Methods and Models of Optimization

Abstract

Most engineering problems, including planning, control and design, have more than one solution. The theory of optimization provides a mathematical basis for the establishing the acceptability conditions that outline the class of acceptable solutions, for the definition of the criterion that provides the measure of goodness of every individual solution, and the optimization procedure (algorithm) that results in finding the optimal solution, i.e. the solution maximizing the value of the goodness criterion. These three components, the class of acceptable solutions, the criterion of goodness, and the optimization procedure are to be present in any definition of the optimization problem.

Solutions

4.1.1 Exercise 4.1: Problem 1

The first constraint reflects the requirement on the total weight of the mixture:

$$ {x}_1+{x}_2+{x}_3+{x}_4+{x}_5+{x}_6=500 $$

The following expressions represent the required concentrations of each chemical ingredient in the final mixture:

$$ Fe=.15\cdot {x}_1+.40\cdot {x}_2+.35\cdot {x}_3+.16\cdot {x}_4+.33\cdot {x}_5+.07\cdot {x}_6\ge\ .20\cdot 500 $$

$$ Zn=.38\cdot {x}_1+.12\cdot {x}_2+.05\cdot {x}_3+.11\cdot {x}_4+.01\cdot {x}_5+.23\cdot {x}_6\ge\ .10\cdot 500 $$

$$ Si{O}_2=.41\cdot {x}_1+.40\cdot {x}_2+.27\cdot {x}_3+.21\cdot {x}_4+.60\cdot {x}_5+.45\cdot {x}_6\le\ .42\cdot 500 $$

$$ Cu=.06\cdot {x}_1+.01\cdot {x}_2+.28\cdot {x}_3+.18\cdot {x}_4+.05\cdot {x}_5+.25\cdot {x}_6\ge\ .05\cdot 500 $$

The feasibility constraints reflect the availability of the materials. The amount of each material used is equal to the percentage of that material multiplied by the total weight of the end mixture. This must be no greater than the available weight of the material.

$$ {x}_1\le 250 $$

$$ {x}_2\le 590 $$

$$ {x}_3\le 1000 $$

$$ {x}_4\le 520 $$

$$ {x}_5\le 2500 $$

$$ {x}_6\le 800 $$

It should be noted that all variables of this problem are non-negative, but this requirement is very common for linear programming problems and is addressed by the solution algorithm.

The “minimum cost” requirement is addressed as follows:

$$ 120\cdot {x}_1+150\cdot {x}_2+211\cdot {x}_3+140\cdot {x}_4+75\cdot {x}_5+214\cdot {x}_6\ \to Min $$

The problem solution is obtained by the use of the linear programming software available in MATLAB. The optimal weight of each material in tons is:

x 1	x 2	x 3	x 4	x 5	x 6	Total	Cost,$
68.2363	0.0000	0.0000	197.5259	234.2378	0.0000	500.0000	53409.80

and the chemical composition of the mixture is

Fe%	Zn%	SiO 2%	Cu%
23.83	10.00	42.00	10.27

4.1.2 Exercise 4.1: Problem 2

For this problem, we were required to use the gradient-based LSM procedure to find the optimal solution of the a coefficients in the following equation.

$$ {y}^{mod}(k)={a}_1\cdot {x}_1(k)+{a}_2\cdot {x}_2(k)+{a}_3\cdot {x}_3(k)+{a}_4\cdot {x}_4(k) $$

The method for the gradient-based LSM is a simple iterative procedure which “moves” the point representing unknown coefficients in four-dimensional space in the direction toward the minimum value of the criterion. In this case, the criterion, Q, is calculated as the sum of squared values of the discrepancy e(k) = y(k)−y ^mod(k):

$$ {A}_{new}=\kern-.2em \left[\begin{array}{c}\hfill A(1)-\gamma \cdot \frac{\varDelta Q\left[A(1),A(2),A(3),A(4)\right]}{\varDelta (1)}\hfill \\ {}\hfill A(2)-\gamma \cdot \frac{\varDelta Q\left[A(1),A(2),A(3),A(4)\right]}{\varDelta (2)}\hfill \\ {}\hfill A(3)-\gamma \cdot \frac{\varDelta Q\left[A(1),A(2),A(3),A(4)\right]}{\varDelta (3)}\hfill \\ {}\hfill A(4)-\gamma \cdot \frac{\varDelta Q\left[A(1),A(2),A(3),A(4)\right]}{\varDelta (4)}\hfill \end{array}\right] $$

where $\frac{\varDelta Q\left[A(1),A(2),A(3),A(4)\right]}{\varDelta (i)}$ are estimated partial derivatives of the LSM criterion Q with respect to particular coefficients (i = 1,2,3,4) chosen to be 0.0001, and γ > 0, is a scaling factor. Initially, γ is chosen to be 0.02, however, in the case of an unsuccessful step leading to an increase of criterion Q criterion instead of a decrease, the magnitude of gamma is cut in half. This ensures that the procedure will converge.

The results of this procedure were reached after 250 iterations, starting with zero initial conditions. This procedure could be less accurate but also faster if the termination conditions were made less strict. For this termination condition, the change between the newly generated Q and the previous Q needs to be less than .0000001 in absolute. The optimal result was:

$$ A=\kern-.2em \left[\begin{array}{c}\hfill 2.0001\hfill \\ {}\hfill 2.9918\hfill \\ {}\hfill \begin{array}{c}\hfill -2.0001\hfill \\ {}\hfill 5.0235\hfill \end{array}\hfill \end{array}\right] $$

Since the coefficient of determination for this model is 0.9996, this is an excellent model of our linear system.

4.1.3 Exercise 4.1: Problem 3

This problem differs from the previous one because of the additional requirement: all model parameters are to be positive. This condition is achieved by the use of penalty functions added to the original LSM criterion Q. In this problem the criterion to be minimized is:

$$ {Q}_1=Q+{\displaystyle \sum_{i=1}^4{P}_i} $$

where $Pi=\left\{\begin{array}{c}\hfill 0\kern0.5em \mathrm{if}\ A(i)\ge 0\hfill \\ {}\hfill {10}^{10}\cdot A{(i)}^2\kern0.5em \mathrm{if}\ A(i)<0\hfill \end{array}\right.$, i = 1,2,3,4

The optimal result, with a coefficient of determination of 0.9348, reached after 300 iterations was:

$$ A=\left[\begin{array}{c}\hfill 1.0790\hfill \\ {}\hfill 2.4456\hfill \\ {}\hfill .0009\hfill \\ {}\hfill 4.4795\hfill \end{array}\right] $$

C omparing the coefficient of determination to the one from Problem 2, 0.9348 < 0.9996 Therefore the coefficients found in Problem 2 are a better representation of the actual system. Since the actual system includes a negative coefficient for a₃, not allowing negative coefficients in Problem 3 impacted the ability of the optimization to get close to the actual coefficient values.

4.1.4 Exercise 4.2: Problem 1

For this problem, we were given the following transfer function of the controlled plant, and were required to use the Simplex code provided to find the optimal coefficients of a PID controller. The PID controller was configured to use the system error as its input for the controller.

As could be seen, the optimization criterion, known as “integral-time-error-squared” was chosen. The Simplex procedure began with zero initial coefficient values, and progressively changed these values, minimizing the criterion. All 30 iteration of the Simplex procedure are shown below.

Iteration	k_P	k_I	k_D	Q
1	0.000	0.000	0.000	50.00000
2	0.094	0.024	0.024	38.18000
3	0.024	0.094	0.024	25.61000
4	0.024	0.024	0.094	41.24000
5	0.094	0.094	0.094	24.15000
6	0.141	0.141	0.141	17.33000
7	0.149	0.149	0.031	16.29000
8	0.212	0.212	0.000	10.76000
9	0.157	0.275	0.086	7.86200
10	0.189	0.401	0.118	4.21000
11	0.338	0.409	0.149	3.87000
12	0.495	0.566	0.212	2.06500
13	0.456	0.644	0.079	1.62100
14	0.613	0.896	0.047	0.86220
15	0.652	1.029	0.251	0.72740
16	0.872	1.438	0.377	0.48830
17	1.131	1.532	0.306	0.39070
18	1.603	2.098	0.401	0.29570
19	1.563	2.388	0.338	0.34080
20	2.079	3.054	0.697	0.29080
21	2.813	4.133	1.021	0.26310
22	3.114	4.308	0.796	0.25040
23	4.235	5.743	1.006	0.21720
24	4.203	5.594	1.281	0.19470
25	5.523	7.197	1.752	0.15960
26	6.778	9.284	2.119	0.14560
27	9.365	12.877	2.978	0.11670
28	9.936	13.079	2.802	0.10360
29	13.498	17.552	3.693	0.08105
30	14.689	19.341	4.609	0.08550

The following plots illustrate gradual improvement of the closed-loop step response of the system, iterations 1, 2, 8, 12, 14, 16 are shown below . Technically, the procedure could be terminated after the 16-th iteration when the design requirements were met.

1	2
8	12
14	16

4.1.5 Exercise 4.2: Problem 2

The task is to estimate parameters a ₁, a ₂, a ₃, a ₄, a ₅ of the following model

$$ {y}^{MOD}={a}_1{x}_1{e}^{a_2{x}_2}+{a_3}^{\left({a}_4{x}_3+{a}_5\right)} $$

based on the input–output data presented in the table below. It could be seen that while the Least Squares Method (LSM) is to be applied, due to the nonlinearity of the model, traditional LSM equation is unusable and the LSM criterion,

$$ Q\left({a}_1,\ {a}_2,\ {a}_3,\ {a}_4,\ {a}_5\right)={\displaystyle \sum_i{\left[y(i)-{y}^{MOD}(i)\right]}^2} $$

could be minimized only by a genetic optimization (GO) procedure.

Assume that a “generation” size is 20. To start the procedure, begin with 20 randomly generated sets of 5 coefficients compiled into in a 20 × 5 matrix, with each row representing a particular set of coefficients (an individual). In each generation these 20 “individuals” will become the source of off-spring and mutants. Each combination of two individuals within a generation results in 5 off-spring. Each individual in the generation originates 5 mutants.

It could be seen that this process results in an immense number of new individuals, however, each newly created individual is subjected to the “fitness test” and only the 20 most fit individuals are included in the next generation. Each “next generation” is subjected to the same treatment until some termination conditions are met.

Below are several successive generation created by GO and accompanying values of the coefficient of determination of the resultant model.

Initial generation

a 1	a 2	a 3	a 4	a 5
2.8985	1.1513	1.3472	0.7978	−1.7259
1.6345	0.3528	−3.1390	−0.4051	−1.6175
4.0328	−0.2817	−0.8104	−0.2262	2.0505
0.5651	−0.3717	−1.4612	2.1874	0.2249
0.0178	−2.6064	0.7837	2.0933	0.9262
−0.4143	1.3217	0.6731	−1.0293	0.6581
0.3766	2.1826	2.5247	−1.7790	−0.3017
1.9690	−1.3128	−0.5600	2.6983	−2.0503
1.5223	−1.7655	−1.0972	−6.0455	1.9243
−1.9364	2.2881	−2.5293	1.9482	−1.2985
−0.1199	3.7093	0.4981	0.0415	−0.1648
−2.6940	2.6746	−2.0002	1.0056	−0.6448
−1.6309	3.0015	−1.4742	2.2930	2.2985
−1.6146	3.2141	3.1189	1.3045	−4.4747
3.7037	−0.5679	0.2711	−1.9776	0.4899
1.3725	−0.5816	0.3215	−2.4576	1.0954
0.9418	2.1572	−0.3685	3.4800	−3.1552
−2.9340	−4.9244	−0.0846	0.5385	−1.4570
3.5418	−0.5850	−0.3802	−4.0273	0.3840
0.5160	−4.4928	−0.1777	3.0757	−1.5804

Generation 1

a 1	a 2	a 3	a 4	a 5	Determination coefficient
2.8985	1.1513	1.3472	0.7978	−1.7259	0.8825
2.2881	1.2683	1.3268	0.3294	−0.2899	0.8756
3.1209	1.2110	1.8546	−1.8284	0.2756	0.8755
2.5022	1.2605	0.6157	−0.9998	1.3549	0.8656
1.8744	1.2679	−0.9757	0.6641	2.0549	0.8601
1.1644	1.4379	−1.7683	−1.0123	−0.8917	0.8532
1.6140	1.2957	−0.5244	−0.5290	1.4760	0.8408
0.7947	1.6388	0.9696	−2.4417	0.6111	0.8397
2.0885	1.1947	0.8106	−0.2046	−0.5338	0.8384
0.8033	1.5508	2.2729	−0.0392	−0.5913	0.8352
2.5520	1.1144	0.4245	−0.3246	0.5987	0.8319
1.1761	1.5435	1.4602	−2.2510	−0.0922	0.8282
3.4343	1.0007	−0.5871	0.2609	1.1236	0.8243
1.6577	1.2627	−0.9837	1.3665	−0.4513	0.8215
1.3407	1.4972	−2.5164	0.8116	−2.9722	0.8175
2.9522	1.0467	0.1449	0.5600	1.8093	0.8157
0.6689	1.5793	−0.7417	2.8365	0.1993	0.8074
2.7650	1.0530	1.1006	0.2516	−1.6950	0.7990
1.8029	1.2170	−1.9619	1.3318	−1.5114	0.7959
2.8907	1.2643	0.7140	0.1415	−2.4496	0.7925

Generation 3

a 1	a 2	a 3	a 4	a 5	Determination coefficient
3.2330	1.1311	0.1180	0.1429	−1.5847	0.9203
3.4089	1.1194	0.1505	0.2626	−1.3844	0.9178
3.0811	1.1600	0.1540	0.3674	−1.2171	0.9152
3.2224	1.1584	0.1650	0.2263	−1.4428	0.9142
3.2893	1.1546	0.1817	0.8279	−0.1388	0.9137
2.5298	1.2368	0.0907	0.5090	−0.1884	0.9114
3.0611	1.1398	0.0935	0.0080	−1.5563	0.9112
2.4017	1.2431	0.2474	0.7796	−0.6685	0.9102
3.3596	1.1058	0.1555	0.1667	−1.3748	0.9095
2.8410	1.1452	0.1536	0.0956	−1.7017	0.9089
2.4652	1.2277	0.0785	0.3539	−0.4969	0.9087
2.2753	1.2568	0.2190	0.5816	−0.8525	0.9073
2.3878	1.2545	0.1447	0.5450	−0.3459	0.9068
2.6861	1.1952	0.2399	0.2100	−1.6736	0.9057
3.1428	1.1634	0.2227	0.1615	−1.6702	0.9056
2.6985	1.1850	0.2020	0.1200	−1.6083	0.9054
2.4388	1.2495	0.0313	0.3191	−0.1237	0.9054
3.2899	1.1293	0.2615	0.2999	−1.4908	0.9051
2.3510	1.2336	0.2180	0.3131	−1.4478	0.9048
2.8936	1.1421	0.1960	0.1882	−1.6563	0.9044

Generation 5

a 1	a 2	a 3	a 4	a 5	Determination coefficient
3.2187	1.1474	0.1081	0.1070	−1.5610	0.9243
3.2307	1.1444	0.1065	0.0930	−1.5475	0.9242
3.1887	1.1535	0.1066	0.0969	−1.5458	0.9241
3.1930	1.1516	0.1153	0.0942	−1.6730	0.9241
3.1678	1.1503	0.1164	0.1062	−1.5840	0.9239
3.3344	1.1401	0.1163	0.1289	−1.5773	0.9239
3.2312	1.1458	0.1056	0.0839	−1.5436	0.9239
3.1755	1.1476	0.1139	0.0927	−1.5828	0.9238
3.1045	1.1533	0.1032	0.0814	−1.5745	0.9238
3.1020	1.1555	0.1101	0.0826	−1.5823	0.9238
3.1053	1.1523	0.1109	0.0850	−1.5895	0.9237
3.2088	1.1392	0.1033	0.0831	−1.5563	0.9237
3.3173	1.1454	0.1160	0.1322	−1.5786	0.9236
3.1123	1.1533	0.1164	0.1001	−1.5813	0.9236
3.2996	1.1295	0.1009	0.0858	−1.5604	0.9236
3.3068	1.1395	0.1328	0.1453	−1.6407	0.9236
3.3615	1.1333	0.1098	0.1346	−1.5013	0.9235
3.2392	1.1462	0.1213	0.1365	−1.5729	0.9235
3.2177	1.1370	0.1105	0.0952	−1.5843	0.9235
3.0982	1.1542	0.1172	0.1026	−1.5828	0.9235

Generation 7

a 1	a 2	a 3	a 4	a 5	Determination coefficient
3.2964	1.1409	0.1069	0.1003	−1.5657	0.9244
3.2921	1.1406	0.1063	0.0987	−1.5740	0.9244
3.2906	1.1400	0.1057	0.1014	−1.5538	0.9244
3.2617	1.1426	0.1080	0.1004	−1.5746	0.9244
3.2937	1.1406	0.1141	0.1063	−1.6061	0.9244
3.2497	1.1445	0.1063	0.1002	−1.5612	0.9244
3.2837	1.1424	0.1048	0.0966	−1.5579	0.9244
3.2246	1.1478	0.1122	0.1016	−1.6072	0.9244
3.2764	1.1418	0.1104	0.1047	−1.5810	0.9244
3.2292	1.1473	0.1077	0.1007	−1.5728	0.9244
3.2272	1.1467	0.1149	0.1005	−1.6283	0.9244
3.2825	1.1422	0.1085	0.0975	−1.5793	0.9244
3.2831	1.1400	0.1081	0.1029	−1.5699	0.9244
3.2317	1.1464	0.1090	0.1021	−1.5838	0.9244
3.2291	1.1462	0.1074	0.0996	−1.5691	0.9244
3.2202	1.1473	0.1083	0.1002	−1.5767	0.9244
3.2817	1.1425	0.1073	0.1012	−1.5615	0.9244
3.2629	1.1422	0.1116	0.1037	−1.5985	0.9244
3.2778	1.1426	0.1187	0.1053	−1.6354	0.9244
3.2414	1.1462	0.1072	0.1030	−1.5708	0.9244

Generation 9

a 1	a 2	a 3	a 4	a 5	Determination coefficient
3.2931	1.1403	0.1067	0.0996	−1.5671	0.9244
3.2946	1.1402	0.1062	0.0997	−1.5652	0.9244
3.2915	1.1403	0.1077	0.1000	−1.5749	0.9244
3.2941	1.1402	0.1060	0.0996	−1.5644	0.9244
3.2921	1.1404	0.1069	0.0999	−1.5686	0.9244
3.2934	1.1404	0.1059	0.0986	−1.5632	0.9244
3.2905	1.1404	0.1056	0.0991	−1.5608	0.9244
3.2936	1.1404	0.1113	0.1011	−1.5986	0.9244
3.2937	1.1404	0.1064	0.0998	−1.5654	0.9244
3.2936	1.1402	0.1061	0.1000	−1.5628	0.9244
3.2917	1.1402	0.1062	0.0996	−1.5629	0.9244
3.2895	1.1405	0.1055	0.0987	−1.5605	0.9244
3.2943	1.1403	0.1063	0.1001	−1.5638	0.9244
3.2900	1.1404	0.1057	0.0986	−1.5612	0.9244
3.2913	1.1403	0.1060	0.0996	−1.5622	0.9244
3.2918	1.1404	0.1064	0.0999	−1.5648	0.9244
3.2902	1.1402	0.1056	0.0985	−1.5612	0.9244
3.2901	1.1406	0.1072	0.0997	−1.5719	0.9244
3.2918	1.1406	0.1057	0.0988	−1.5631	0.9244
3.2933	1.1405	0.1065	0.0999	−1.5674	0.9244

After 9 iterations of the GO procedure stopped, and the resultant value of the coefficient of determination was 0.9244 for coefficients [3.2931, 1.1403, 0.1067, 0.0996, −1.5671], and the model expression is: y ^MOD = $3.2931{x}_1{e}^{1.1403{x}_2}+{0.1067}^{0.0996{x}_3-1.5671}$

4.1.6 Exercise 4.3: Problem 1

Conditional optimization of Process IV:

If ${x}_4=\left[10,40\right]$

• Choose u ₄ = 2

• Cost = 4

If ${x}_4=\left(40,70\right]$

• Choose u ₄ = 3

• Cost = 4

If ${x}_4=\left(70,100\right]$

• Choose u ₄ = 3

• Cost = 3

Conditional optimization of Process III and Process IV:

If ${x}_3=\left[10,40\right]$

• If u ₃ = 1, cost = 16 + cost(x ₄ = 13) = 16 + 4 = 20

• If u ₃ = 2, cost = 18 + cost(x ₄ = 45) = 18 + 4 = 22

• If u ₃ = 3, cost = 9 + cost(x ₄ = 92) = 9 + 3 = 12 (optimal)

• Choose u ₃ = 3

• Cost = 12

If ${x}_3=\left(40,70\right]$

• If u ₃ = 1, cost = 13 + cost(x ₄ = 48) = 13 + 4 = 17

• If u ₃ = 2, cost = 17 + cost(x ₄ = 18) = 17 + 4 = 21

• If u ₃ = 3, cost = 8 + cost(x ₄ = 68) = 8 + 4 = 12 (optimal)

• Choose u ₃ = 3

• Cost = 12

If ${x}_3=\left(70,100\right]$

• If u ₃ = 1, cost = 10 + cost(x ₄ = 81) = 10 + 3 = 13

• If u ₃ = 2, cost = 14 + cost(x ₄ = 66) = 14 + 4 = 18

• If u ₃ = 3, cost = 6 + cost(x ₄ = 21) = 6 + 4 = 10 (optimal)

• Choose u ₃ = 3

• Cost = 10

Conditional optimization of Process II, Process III and Process IV:

If ${x}_2=\left[10,40\right]$

• If u ₂ = 1, cost = 13 + cost(x ₃ = 65) = 13 + 12 = 25 (optimal)

• If u ₂ = 2, cost = 21 + cost(x ₃ = 44) = 21 + 12 = 33

• If u ₂ = 3, cost = 33 + cost(x ₃ = 74) = 33 + 10 = 43

• Choose u ₂ = 1

• Cost = 25

If ${x}_2=\left(40,70\right]$

• If u ₂ = 1, cost = 15 + cost(x ₃ = 66) = 15 + 12 = 27 (optimal)

• If u ₂ = 2, cost = 22 + cost(x ₃ = 50) = 22 + 12 = 33

• If u ₂ = 3, cost = 37 + cost(x ₃ = 81) = 37 + 10 = 47

• Choose u ₂ = 1

• Cost = 27

If ${x}_2=\left(70,100\right]$

• If u ₂ = 1, cost = 18 + cost(x ₃ = 78) = 18 + 10 = 28 (optimal)

• If u ₂ = 2, cost = 28 + cost(x ₃ = 62) = 28 + 12 = 40

• If u ₂ = 3, cost = 40 + cost(x ₃ = 96) = 40 + 10 = 50

• Choose u ₂ = 1

• Cost = 28

Conditional optimization of Process I, Process II, Process III, and Process IV:

If ${x}_1=\left[10,40\right]$

• If u ₁ = 1, cost = 25 + cost(x ₂ = 25) = 25 + 25 = 50 (optimal)

• If u ₁ = 2, cost = 28 + cost(x ₂ = 45) = 28 + 27 = 55

• If u ₁ = 3, cost = 25 + cost(x ₂ = 55) = 25 + 27 = 52

• Choose u ₁ = 1

• PATH: u ₁ = 1 $\to\ {u}_2$ = 1 $\to\ {u}_3$ = 3 $\to\ {u}_4$ = 3

• Cost = 50

If ${x}_1=\left(40,70\right]$

• If u ₁ = 1, cost = 27 + cost(x ₂ = 37) = 27 + 25 = 52 (optimal)

• If u ₁ = 2, cost = 33 + cost(x ₂ = 48) = 33 + 27 = 60

• If u ₁ = 3, cost = 27 + cost(x ₂ = 63) = 27 + 28 = 55

• Choose u ₁ = 1

• PATH: u ₁ = 1 $\to\ {u}_2$ = 1 $\to\ {u}_3$ = 3 $\to\ {u}_4$ = 3

• Cost = 52

If ${x}_1=\left(70,100\right]$

• If u ₁ = 1, cost = 22 + cost(x ₂ = 45) = 22 + 27 = 49 (optimal)

• If u ₁ = 2, cost = 24 + cost(x ₂ = 58) = 24 + 27 = 51

• If u ₁ = 3, cost = 25 + cost(x ₂ = 79) = 25 + 28 = 53

• Choose u ₁ = 1

• PATH: u ₁ = 1 $\to\ {u}_2$ = 1 $\to\ {u}_3$ = 3 $\to\ {u}_4$ = 3

• Cost = 49

4.1.6.1 Optimal Plan

Since x = 37, the optimal path is u ₁ = 1 $\to\ {u}_2$ = 1 $\to\ {u}_3$ = 3 $\to\ {u}_4$ = 3 and the cost = 50.

4.1.7 Exercise 4.3: Problem 2

Conditional optimization of particular stages of the process starting from the last stage:

Stage 5 → 6

• If at A5 → A6 − Cost = 8 (optimal)

• If at B5 → A6 − Cost = 7 (optimal)

Stage 4 → 5

• If at A4 → A5 – Cost = 15 + 8 = 23

•     → B5 − Cost = 10 + 7 = 17 (optimal)

• If at B4 → A5 – Cost = 19 + 8 = 27

•     → B5 − Cost = 14 + 7 = 21 (optimal)

• If at C4 → A5 – Cost = 16 + 8 = 24

•     → B5 − Cost = 13 + 7 = 20 (optimal)

Stage 3 → 4

• If at A3 → A4 – Cost = 6 + 17 = 23

•     → B4 – Cost = 4 + 21 = 25

•     → C4 − Cost = 2 + 20 = 22 (optimal)

• If at B3 → A4 − Cost = 7 + 17 = 24 (optimal)

•     → B4 – Cost = 4 + 21 = 25

•     → C4 – Cost = 12 + 20 = 32

• If at C3 → A4 − Cost = 5 + 17 = 22 (optimal)

•     → B4 − Cost = 3 + 21 = 24

•     → C4 − Cost = 7 + 20 = 27

Stage 2 → 3

• If at A2 → A3 – Cost = 2 + 22 = 24

•     → B3 – Cost = 3 + 24 = 27

•     → C3 − Cost = 1 + 22 = 23 (optimal)

• If at B2 → A3 – Cost = 9 + 22 = 31

•     → B3 – Cost = 2 + 24 = 26

•     → C3 − Cost = 2 + 22 = 24 (optimal)

• If at C2 → A3 – Cost = 6 + 22 = 28

•     → B3 – Cost = 4 + 24 = 28

•     → C3 − Cost = 3 + 22 = 25 (optimal)

• If at D2 → A3 − Cost = 3 + 22 = 25 (optimal)

•     → B3 – Cost = 3 + 24 = 27

•     → C3 – Cost = 4 + 22 = 26

Stage 1 → 2

• If at A1 → A2 – Cost = 7 + 23 = 30

•     → B2 − Cost = 4 + 24 = 28 (optimal)

•     → C2 – Cost = 5 + 25 = 30

•     → D2 – Cost = 6 + 25 = 31

4.1.7.1 Optimal Path

A1 → B2 → C3 → A4 → B5 → A6, Cost: 28