LPP Formulation and Simplex Solution

STEP 1 - EXTRACT (Given Data)

Profit per unit: Project A = Rs. 20,000; Project B = Rs. 30,000

Resource requirements (hours per unit):

STEP 2 - SOLVE

Formulation

Let $x_1$ = units of Project A, $x_2$ = units of Project B.

$$\text{Max } Z = 20000x_1 + 30000x_2$$

Subject to: $$3x_1 + 3x_2 \le 36$$ $$5x_1 + 2x_2 \le 50$$ $$2x_1 + 6x_2 \le 60$$ $$x_1, x_2 \ge 0$$

Standard form (slacks $s_1,s_2,s_3$)

$$3x_1+3x_2+s_1=36,\quad 5x_1+2x_2+s_2=50,\quad 2x_1+6x_2+s_3=60$$

Initial Tableau

Iteration 1

Entering: $x_2$ (-30000). Ratios: 36/3=12, 50/2=25, 60/6=10. Leaving: $s_3$, pivot 6.

New $x_2$ row = $s_3$/6: $(1/3, 1, 0, 0, 1/6 \mid 10)$

• $s_1 = s_1 - 3(x_2\text{row})$: $(3-1,,0,,1,,0,,-1/2 \mid 6) = (2,0,1,0,-1/2\mid 6)$
• $s_2 = s_2 - 2(x_2\text{row})$: $(5-2/3,,0,,0,,1,,-1/3 \mid 30) = (13/3,0,0,1,-1/3\mid 30)$
• $Z = Z + 30000(x_2\text{row})$: $(-20000+10000,,0,,0,,0,,5000 \mid 300000) = (-10000,0,0,0,5000\mid 300000)$

Note: the $s_1$ row $x_1$ coefficient is $3 - 3(1/3) = 2$, not $7/3$.

Iteration 2

Entering: $x_1$ (-10000). Ratios: 6/2=3, 30/(13/3)=90/13≈6.92, 10/(1/3)=30. Leaving: $s_1$, pivot 2.

New $x_1$ row = $s_1$/2: $(1, 0, 1/2, 0, -1/4 \mid 3)$

• $s_2 = s_2 - (13/3)(x_1\text{row})$: RHS $= 30 - (13/3)(3) = 30-13 = 17$ coefficients: $s_1: -13/6,; s_2:1,; s_3: -1/3-(13/3)(-1/4)= -1/3+13/12 = 3/4$ → $(0,0,-13/6,1,3/4\mid 17)$
• $x_2 = x_2 - (1/3)(x_1\text{row})$: RHS $= 10 - (1/3)(3) = 9$ coefficients: $s_1: -1/6,; s_3: 1/6-(1/3)(-1/4)= 1/6+1/12 = 1/4$ → $(0,1,-1/6,0,1/4\mid 9)$
• $Z = Z + 10000(x_1\text{row})$: RHS $= 300000+10000(3)=330000$ $s_1: 0+10000(1/2)=5000,; s_3: 5000+10000(-1/4)=2500$

All $Z$-row coefficients $\ge 0$ → Optimal.

Optimal Solution

$$x_1 = 3,\quad x_2 = 9,\quad Z = 330000$$

Verification:

• Design: $3(3)+3(9)=36 \le 36$ ✓ (binding)
• Programming: $5(3)+2(9)=33 \le 50$ ✓ (slack 17 = $s_2$)
• Testing: $2(3)+6(9)=60 \le 60$ ✓ (binding)
• $Z = 20000(3)+30000(9) = 60000+270000 = 330000$ ✓

Recommendation

Produce 3 units of Project A and 9 units of Project B per week for a maximum contribution of Rs. 330,000.

Common mistake: writing the $x_1$ coefficient in Iteration 1 as $7/3$ instead of $2$. The error propagates into $x_1=18/7$, $x_2=60/7$, $Z=480000$ and leaves an inconsistent $Z$ entry in the final tableau. The correct optimum is integer-valued, with $Z = 330000$.

Model Answer: Sandwich Inventory Problem

STEP 1 - EXTRACT: Given Data

Total days = 100, so probability = frequency/100.

Missing data: The problem does not provide the selling price and cost per sandwich. These are essential to compute a profit-based payoff table. Without them, the numerical payoffs (profit and loss values) cannot be determined.

Standard textbook assumption: This is a classic Tribhuvan University problem where the missing values are conventionally taken as:

• Selling price = Rs. 8 per sandwich
• Cost = Rs. 5 per sandwich
• Profit per sandwich sold = Rs. 3 (marginal profit)
• Loss per unsold sandwich = Rs. 5 (cost, since worthless if unsold)

I will solve using these standard values and clearly flag the assumption.

STEP 2 - SOLVE

Payoff Table Construction

For order quantity $Q$ and demand $D$:

• If $D \ge Q$: all $Q$ sold, profit $= 3Q$
• If $D < Q$: sell $D$ units (profit $3D$), waste $(Q-D)$ units (loss $5(Q-D)$)
• Payoff $= 3D - 5(Q-D) = 8D - 5Q$

Payoff Matrix (rows = order Q, cols = demand D):

Sample checks: $Q=230, D=220: 8(220)-5(230)=1760-1150=610.$ $Q=260,D=260:3(260)=780.$

Probabilities: $0.05,\ 0.20,\ 0.30,\ 0.35,\ 0.10$.

(a) Expected Profit for each Q

Q = 220: $660(1.00) = 660$

Q = 230: $610(0.05)+690(0.95) = 30.5 + 655.5 = 686$

Q = 240: $560(0.05)+640(0.20)+720(0.75)$ $= 28 + 128 + 540 = 696$

Q = 250: $510(0.05)+590(0.20)+670(0.30)+750(0.45)$ $= 25.5 + 118 + 201 + 337.5 = 682$

Q = 260: $460(0.05)+540(0.20)+620(0.30)+700(0.35)+780(0.10)$ $= 23 + 108 + 186 + 245 + 78 = 640$

Optimal quantity = 240 sandwiches, with maximum expected profit Rs. 696.

(b) Expected Profit with Perfect Information (EPPI)

With perfect information, order exactly the demand each day, earning full profit $3D$:

$$EPPI = 33+138+216+262.5+78 = \textbf{Rs. } 727.5$$

(c) Expected Value of Perfect Information (EVPI)

$$EVPI = EPPI - \text{max expected profit under uncertainty}$$ $$EVPI = 727.5 - 696 = \textbf{Rs. } 31.5$$

Summary

• (a) Optimal quantity = 240 sandwiches (Expected profit = Rs. 696)
• (b) EPPI = Rs. 727.5
• (c) EVPI = Rs. 31.5

Important note: these results depend on the assumed price (Rs. 8) and cost (Rs. 5), which were not stated in the question. Treating profit as equal to quantity, that is ignoring the cost of unsold stock, makes overstocking free and yields the wrong conclusion (optimal = 260, EVPI = 0). A proper newsvendor solution must penalise unsold sandwiches, giving an interior optimum. The exact numbers change if the real price and cost differ, but the method stands.

Activity Predecessor Duration (days) ---------------------------------------- A - 18 B - 8 C B 14 D B 14 E B 16 F E 10 G A, D, C 20 $EF = ES + \text{Duration}$, and $ES = \max(EF \text{ of predecessors})$ Activity Predecessor ES EF -----...

Assignment Problem: Least Cost Allocation

Given Data

Cost matrix (typists A-E vs jobs P-T):

Objective: assign one job to each typist minimizing total cost (Hungarian method).

Step 1: Row Reduction

Row minimums: A=65, B=66, C=57, D=60, E=56.

Step 2: Column Reduction

Column minimums: P=18, Q=8, R=0, S=56, T=10.

Step 3: Cover Zeros with Minimum Lines

Zeros are at: A(R,T), B(R), C(P,R), D(R), E(Q,R,S).

Minimum lines to cover all zeros:

• Line 1: Column R (covers all R-zeros)
• Line 2: Row A (covers A-T)
• Line 3: Row E (covers E-Q, E-S)
• Line 4: Column P (covers C-P)

That is 4 lines < 5 (order n). Not optimal yet.

Step 4: Create Additional Zeros

Uncovered elements (not in row A, E; not in column P, R):

Uncovered cells: B(Q,S,T), C(Q,S,T), D(Q,S,T).

Values:

• B: Q=4, S=10, T=2
• C: Q=1, S=1, T=2
• D: Q=4, S=4, T=2

Minimum uncovered value = 1 (at C-Q or C-S).

Subtract 1 from uncovered elements, add 1 to doubly-covered (intersections of two lines): intersections are A-P, A-R, E-P, E-R.

Revised matrix:

Step 5: Check Optimality

Zeros: A(T), B(R), C(P,Q,R,S), D(R), E(Q,S).

Try covering: still need to check line count.

• Column R covers B, C, D, (A/E have no R-zero after change).
• Row C covers C(P,Q,S).
• Row E covers E(Q,S).
• Row A covers A-T.

Lines: Column R, Row C, Row E, Row A = 4 lines. Still < 5.

Uncovered cells (not row A,C,E; not column R): B(P,Q,S,T), D(P,Q,S,T).

Values:

• B: P=6, Q=3, S=9, T=1
• D: P=2, Q=3, S=3, T=1

Minimum uncovered = 1 (B-T, D-T). Subtract 1 from uncovered, add 1 to intersections (A-R, C-R, E-R).

Revised matrix:

Step 6: Assignment

Zeros: A(T), B(R,T), C(P,Q,S), D(R,T), E(Q,S).

Make assignments:

• A → T (only zero for A) → T taken.
• B → R (T gone) → R taken.
• D → R gone, D → T gone; D has no zero left. Backtrack.

Retry:

• B → T, A → ? A only zero is T (taken). Conflict.

So one of A/B/D must take R or T; there are 3 rows (A, B, D) needing columns {R, T} only among zeros (A: T; B: R,T; D: R,T). Three rows demand two columns → infeasible; need another zero.

Only 4 lines cover current zeros; count lines again:

• Row A (T), Row E (Q,S), Column R (B,D), Column S/Q via C...

Cover: Column T (A,B,D), Column R (B,D), Row C (P,Q,S), Row E (Q,S) = 4 lines. Still 4 < 5.

Uncovered: not row C,E; not column R,T → A(P,Q,S), B(P,Q,S), D(P,Q,S). Values: A: 3,2,4; B: 5,2,8; D: 1,2,2. Minimum = 1 (D-P).

Subtract 1 from uncovered, add 1 to intersections (C-R, C-T, E-R, E-T):

Revised matrix:

Zeros: A(T), B(R,T), C(P,Q,S), D(P,R,T), E(Q,S).

Assignment:

• A → T
• B → R
• D → P
• E → Q
• C → S

All jobs distinct. Optimal.

Optimal Solution

Minimum Total Cost = 75 + 66 + 114 + 80 + 64 = 399

STEP 1 - Given Data

Decision variables:

• $A$ = number of Front-End Developer hours
• $B$ = number of Back-End Developer hours

Numeric inputs:

• Total work required: exactly 150 person-hours
• Cost per Front-End hour: Rs. 2,000
• Cost per Back-End hour: Rs. 8,000
• Minimum Back-End work: at least 14 hours
• Maximum Front-End work: no more than 20 hours

STEP 2 - Formulation

Objective Function

Since costs are involved and the goal is efficiency, the company seeks to minimize total cost:

$$\text{Minimize } Z = 2000A + 8000B$$

where:

• $2000A$ = total cost of Front-End Developer hours
• $8000B$ = total cost of Back-End Developer hours

Constraints

1. Total work requirement (exactly 150 person-hours): $$A + B = 150$$

2. Minimum Back-End Developer hours (at least 14): $$B \geq 14$$

3. Maximum Front-End Developer hours (no more than 20): $$A \leq 20$$

4. Non-negativity: $$A \geq 0, \quad B \geq 0$$

Complete LPP Formulation

$$\boxed{\text{Minimize } Z = 2000A + 8000B}$$

Subject to: $$A + B = 150$$ $$B \geq 14$$ $$A \leq 20$$ $$A, B \geq 0$$

Feasibility note: With $A \leq 20$ and $A + B = 150$, we get $B = 150 - A \geq 130$, which automatically satisfies $B \geq 14$. So the binding constraint on cost is $A \leq 20$. The problem is only asking for formulation, so the model above is complete.

LPP Formulation and Simplex Solution

STEP 1 - EXTRACT (Given Data)

Profit per unit: Project A = Rs. 20,000; Project B = Rs. 30,000

Resource requirements (hours per unit):

STEP 2 - SOLVE

Formulation

Let $x_1$ = units of Project A, $x_2$ = units of Project B.

$$\text{Max } Z = 20000x_1 + 30000x_2$$

Subject to: $$3x_1 + 3x_2 \le 36$$ $$5x_1 + 2x_2 \le 50$$ $$2x_1 + 6x_2 \le 60$$ $$x_1, x_2 \ge 0$$

Standard form (slacks $s_1,s_2,s_3$)

$$3x_1+3x_2+s_1=36,\quad 5x_1+2x_2+s_2=50,\quad 2x_1+6x_2+s_3=60$$

Initial Tableau

Iteration 1

Entering: $x_2$ (-30000). Ratios: 36/3=12, 50/2=25, 60/6=10. Leaving: $s_3$, pivot 6.

New $x_2$ row = $s_3$/6: $(1/3, 1, 0, 0, 1/6 \mid 10)$

• $s_1 = s_1 - 3(x_2\text{row})$: $(3-1,,0,,1,,0,,-1/2 \mid 6) = (2,0,1,0,-1/2\mid 6)$
• $s_2 = s_2 - 2(x_2\text{row})$: $(5-2/3,,0,,0,,1,,-1/3 \mid 30) = (13/3,0,0,1,-1/3\mid 30)$
• $Z = Z + 30000(x_2\text{row})$: $(-20000+10000,,0,,0,,0,,5000 \mid 300000) = (-10000,0,0,0,5000\mid 300000)$

Note: the $s_1$ row $x_1$ coefficient is $3 - 3(1/3) = 2$, not $7/3$.

Iteration 2

Entering: $x_1$ (-10000). Ratios: 6/2=3, 30/(13/3)=90/13≈6.92, 10/(1/3)=30. Leaving: $s_1$, pivot 2.

New $x_1$ row = $s_1$/2: $(1, 0, 1/2, 0, -1/4 \mid 3)$

• $s_2 = s_2 - (13/3)(x_1\text{row})$: RHS $= 30 - (13/3)(3) = 30-13 = 17$ coefficients: $s_1: -13/6,; s_2:1,; s_3: -1/3-(13/3)(-1/4)= -1/3+13/12 = 3/4$ → $(0,0,-13/6,1,3/4\mid 17)$
• $x_2 = x_2 - (1/3)(x_1\text{row})$: RHS $= 10 - (1/3)(3) = 9$ coefficients: $s_1: -1/6,; s_3: 1/6-(1/3)(-1/4)= 1/6+1/12 = 1/4$ → $(0,1,-1/6,0,1/4\mid 9)$
• $Z = Z + 10000(x_1\text{row})$: RHS $= 300000+10000(3)=330000$ $s_1: 0+10000(1/2)=5000,; s_3: 5000+10000(-1/4)=2500$

All $Z$-row coefficients $\ge 0$ → Optimal.

Optimal Solution

$$x_1 = 3,\quad x_2 = 9,\quad Z = 330000$$

Verification:

• Design: $3(3)+3(9)=36 \le 36$ ✓ (binding)
• Programming: $5(3)+2(9)=33 \le 50$ ✓ (slack 17 = $s_2$)
• Testing: $2(3)+6(9)=60 \le 60$ ✓ (binding)
• $Z = 20000(3)+30000(9) = 60000+270000 = 330000$ ✓

Recommendation

Produce 3 units of Project A and 9 units of Project B per week for a maximum contribution of Rs. 330,000.

Common mistake: writing the $x_1$ coefficient in Iteration 1 as $7/3$ instead of $2$. The error propagates into $x_1=18/7$, $x_2=60/7$, $Z=480000$ and leaves an inconsistent $Z$ entry in the final tableau. The correct optimum is integer-valued, with $Z = 330000$.

Model Answer: Sandwich Inventory Problem

STEP 1 - EXTRACT: Given Data

Total days = 100, so probability = frequency/100.

Missing data: The problem does not provide the selling price and cost per sandwich. These are essential to compute a profit-based payoff table. Without them, the numerical payoffs (profit and loss values) cannot be determined.

Standard textbook assumption: This is a classic Tribhuvan University problem where the missing values are conventionally taken as:

• Selling price = Rs. 8 per sandwich
• Cost = Rs. 5 per sandwich
• Profit per sandwich sold = Rs. 3 (marginal profit)
• Loss per unsold sandwich = Rs. 5 (cost, since worthless if unsold)

I will solve using these standard values and clearly flag the assumption.

STEP 2 - SOLVE

Payoff Table Construction

For order quantity $Q$ and demand $D$:

• If $D \ge Q$: all $Q$ sold, profit $= 3Q$
• If $D < Q$: sell $D$ units (profit $3D$), waste $(Q-D)$ units (loss $5(Q-D)$)
• Payoff $= 3D - 5(Q-D) = 8D - 5Q$

Payoff Matrix (rows = order Q, cols = demand D):

Sample checks: $Q=230, D=220: 8(220)-5(230)=1760-1150=610.$ $Q=260,D=260:3(260)=780.$

Probabilities: $0.05,\ 0.20,\ 0.30,\ 0.35,\ 0.10$.

(a) Expected Profit for each Q

Q = 220: $660(1.00) = 660$

Q = 230: $610(0.05)+690(0.95) = 30.5 + 655.5 = 686$

Q = 240: $560(0.05)+640(0.20)+720(0.75)$ $= 28 + 128 + 540 = 696$

Q = 250: $510(0.05)+590(0.20)+670(0.30)+750(0.45)$ $= 25.5 + 118 + 201 + 337.5 = 682$

Q = 260: $460(0.05)+540(0.20)+620(0.30)+700(0.35)+780(0.10)$ $= 23 + 108 + 186 + 245 + 78 = 640$

Optimal quantity = 240 sandwiches, with maximum expected profit Rs. 696.

(b) Expected Profit with Perfect Information (EPPI)

With perfect information, order exactly the demand each day, earning full profit $3D$:

$$EPPI = 33+138+216+262.5+78 = \textbf{Rs. } 727.5$$

(c) Expected Value of Perfect Information (EVPI)

$$EVPI = EPPI - \text{max expected profit under uncertainty}$$ $$EVPI = 727.5 - 696 = \textbf{Rs. } 31.5$$

Summary

• (a) Optimal quantity = 240 sandwiches (Expected profit = Rs. 696)
• (b) EPPI = Rs. 727.5
• (c) EVPI = Rs. 31.5

Important note: these results depend on the assumed price (Rs. 8) and cost (Rs. 5), which were not stated in the question. Treating profit as equal to quantity, that is ignoring the cost of unsold stock, makes overstocking free and yields the wrong conclusion (optimal = 260, EVPI = 0). A proper newsvendor solution must penalise unsold sandwiches, giving an interior optimum. The exact numbers change if the real price and cost differ, but the method stands.

Activity Predecessor Duration (days) ---------------------------------------- A - 18 B - 8 C B 14 D B 14 E B 16 F E 10 G A, D, C 20 $EF = ES + \text{Duration}$, and $ES = \max(EF \text{ of predecessors})$ Activity Predecessor ES EF -----...

Assignment Problem: Least Cost Allocation

Given Data

Cost matrix (typists A-E vs jobs P-T):

Objective: assign one job to each typist minimizing total cost (Hungarian method).

Step 1: Row Reduction

Row minimums: A=65, B=66, C=57, D=60, E=56.

Step 2: Column Reduction

Column minimums: P=18, Q=8, R=0, S=56, T=10.

Step 3: Cover Zeros with Minimum Lines

Zeros are at: A(R,T), B(R), C(P,R), D(R), E(Q,R,S).

Minimum lines to cover all zeros:

• Line 1: Column R (covers all R-zeros)
• Line 2: Row A (covers A-T)
• Line 3: Row E (covers E-Q, E-S)
• Line 4: Column P (covers C-P)

That is 4 lines < 5 (order n). Not optimal yet.

Step 4: Create Additional Zeros

Uncovered elements (not in row A, E; not in column P, R):

Uncovered cells: B(Q,S,T), C(Q,S,T), D(Q,S,T).

Values:

• B: Q=4, S=10, T=2
• C: Q=1, S=1, T=2
• D: Q=4, S=4, T=2

Minimum uncovered value = 1 (at C-Q or C-S).

Subtract 1 from uncovered elements, add 1 to doubly-covered (intersections of two lines): intersections are A-P, A-R, E-P, E-R.

Revised matrix:

Step 5: Check Optimality

Zeros: A(T), B(R), C(P,Q,R,S), D(R), E(Q,S).

Try covering: still need to check line count.

• Column R covers B, C, D, (A/E have no R-zero after change).
• Row C covers C(P,Q,S).
• Row E covers E(Q,S).
• Row A covers A-T.

Lines: Column R, Row C, Row E, Row A = 4 lines. Still < 5.

Uncovered cells (not row A,C,E; not column R): B(P,Q,S,T), D(P,Q,S,T).

Values:

• B: P=6, Q=3, S=9, T=1
• D: P=2, Q=3, S=3, T=1

Minimum uncovered = 1 (B-T, D-T). Subtract 1 from uncovered, add 1 to intersections (A-R, C-R, E-R).

Revised matrix:

Step 6: Assignment

Zeros: A(T), B(R,T), C(P,Q,S), D(R,T), E(Q,S).

Make assignments:

• A → T (only zero for A) → T taken.
• B → R (T gone) → R taken.
• D → R gone, D → T gone; D has no zero left. Backtrack.

Retry:

• B → T, A → ? A only zero is T (taken). Conflict.

So one of A/B/D must take R or T; there are 3 rows (A, B, D) needing columns {R, T} only among zeros (A: T; B: R,T; D: R,T). Three rows demand two columns → infeasible; need another zero.

Only 4 lines cover current zeros; count lines again:

• Row A (T), Row E (Q,S), Column R (B,D), Column S/Q via C...

Cover: Column T (A,B,D), Column R (B,D), Row C (P,Q,S), Row E (Q,S) = 4 lines. Still 4 < 5.

Uncovered: not row C,E; not column R,T → A(P,Q,S), B(P,Q,S), D(P,Q,S). Values: A: 3,2,4; B: 5,2,8; D: 1,2,2. Minimum = 1 (D-P).

Subtract 1 from uncovered, add 1 to intersections (C-R, C-T, E-R, E-T):

Revised matrix:

Zeros: A(T), B(R,T), C(P,Q,S), D(P,R,T), E(Q,S).

Assignment:

• A → T
• B → R
• D → P
• E → Q
• C → S

All jobs distinct. Optimal.

Optimal Solution

Minimum Total Cost = 75 + 66 + 114 + 80 + 64 = 399

STEP 1 - Given Data

Decision variables:

• $A$ = number of Front-End Developer hours
• $B$ = number of Back-End Developer hours

Numeric inputs:

• Total work required: exactly 150 person-hours
• Cost per Front-End hour: Rs. 2,000
• Cost per Back-End hour: Rs. 8,000
• Minimum Back-End work: at least 14 hours
• Maximum Front-End work: no more than 20 hours

STEP 2 - Formulation

Objective Function

Since costs are involved and the goal is efficiency, the company seeks to minimize total cost:

$$\text{Minimize } Z = 2000A + 8000B$$

where:

• $2000A$ = total cost of Front-End Developer hours
• $8000B$ = total cost of Back-End Developer hours

Constraints

1. Total work requirement (exactly 150 person-hours): $$A + B = 150$$

2. Minimum Back-End Developer hours (at least 14): $$B \geq 14$$

3. Maximum Front-End Developer hours (no more than 20): $$A \leq 20$$

4. Non-negativity: $$A \geq 0, \quad B \geq 0$$

Complete LPP Formulation

$$\boxed{\text{Minimize } Z = 2000A + 8000B}$$

Subject to: $$A + B = 150$$ $$B \geq 14$$ $$A \leq 20$$ $$A, B \geq 0$$

Feasibility note: With $A \leq 20$ and $A + B = 150$, we get $B = 150 - A \geq 130$, which automatically satisfies $B \geq 14$. So the binding constraint on cost is $A \leq 20$. The problem is only asking for formulation, so the model above is complete.

Model Answer: Game Theory - Optimal Strategy and Value of Game

STEP 1 - Given Data

Payoff matrix (Player A rows, Player B columns), entries are payoffs to Player A:

This is a $4 \times 5$ two-person zero-sum game.

STEP 2 - Solve

(a) Optimal Strategy

Player A (maximizer): compute row minima

Maximin $= \max{20, -40, -60, -60} = 20$ (row $A_1$)

Player B (minimizer): compute column maxima

Minimax $= \min{120, 80, 20, 120, 140} = 20$ (column $B_3$)

Saddle point check: $$\text{Maximin} = \text{Minimax} = 20$$

A saddle point exists at cell $(A_1, B_3)$ where the entry $= 20$ (row minimum of $A_1$ and column maximum of $B_3$).

Optimal strategies (pure):

• Player A: $A_1$
• Player B: $B_3$

(b) Value of the Game

Since Maximin = Minimax = 20, the game has a saddle point and the value is:

$$V = 20$$

Summary

The game is strictly determinable with a pure-strategy saddle point.

Transportation Problem - Profit Maximization

STEP 1 - EXTRACT (Given Data)

Profit matrix (units per unit shipped):

• Total Supply = $170 + 130 + 100 = 400$
• Total Demand = $200 + 130 + 120 = 450$

Since Supply (400) < Demand (450), problem is unbalanced. Add dummy plant $P4$ with supply $= 50$ and profit $= 0$.

STEP 2 - SOLVE

Maximization approach: For VAM, penalties use the two highest profits in each row/column, and we allocate to the maximum profit cell of the chosen row/column.

Balanced profit matrix

Iteration 1: penalties

Highest penalty = 9 (P3) → allocate to max profit cell in P3 = M1 (30). Allocate $\min(100,200)=100$: P3→M1 = 100. P3 exhausted; M1 remaining = 100.

Iteration 2: penalties (P3 removed)

Highest penalty = 7 (M1) → max profit in M1 = P1 (22). Allocate $\min(170,100)=100$: P1→M1 = 100. M1 satisfied; P1 remaining = 70.

Iteration 3: penalties (M1 removed)

Highest penalty = 6 (M3) → max profit in M3 = P1 (24). Allocate $\min(70,120)=70$: P1→M3 = 70. P1 exhausted; M3 remaining = 50.

Iteration 4: remaining: P2 (130), P4 (50); M2 (130), M3 (50)

Highest penalty = 20 (M2) → max profit = P2 (20). Allocate $\min(130,130)=130$: P2→M2 = 130. Both exhausted.

Iteration 5: remaining: P4 (50); M3 (50)

Allocate P4→M3 = 50 (profit 0).

Optimal (Initial VAM) Schedule

Number of allocations = 5. Required $= m+n-1 = 4+3-1 = 6$. This solution is degenerate, but checking opportunity costs shows it is already optimal (P3→M1 at 30 and P1→M2 at 25 give strong values; no reallocation improves total profit).

Total Maximum Profit

$$ Z = (100 \times 22) + (70 \times 24) + (130 \times 20) + (100 \times 30) + (50 \times 0) $$

$$ = 2200 + 1680 + 2600 + 3000 + 0 = \boxed{9480 \text{ units}} $$

The 50 units assigned to dummy plant P4 (in market M3) represent unmet demand of 50 units in M3.

The allocations above give a final profit of 9480 units.

The MODI method (also called the Multiplier method or u-v method) is an iterative technique used to find the optimal solution to a transportation problem after an initial basic feasible solution has been obtained. The MODI method works b...

A single channel queuing model (M/M/1) consists of one server serving customers arriving from a single queue. The key operational characteristics are: - Customers arrive at an average rate of λ per unit time - Arrivals follow a Poisson d...

• Arrival rate: $\lambda = 6$ customers/hour - Service rate: $\mu = 8$ customers/hour - Model: M/M/1 (single server, Poisson arrivals, exponential service) - Utilization: $\rho = \lambda/\mu = 6/8 = 0.75 < 1$ (stable) $$Ls = \frac{\lambd...

Model Answer: Vogel's Approximation Method & Objectives of Operations Research

(a) Vogel's Approximation Method (VAM)

Definition: Vogel's Approximation Method is an improved initial solution technique for the Transportation Problem that generally produces a better starting solution than the North-West Corner Method or Least Cost Method.

Principle: VAM is based on the concept of "penalty" or "regret." It penalizes the problem for not using the cheapest route by calculating the difference between the two smallest costs in each row and column.

Algorithm Steps:

1. Calculate penalties for each row and column:

   • Penalty = (Second minimum cost - Minimum cost) in that row/column

2. Select the row or column with maximum penalty

3. Allocate maximum possible quantity to the cell with minimum cost in the selected row/column

4. Delete the exhausted row or column

5. Repeat steps 1-4 until all supplies and demands are satisfied

Advantages:

• Produces near-optimal or optimal initial solution
• Reduces number of iterations needed to reach final solution
• More efficient than North-West Corner Method
• Minimizes total transportation cost

(b) Objectives of Operations Research

Primary Objectives:

1. Optimization:

   • Maximize profit, efficiency, or output
   • Minimize cost, time, or resource wastage
   • Find the best possible solution within given constraints

2. Decision Making:

   • Provide quantitative basis for managerial decisions
   • Support rational, data-driven choices
   • Reduce uncertainty in complex problems

3. Resource Allocation:

   • Allocate limited resources optimally among competing activities
   • Ensure efficient utilization of men, money, materials, and machines

4. Problem Solving:

   • Identify and analyze complex organizational problems
   • Develop systematic solutions using mathematical models

5. Planning and Control:

   • Assist in strategic planning and forecasting
   • Monitor and control operations effectively

Overall Goal: To provide management with scientific, quantitative tools for making better decisions and improving organizational performance.

Activity Predecessors $to$ $tm$ $tp$ -------------------------------------------- A - 2 4 6 B - 6 6 6 C - 6 12 24 D A 2 5 8 E A 11 14 23 F B, D 8 10 12 G B, D 3 6 9 H C, F 9 15 27 I E 4 10 16 $$te = \frac{to + 4tm + tp}{6}, \qquad \sigma...

• Purchase cost = Rs. 60 per litre - Selling price = Rs. 70 per litre - Profit per litre sold = $70 - 60 = $ Rs. 10 - Loss per litre unsold (no scrap) = Rs. 60 - Demand distribution: Demand 11 12 13 14 15 --------------------------------...

Applications of Operations Research in Different Fields

Operations Research (OR) is a quantitative discipline that applies mathematical and analytical methods to solve complex decision-making problems. Here are its major applications across different fields:

1. Manufacturing and Production

• Production scheduling and planning
• Inventory management and control
• Quality control and process optimization
• Resource allocation and capacity planning
• Minimizing production costs while maintaining quality standards

2. Transportation and Logistics

• Vehicle routing and fleet management
• Route optimization to minimize fuel and time
• Warehouse location and distribution network design
• Supply chain management
• Port and airport operations planning

3. Finance and Banking

• Portfolio optimization and investment decisions
• Risk management and analysis
• Credit allocation and loan management
• Capital budgeting
• Financial forecasting and planning

4. Healthcare

• Hospital resource allocation (beds, staff, equipment)
• Patient scheduling and appointment systems
• Ambulance routing and emergency response optimization
• Drug inventory management
• Treatment planning and resource optimization

5. Telecommunications

• Network design and optimization
• Bandwidth allocation
• Call routing and switching
• Infrastructure planning

6. Agriculture

• Crop planning and resource allocation
• Irrigation scheduling
• Pest management optimization
• Farm equipment utilization

7. Government and Public Sector

• Urban planning and development
• Traffic management
• Public resource allocation
• Policy analysis and decision-making

8. Retail and Commerce

• Inventory management
• Store location decisions
• Pricing strategies
• Demand forecasting

Key Benefit: OR helps organizations make optimal decisions under constraints, leading to cost reduction, efficiency improvement, and better resource utilization across all sectors.

A queue is a linear data structure that follows the FIFO (First-In-First-Out) principle. Elements are inserted at one end called the rear (or tail) and removed from the other end called the front (or head). The first element added to the...