Question 1 of 9: Programming — Sphere Buoyancy and a Subsystem Trade Study

Question 1: Programming — Sphere Buoyancy and a Subsystem Trade Study (20 marks: (a) 10, (b) 10)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. For (a), the specific weight of water $\gamma=62.4\ \text{lb/ft}^3$ and the volume law $V=(4/3)\pi r^3$ for a sphere of radius $r$; the program itself reads the weight $W$ (lb) and radius $r$ (ft). For (b), four subsystems in series, each in two variants, with the manufacturing cost and MTBF of every variant.

Find. (a) Whether a given sphere sinks or floats in water. (b) The cheapest, most reliable, and best-value whole-car configuration out of all combinations of subsystem variants.

Approach. (a) Compute the buoyant force directly from the volume formula and compare it against the input weight. (b) Enumerate all $2^4=16$ combinations with four nested loops, since the design space is tiny; track the three running winners (cheapest cost, longest MTBF, best cost per failure-free year) in one pass.

(a) Sink or float? (10 marks)

1. Write the buoyant force directly from the given formulas. $$F_b = V\gamma = \frac{4}{3}\pi r^3 \times 62.4$$ There is no iteration or search here — both quantities are computed by one closed-form expression each, so the program is a straight-line calculation followed by one comparison.

   #include <stdio.h>
   #include <math.h>
   
   #define GAMMA_WATER 62.4   /* lb/ft^3, specific weight of water */
   
   int main(void)
   {
       double weight, radius, volume, buoyant_force;
   
       printf("Enter the weight (lb) and radius (ft) of the sphere: ");
       if (scanf("%lf %lf", &weight, &radius) != 2 || weight < 0.0 || radius < 0.0) {
           printf("Invalid input.\n");
           return 1;
       }
   
       volume        = (4.0 / 3.0) * M_PI * pow(radius, 3);
       buoyant_force = volume * GAMMA_WATER;
   
       if (buoyant_force >= weight)
           printf("The sphere will float (buoyant force = %.2f lb).\n", buoyant_force);
       else
           printf("The sphere will sink (buoyant force = %.2f lb).\n", buoyant_force);
   
       return 0;
   }

2. Trace a sphere that floats. For $r=1\ \text{ft}$, $$V = \frac{4}{3}\pi(1)^3 = 4.18879\ \text{ft}^3, \qquad F_b = 4.18879 \times 62.4 = 261.38\ \text{lb}$$ $$\boxed{F_b(r{=}1) = 261.38\ \text{lb}}$$ so a 150 lb sphere of this radius floats, since $261.38 \ge 150$.
3. Trace a sphere that sinks. For $r=0.5\ \text{ft}$, $$V = \frac{4}{3}\pi(0.5)^3 = 0.52360\ \text{ft}^3, \qquad F_b = 0.52360 \times 62.4 = 32.67\ \text{lb}$$ $$\boxed{F_b(r{=}0.5) = 32.67\ \text{lb}}$$ so a 50 lb sphere of this smaller radius sinks, since $32.67 < 50$: the same material at a smaller radius displaces far less water (volume scales as $r^3$), and a modest weight is then enough to overcome it.

(b) Cheapest, most reliable, and best-value car (10 marks)

1. Recognise the series structure. The failure of any subsystem dooms the car, so under the constant-hazard model failure rates add: $$\lambda_{total} = \sum_i \lambda_i = \sum_i \frac{1}{MTBF_i}$$ Four subsystems × two variants each gives $N = 2^4 = 16$ whole cars — small enough that exhaustive enumeration is not just acceptable, it is the correct engineering answer.
2. Define the three figures of merit and enumerate. $$C=\sum_i c_{i,v_i}, \qquad MTBF=\Big(\sum_i \tfrac{1}{m_{i,v_i}}\Big)^{-1}, \qquad R = C\sum_i\tfrac{1}{m_{i,v_i}}$$ $R$ (cost per failure-free year) is computed as $C$ times the rate sum rather than $C/MTBF$, avoiding a division.

   #include <stdio.h>
   
   #define NSUB 4
   #define NOPT 2
   
   int main(void)
   {
       const char *name[NSUB][NOPT] = {
           {"disc",     "drum"},
           {"wankle",   "conventional"},
           {"air",      "oil"},
           {"computer", "standard"}
       };
       const double mtbf[NSUB][NOPT] = {{4, 5}, {3, 6}, {9, 7}, {2, 4}};
       const double cost[NSUB][NOPT] = {{  15.00,   25.00},
                                        {1067.00, 1850.00},
                                        { 430.00,  320.00},
                                        { 130.00,   40.00}};
       int b, e, s, l;
       double cheapC = -1.0, longM = -1.0, bestR = -1.0;
   
       for (b = 0; b < NOPT; b++)
         for (e = 0; e < NOPT; e++)
           for (s = 0; s < NOPT; s++)
             for (l = 0; l < NOPT; l++) {
                 int pick[NSUB] = {b, e, s, l};
                 double c = 0.0, inv = 0.0;
                 int k;
                 for (k = 0; k < NSUB; k++) {
                     c   += cost[k][pick[k]];
                     inv += 1.0 / mtbf[k][pick[k]];
                 }
                 double m = 1.0 / inv, perYear = c * inv;
                 if (cheapC < 0.0 || c < cheapC) cheapC = c;
                 if (longM  < 0.0 || m > longM ) longM  = m;
                 if (bestR  < 0.0 || perYear < bestR) bestR = perYear;
             }
   
       printf("cheapest     : %.2f\n", cheapC);
       printf("longest MTBF : %.4f years\n", longM);
       printf("best value   : %.2f per failure-free year\n", bestR);
       return 0;
   }

3. Cheapest car takes every subsystem's cheaper variant independently (cost is a plain sum, hence separable): disc, wankle, oil, standard. $$\boxed{C_{\min} = 15.00+1067.00+320.00+40.00 = 1442.00\ \text{dollars}}$$ with MTBF $=1.0244$ years.
4. Most reliable car takes every subsystem's largest MTBF (the rate sum is likewise separable): drum, conventional, air, standard. $$\frac{1}{MTBF}=\frac{1}{5}+\frac{1}{6}+\frac{1}{9}+\frac{1}{4} =\frac{36+30+20+45}{180}=\frac{131}{180}$$ $$\boxed{MTBF_{\max}=\frac{180}{131}=1.3740\ \text{years}}$$ at a cost of $2,345.00.
5. Best-value car is neither of the above, because $R$ couples cost and reliability and does not separate — it must genuinely be searched. The winner is drum, wankle, oil, standard: $$R=\frac{1452.00}{1.0797}$$ $$\boxed{R_{\min}=1344.83\ \text{dollars per failure-free year}}$$ For just $10.00 more than the cheapest car, drum brakes buy enough extra life to beat it on value.