Question

Problem 4-03 (Algorithmic)

The employee credit union at State University is planning the allocation of funds for the coming year. The credit union makes four types of loans to its members. In addition, the credit union invests in risk-free securities to stabilize income. The various revenueproducing investments together with annual rates of return are as follows:

Type of Loan/Investment	Annual Rate of Return (%)
Automobile loans	8
Furniture loans	12
Other secured loans	14
Signature loans	13
Risk-free securities	9

The credit union will have $1.7 million available for investment during the coming year. State laws and credit union policies impose the following restrictions on the composition of the loans and investments.

- Risk-free securities may not exceed 30 %of the total funds available for investment.

- Signature loans may not exceed 10 %of the funds invested in all loans (automobile, furniture, other secured, and signature loans).

- Furniture loans plus other secured loans may not exceed the automobile loans.

- Other secured loans plus signature loans may not exceed the funds invested in risk-free securities.

How should the $1.7 million be allocated to each of the loan/investment alternatives to maximize total annual return? Round your answers to the nearest dollar.

What is the projected total annual return? Round your answer to the nearest dollar.

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Answer 1

Let, $%, be the automobile leans. $ be the furniture loans. $ 33 be the other secured loans. & Hy be the signature loans and $25 be the risk free security

". Annual return.. 8x + 12x2 + 14x2 + 1384 +995, 100 100 Too Too Too Given that, 2n+262 +263 + 24 +K5 2 17 000 TO constraint on risk free seewrity, 365 < 30 x 17 00000 zy og < 510000 constraint on signature loan, H4 5 10 (X, +U2 + X3; +24) .2% 26, +42 +33 -9x >0 constraint on fueniture loans + other secured loans, X + Uz sa 27 21-12-U970 constraint on other secured loans + signature loans Az + xy S5 27 Xz+ky -U550. Hi totiz 1,2,3,4,5.

Hiter applying simplen, ue hace, N, 2 $53 5500 U2z$ 144500 Xz z $391000 Hy 2 $ 119000 26 2$ 510000 -Max annual return $17629.000

The simplex Calculation:

Problem is Max Z = 8X1 + 12 X2 + 14X3 + 13x4 + 9x5 subject to X1 + x2 + x3 + x4 + X5 = 1700000 X5 s 510000 X1 + x2 + x3 - 9X4 20 X1 - X2 - X3 2 0 X3 + X4 - X5 50 and X1, X2, X3, X4, X5 2 0; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type' =' we should add artificial variable A, 2. As the constraint-2 is of type's' we should add slack variable S, 3. As the constraint-3 is of type'>' we should subtract surplus variable S, and add artificial variable A2 4. As the constraint-4 is of type'z' we should subtract surplus variable S, and add artificial variable A3 5. As the constraint-5 is of type's' we should add slack variable S4

After introducing slack, surplus,artificial variables Max Z = 8x1 + 12xy + 14x3 + 13x4 + 9x5 + 05, +0S2 + 053 +054 - MA - MA2 - MA3 subject to X1 + x2 + x3 + x4 + X5 + A, = 1700000 X5 + S1 = 510000 X1 + X2 + X3 - 9X4 + A = 0 X1 - X2 - X3 X3 + x4 - Xs +S4 = 0 and X1, X2, X3, X4, X5, S1, S2, S3, S4, A4, A2, A, 20 si S2 S3 + Az=0 Iteration-1 13 0 0 0 0 -M -M -M MinRatio s sz Sz sa Az Az 1700000 0 0 0 0 1 1700000 = 1700000 0 510000 000 0 0 1 1 -1 0 0 - 0 -M Z = - 1700000M -3M -M 8M -M 0 0 1 0 MOM M 0 0 M 0 -M 0 Z;-C; -3M-81 -M-12 -M-14 8M - 13 -M-9 0 M 0

Negative minimum 2;-Cis -3M - 8 and its column index is 1. So, the entering variable is X1. Minimum ratio is 0 and its row index is 3. So, the leaving basis variable is Az- :. The pivot element is 1. Entering = x1, Departing = Az, Key Element = 1 + R3(new) = Rz(old) + R (new) = R (old) - R3(new) + R (new) = R (old) + R (new) = R (old) - R3(new) + Rs(new) = Rs(old)

Iteration-2 OM Min Ratio 1700000 1700000 001 10 = 170000 510000 0 oo 1 0 1 -2 0 -2 o=0 SA 0 0 2; Z-C; 0 8 Z = - 1700000M 0 2M +8 2 M-4 2M +8 2M-6 -19M - 72 -19M - 85 M -M-9 0 0 -2M-8 -2M-8 M M 0 0 M 0 -M 0 0 Negative minimum Z- is - 19M - 85 and its column index is 4. So, the entering variable is x4 Minimum ratio is 0 and its row index is 4. So, the leaving basis variable is Az. :. The pivot element is 9. Entering = x4, Departing = Az, Key Element = 9

Entering = x4, Departing = Az, Key Element = 9 + R_(new) = R (old) = 9 + R(new) = R, (old) - 10R_(new) + R_(new) = R (old) + Rz(new) = Rs(old) + 9R_(new) + R$(new) = Rs(old) - R (new)

Iteration-3 9 0 -M Min Ratio SAA 1700000 2.2222 2.2222 1 0 -0.1111 1.1111 1700000 2.2222 = 765000 510000 0 0 1 0 0 -1 0 0 -1 00 0 0 -0.2222 -0.2222 0.1111 -0.1111 00 - 10 0.2222 (1.2222) -0.1111 0.1111 1.2222 = 0 - Z = - 1700000M 2 -2.2222M - 10.8889 -2.2222M - 10.8889 0.1111M +1.4444 - 1.1111M - 9.4444 8 2- 60 13 0 -M -M-9 0 0 -2.2222M - 22.8889 -2.2222M - 24.8889 0.1111M +1.4444 -1.1111M -9.4444 0 0 Negative minimum Z;- is -2.2222M - 24.8889 and its column index is 3. So, the entering variable is xa- Minimum ratio is 0 and its row index is 5. So, the leaving basis variable is 54- .: The pivot element is 1.2222.

Entering = xz, Departing = 54, Key Element = 1.2222 + Rs(new) = Rs(old) = 1.2222 + R, (new) = R(old) - 2.2222R_(new) + R (new) = R (old) + Rz(new) = Rs(old) + Rs(new) + R_(new) = R (old) + 0.2222R=(new)

Iteration-4 C 8 12 14 13 XB x1 MinRatio XB Sz -M 17000000 1.8182 2.8182 0.0909 0.9091 -1.8182 1700000 - = 603225.8065 2.8182 510000 0 (1) 510000 - = 510000 - -0.8182 -0.0909 -0.9091 0.8182 8 13 14 -0.8182 -0.1818 -0.1818 0.0909 -0.0909 X4 X3 0.1818 0 0 0 Zi 1 0 0 8 0.1818 -0.8182 -0.0909 0.0909 0.8182 Z = - 1700000M -1.8182M - 6.3636 14 13 -2.8182M - 20.3636 -0.0909M -0.8182 -0.9091M -7.1818 1.8182M +20.3636 Z;-C 0 -1.8182M - 18.3636 0 0 -2.8182M - 29.3636 0 -0.0909M - 0.8182 -0.9091M - 7.1818 1.8182M + 20.3636 0 Negative minimum 2 - Cis -2.8182M - 29.3636 and its column index is 5. So, the entering variable is xs. Minimum ratio is 510000 and its row index is 2. So, the leaving basis variable is S. The pivot element is 1.

Entering = xs, Departing = S, Key Element = 1 + R (new) = R (old) + R, (new) = R (old) - 2.8182R_(new) + R3(new) = R (old) + 0.8182R, (new) + R (new) = R.(old) + 0.1818R,(new) + Rs(new) = Rs(old) + 0.8182R_(new)

Iteration-5 ci 8 0 MinRatio I XB -M 2627272727 0 (1.8182) 0 0 0 -2.8182 0.0909 0.9091 -1.8182 262727.2727 - = 144500 - 1.8182 5100000 0 -0.8182 0.8182 0 0 1 -0.0909 1 0 0 8 13 -0.9091 417272.7273 92727.2727 0.8182 1 0 -0.1818 0.1818 0.0909 -0.0909 0.1818 417272.7273 14 417272.7273 0 0.1818 1 0 0 0.8182 -0.0909 0.0909 0.8182 0.1818 = 229500000 Z = -262727.2727M + 14975454.5455 Zi Z;-C; 8 0 -1.8182M - 6.3636 -1.8182M - 18.3636 14 13 o o 9 o 2.8182M + 29.3636 2.8182M + 29.3636 -0.0909M -0.8182 -0.0909M -0.8182 -0.9091M - 7.1818 -0.9091M -7.1818 1.8182M + 20.3636 1.8182M +20.3636 -M 0 Negative minimum z; - Cis - 1.8182M - 18.3636 and its column index is 2. So, the entering variable is x, Minimum ratio is 144500 and its row index is 1. So, the leaving basis variable is A1- The pivot element is 1.8182 Entering = xz, Departing = A, Key Element = 1.8182

: The pivot element is 1.8182. Entering = x2, Departing = A, Key Element = 1.8182 + R, (new) = R(old) = 1.8182 + R (new) = R (old) + R, (new) = R(old) + 0.8182R, (new) + R (new) = R (old) + 0.1818R, (new) + R$(new) = Rs(old) - 0.1818R, (new)

Iteration-6 9 MinRatio 0 0 0 0 si SSS -1.55 0.05 0.5 -1 i 0 0 x2 x5 12 9 C 8 12 14 XBX X2 X3 144500010 510000 0 0 535500 1 0 119000 1 0 0 0 391000001 281214 2-0 0 0 0 -0.45 -0.05 X4 -0.1 0.1 13 14 1.1 -0.1 Z = 17629000 9 0.9 1 0. 2 0 0.9 0.1 2 Since all Z: -620 Hence, optimal solution is arrived with value of variables as : X1 = 535500, x2 = 144500, x3 = 391000, x4 = 119000,X5 = 510000 Max Z = 17629000

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Answer 2

Solucion Given that Type of hoan / Investment 1 Annual date of return (%) Automobile lean furniture loans Other secured leang . Signature loans Rick free securites Amount x invested " by ," Automible furniture loans . a Ivans = , 2 other Sccured loans .. " Signature loans . xq Risle free Securites 2 Maximize (2) 8% at 12% 22 of 14%*3 of 13%84+ 99 0.08% to 12x2 + 0.1403 + 0.13 24 +0.0985 Then value of att2 t as txq 425 : 7,00, DVO. ats 45 < 30% € 30 of 1700,000 x 17,00,000

KS < 0.3 x .17,00, 000 75 € 5,10,000 . . of Signature loans does not exceed funds invested in remaing all 24.ccom (,772 +23+x4) 10% loans. And the furniture & loans does not secured loans. i exceed automobile X. traça » Other exceed secured the loans funds plus signature invested in loans may na risk free Securities. . Xzt xa cx5 xt net tgt dat dy = 17,00,000 ng . 5,19,000 X4 = 10% Cvrtka t 25+*4) © X2 tagax, 13 + x4 ans - ztraq 5, 10,600]

substitute the Xz + xq & dig values in Then egn 24 = 10% ( 6,80,000 + 5, 10,000) M4 - 100 (1,90,000) Hq - 119.000] Substitecte eq xat x4 x5 23 + 1,19,000 2 5,10,000 Xpz 5,10,000 -1,19,000 [X3 3,91,000 72+23 R . Ny of 3,91,600 - : . . + a = - 3,7,000 Adding equations - Riftixas i-bist XL 6,80,000 3,91,000 2.323 . 289,000

22 = 89,000 2. og 1,44,000 89" @ 24+*3-*1 1,44,000 + 3,91,000.-21 . 2. 18, 36,00 X az de = nac $ 18, 36,000 1,44,500 $ 3,91,000 $ 119,000 The Final Anual return 7: egg, of 0.1242 +01473 + 01874 0.092 3008 (; 34,000) +0.12 ( 1,44; 500) + 01445,91,000) + 01311,19,000) + 0.00 9/ 5,10,000) - 146880 + 17,340 54,740 f15,400 t 45,900 2,60, 330 The projected annual return a 4 2,50, 330/