------------------------------------------------------------| | Minimized OBJ = 109977 Iteration = 10 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| | < PageDown > < PageUp > < Hardcopy > < Cancel > | |----------------------------------------------------------------------------|
Solution with LINDO (1) LP OPTIMUM FOUND AT STEP 6 OBJECTIVE FUNCTION VALUE 1) 109977.0 VARIABLE VALUE REDUCED COST XA1 600.000000 0.000000 XA2 950.000000 0.000000 XA3 1080.000000 0.000000 XB1 680.000000 0.000000 XB2 600.000000 0.000000 XB3 1010.000000 0.000000 IA1 0.000000 0.400000 IA2 50.000000 0.000000 IA3 130.000000 0.000000 IB1 0.000000 0.500000 IB2 0.000000 0.063636 IB3 110.000000 0.000000 ROW SLACK OR SURPLUS 1) 0.000000 2) 0.000000 3) 0.000000 4) 0.000000 5) 0.000000 6) 0.000000 7) 0.000000 8) 0.000000 9) 1012.000000 10) 415.000000 11) 364.000000 12) 1024.000000 13) 535.000000 14) 0.000000 NO. ITERATIONS= 6
DUAL PRICES -20.000000 -25.000000 -20.000000 -25.000000 -20.400000 -25.436363 -20.799999 -25.936363 0.000000 0.000000 0.000000 0.000000 0.000000 0.363636
Solution with LINDO (2) RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE XA1 20.000000 INFINITY 0.400000 XA2 20.000000 0.058334 0.400000 XA3 20.000000 0.400000 0.058334 XB1 25.000000 INFINITY 0.500000 XB2 25.000000 0.500000 0.063637 XB3 25.000000 0.063637 25.936363 IA1 0.400000 INFINITY 0.400000 IA2 0.400000 0.058334 0.400000 IA3 0.400000 INFINITY 20.799999 IB1 0.500000 INFINITY 0.500000 IB2 0.500000 INFINITY 0.063637 IB3 0.500000 INFINITY 25.936363 RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE 1 600.000000 674.666687 600.000000 2 680.000000 632.500000 680.000000 3 900.000000 276.666656 950.000000 4 600.000000 259.375000 600.000000 5 1000.000000 276.666656 50.000000 6 900.000000 253.611115 45.833332 7 130.000000 276.666656 50.000000 8 110.000000 253.611115 45.833332 9 3000.000000 INFINITY 1012.000000 10 2800.000000 INFINITY 415.000000 11 3600.000000 INFINITY 364.000000 12 2500.000000 INFINITY 1024.000000 13 2300.000000 INFINITY 535.000000 14 2400.000000 55.000000 304.333344
Case study 8 Diet problem Identification of a diet or of a prescription meeting specific dietary requirements
Criterion: minimum cost
X j = the amount of ingredient j for the production of one unit of animal f eed
Problem’s data Required nutritional ingredient
Vitamin A Vitamin C Vitamin E Proteins Calories Unit cost
Number of nutritional ingredients per ingredient unit Ingredient 1 2 3 4 80 115 100 90 110 90 85 100 50 70 105 80 250 300 210 240 480 510 470 530 180 160 145 200
Nutritional requirement per animal feed unit
>= 80 >= 100 >= 60 >= 260 <= 2300
Mathematical model Min z = 180x1 + 160x2 + 145x3 + 200x4 with constraints: 1.
80x1 +
115x2 +
100x3 +
90x4 + >=
80
2.
110x1 +
90x2 +
85x3 +
100x4 + >=
100
3.
50x1 +
70x2 +
105x3 +
80x4 + >=
60
4.
250x1 +
300x2 +
210x3 +
240x4 + >=
260
5.
480x1 +
510x2 +
470x3 +
530x4 + >=
2300
6.
x1 +
x2 +
x3 +
x4 + >=
1
x j >=
0,
for
j = 1,2,3,4
Summarized results Optimal solution results Variable (1) x1 x2 x3 x4 Total
Value (2) 0.537 0.317 0.146 0 1
Unit cost (3) 180 160 145 200
Total variable cost (4) = (2) x (3) 96.66 50.72 21.17 0 168.55
Solution with QSB
QSB+ |----------------- Solution Summary for case study 8 ------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | X1 | .5365854 | 0 | 151.6667 | 180 | 220| | 2 | X2 | .3170732 | 0 | 152 | 160 | 223.75| | 3 | X3 | .1463415 | 0 | -M | 145 | 155| | 4 | X4 | 0 | 33.41463 | 166.5854 | 200 | M| |----------------------------------------------------------------------------| | Minimized OBJ = 168.5366 Iteration = 8 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------| |----------------- Constraint Summary for Ενότητα 3.9 -----------------------| |----------------------------------------------------------------------------| |Constraint|Constraint| Shadow | Surplus | Minimum | Current | Maximum | | Number | Status | Price | | R. H. S. | R. H. S. | R. H. S. | |----------+----------+----------+----------+----------+----------+----------| | 1 |Loose (>)| 0 | 14.02439 | -M | 80 | 94.02439 | | 2 |Tight (>)| 1.243902 | 0 | 87.77778 | 100 | 102.5352 | | 3 |Loose (>)| 0 | 4.390244 | -M | 60 | 64.39024 | | 4 |Tight (>)| 0.097560 | 0 | 234 | 260 | 275 | | 5 |Loose (<)| 0 | 1811.951 | 488.0488 | 2300 | M | | 6 |Tight (=)| 18.78049 | 0 | .9860031 | 1 | 1.166667 | |----------------------------------------------------------------------------| | Minimized OBJ = 168.5366 Iteration = 8 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------|
Solution with LINDO (1) LP OPTIMUM FOUND AT STEP 5 OBJECTIVE FUNCTION VALUE 1) 168.5366 VARIABLE VALUE REDUCED COST X1 0.536585 0.000000 X2 0.317073 0.000000 X3 0.146341 0.000000 X4 0.000000 33.414635 ROW 1) 2) 3) 4) 5) 6)
SLACK OR SURPLUS 14.024390 0.000000 4.390244 0.000000 1811.951172 0.000000
NO. ITERATIONS=
5
DUAL PRICES 0.000000 -1.243902 0.000000 -0.097561 0.000000 -18.780487
Solution with LINDO (2) RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X1 180.000000 40.000000 28.333334 X2 160.000000 63.750000 8.000000 X3 145.000000 10.000000 INFINITY X4 200.000000 INFINITY 33.414635
ROW 1 2 3 4 5 6
CURRENT RHS 80.000000 100.000000 60.000000 260.000000 2300.000000 1.000000
RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 14.024390 INFINITY 2.535211 12.222223 4.390244 INFINITY 14.999999 26.000000 INFINITY 1811.951172 0.166667 0.013997
Case study 9 Mix problem
Determination of the best mix program of raw material for the production of final products
Three main ingredients Α, Β and C
Three products: super fuel, unleaded, super unleaded
Minimum required octane number
Maximization of the total daily profit
Available quantities of main ingredients
Minimum required product quantities
Problem’s data Main ingredient
Octane number
Cost per ton (€)
A B C
120 90 130
38 42 105
Maximum daily available quantity (tones) 1000 1200 700
Demand data Fuel A B C
Octane number 94 92 96
Cost per ton (€) 85 80 88
Daily demand (tones) 800 1100 500
Decision variables Xij = quantity of ingredient i mixed for the production of one tone of product j for i = A, B, C and j = 1, 2, 3 j = 1 – fuel super j = 2 – unleaded fuel j = 3 – super unleaded fuel
Consumption of resources and produced quantities X A1 + X A2 + X A3 = daily used quantity of ingredient A XB1 + XB2 + XB3 = daily used quantity of ingredient B XC1 + XC2 + XC3 = daily used quantity of ingredient C X A1 + XB1 + XC1 = daily poduced quantity of super fuel X A2 + XB2 + XC2 = daily poduced quantity of unleaded fuel X A3 + XB3 + XC3 = daily poduced quantity of super unleaded fuel
Income and cost functions Daily income = 85(X A1 + XB1 + XC1) + 80(X A2 + XB2 + XC2) + 88(X A3 + XB3 + XC3) Daily relative cost = 38(X A1 + X A2 + X A3) + 42(XB1 + XB2 + XB3) + 105(X C1 + XC2 + XC3)
Objective function Z=
(85 – 38)X A1 + (80 – 38)X A2 + (88 – 38)X A3 + (85 – 42)X B1 + (80 – 42)X B2 + (88 – 42)X B3 + (85 – 105)X C1 + (80 – 105)X C2 + (88 – 105)X C3,
Max z = 47X A1 +42X A2 +50X A3 + 43XB1 +38XB2 + 46XB3 – 20XC1 – 25XC2 – 17XC3
Proportion (mix) constraints Octanes of super fuel: (120X A1 + 90XB1 + 130XC1) / (X A1 + XB1 + XC1) >= 94 Octanes of unleaded fuel: (120X A1 + 90XB1 + 130XC1) / (X A2 + XB2 + XC2) >= 92 Octanes of super unleaded: (120X A1 + 90XB1 + 130XC1) / (X A3 + XB3 + XC3) >= 96
All constraints 1. 2. 3.
26X A1 28X A2 24X A3
- 4XB1 - 2XB2 + 6XB3
+ + +
36XC1 38XC2 34XC3
>= >= >=
0 0 0
4. 5. 6.
X A1 XB1 XC1
+ X A2 + XB2 + XC2
+ + +
X A3 XB3 XC3
<= <= <=
1000 1200 700
7. 8. 9.
X A1 X A2 X A3
+ XB1 + XB2 + XB3
+ + +
XC1 XC2 XC3
>= >= >=
800 1100 500
Xi
>= 0,
i = A,B,C and
Minimum number of octanes
Ingredients’ consumption
Products’ demand
j = 1,2,3
Summarized results Optimal solution results Variable
Value (tones)
Unit contribution to the profit (€) (3)
(1) X A1
(2) 0
47
0
X A2
500
42
21000
X A3
500
50
25000
XB1
720
43
30960
XB2
480
38
18240
XB3
0
46
0
XC1
80
-20
-1600
XC2
120
-25
-3000
XC3
0
-17
0 Total
Total contribution (€) (4) = (2) x (3)
90600
Solutions with QSB
QSB+ (1) |---------------- Solution Summary for case study 9 -------------------------| |----------------------------------------------------------------------------| | Variable | Variable | |Opportuni-| Minimum | Current | Maximum | | Number | Name | Solution | ty Cost |Obj. Coef.|Obj. Coef.|Obj. Coef.| |----------+----------+----------+----------+----------+----------+----------| | 1 | XA1 | 0 | 0 | -M | 47 | 47| | 2 | XA2 | 500 | 0 | 42 | 42 | 42| | 3 | XA3 | 500 | 0 | 50 | 50 | 67| | 4 | XB1 | 720 | 0 | 43 | 43 | 65.22222| | 5 | XB2 | 480 | 0 | 38 | 38 | 38| | 6 | XB3 | 0 | 0 | -M | 46 | 46| | 7 | XC1 | 80 | 0 | -20 | -20 | -20| | 8 | XC2 | 120 | 0 | -25 | -25 | -25| | 9 | XC3 | 0 | 0 | -M | -17 | -17| |----------------------------------------------------------------------------| | Maximized OBJ = 90600 Iteration = 7 Elapsed CPU seconds = 0 | |----------------------------------------------------------------------------|